Mixing
Not getting the right answer with alternate completing the square method on...
$begingroup$
So I've looked up how to do this problem and when they complete the square it's using the $(frac{-b}{a})^2$ method where, in the denominator, they factor out the $4$, and then make:
$$x^2-x=x^2-x+{1over4}$$ and then
$$x^2-x+{1over4}=(x-frac{1}{2})^2$$
which I understand, but is not the first option that popped into my head.
What I don't understand is why all the tutorials I've looked up are not completing the square in the traditional method. Here's my work:
$intfrac{x^2}{sqrt{3+4x-4x^2}^3}dx$
denominator:
$$3+4x-4x^2$$
$$(-4x^2+4x-1)+4$$
$$-(4x^2-4x+1)+4$$
$$-(2x-1)^2+4$$
So the integral is now:
$$intfrac{x^2}{(4-u^2)^frac{3}{2}}$$
So I do the rest of the work by using trig substitution now:
$u=2x-1$
$x=frac{u+1}{2}$
Now I subbed in the trig identities:
$u=2sintheta$
$du=2costheta dtheta$
$$intfrac{(frac{u+1}{2})^22costheta}{sqrt{4-4sin^2theta}^3}dtheta$$
$$intfrac{frac{(2sintheta+1)^2}{4}2costheta}{sqrt{4cos^2theta}^3}dtheta$$
$$intfrac{frac{(2sintheta+1)^2}{4}2costheta}{(2costheta)^3}dtheta$$
$$intfrac{frac{(2sintheta+1)^2}{4}2costheta}{8cos^3theta}dtheta$$
$$intfrac{frac{(4sin^2theta+4sintheta+1)}{4}2costheta}{8cos^3theta}dtheta$$
$$intfrac{(frac{4sin^2theta}{4}+frac{4sintheta}{4}+frac{1}{4})2costheta}{8cos^3theta}dtheta$$
$$intfrac{(sin^2theta+sintheta+frac{1}{4})2costheta}{8cos^3theta}dtheta$$
$$intfrac{(sin^2theta+sintheta+frac{1}{4})}{8cos^3theta}*frac{2costheta}{1}dtheta$$
$$intfrac{(sin^2theta+sintheta+frac{1}{4})}{4cos^2theta}dtheta$$
$$intfrac{sin^2theta}{4cos^2theta}dtheta+intfrac{sintheta}{4cos^2theta}dtheta+intfrac{frac{1}{4}}{4cos^2theta}dtheta$$
$$frac{1}{4}int tan^2theta dtheta+frac{1}{4}intfrac{sintheta}{cos^2theta}dtheta+intfrac{1}{16cos^2theta}dtheta$$
For the second integral, I subbed $u=costheta$ and $-du=sintheta dtheta$
$$frac{1}{4}int sec^2theta-1dtheta-frac{1}{4}intfrac{du}{u^2}+frac{1}{16}int sec^2theta dtheta$$
$$[frac{tantheta}{4}-frac{theta}{4}]+[frac{1}{4costheta}]+[frac{tantheta}{16}]+C$$
On a right triangle where $sintheta=frac{2x-1}{2}$ and $costheta=frac{sqrt{4-(2x-1)^2}}{2}$, I subbed these values back in to make the integral in terms of $x$:
$$frac{2x-1}{4sqrt{4-(2x-1)^2}}-frac{arcsin(frac{2x-1}{2})}{4}+frac{1}{4(frac{sqrt{4-(2x-1)^2}}{2})}+frac{2x-1}{16sqrt{4-(2x-1)^2}}+C$$
This does not seem to be the right answer when I compare it to a calculator answer, so I'm guessing I either completed the square wrong or made an algebraic mistake? I've done this problem three or four times now and am getting a little frustrated over what I'm guessing is a easy fix. If anyone can help, I'd really appreciate it.
calculus indefinite-integrals trigonometric-integrals
edited Jan 9 at 8:52
Mostafa Ayaz
15.4k3939
asked Jan 9 at 7:37
Ben DreslinskiBen Dreslinski
224
$endgroup$
add a comment |
$begingroup$
So I've looked up how to do this problem and when they complete the square it's using the $(frac{-b}{a})^2$ method where, in the denominator, they factor out the $4$, and then make:
$$x^2-x=x^2-x+{1over4}$$ and then
$$x^2-x+{1over4}=(x-frac{1}{2})^2$$
which I understand, but is not the first option that popped into my head.
What I don't understand is why all the tutorials I've looked up are not completing the square in the traditional method. Here's my work:
$intfrac{x^2}{sqrt{3+4x-4x^2}^3}dx$
denominator:
$$3+4x-4x^2$$
$$(-4x^2+4x-1)+4$$
$$-(4x^2-4x+1)+4$$
$$-(2x-1)^2+4$$
So the integral is now:
$$intfrac{x^2}{(4-u^2)^frac{3}{2}}$$
So I do the rest of the work by using trig substitution now:
$u=2x-1$
$x=frac{u+1}{2}$
Now I subbed in the trig identities:
$u=2sintheta$
$du=2costheta dtheta$
$$intfrac{(frac{u+1}{2})^22costheta}{sqrt{4-4sin^2theta}^3}dtheta$$
$$intfrac{frac{(2sintheta+1)^2}{4}2costheta}{sqrt{4cos^2theta}^3}dtheta$$
$$intfrac{frac{(2sintheta+1)^2}{4}2costheta}{(2costheta)^3}dtheta$$
$$intfrac{frac{(2sintheta+1)^2}{4}2costheta}{8cos^3theta}dtheta$$
$$intfrac{frac{(4sin^2theta+4sintheta+1)}{4}2costheta}{8cos^3theta}dtheta$$
$$intfrac{(frac{4sin^2theta}{4}+frac{4sintheta}{4}+frac{1}{4})2costheta}{8cos^3theta}dtheta$$
$$intfrac{(sin^2theta+sintheta+frac{1}{4})2costheta}{8cos^3theta}dtheta$$
$$intfrac{(sin^2theta+sintheta+frac{1}{4})}{8cos^3theta}*frac{2costheta}{1}dtheta$$
$$intfrac{(sin^2theta+sintheta+frac{1}{4})}{4cos^2theta}dtheta$$
$$intfrac{sin^2theta}{4cos^2theta}dtheta+intfrac{sintheta}{4cos^2theta}dtheta+intfrac{frac{1}{4}}{4cos^2theta}dtheta$$
$$frac{1}{4}int tan^2theta dtheta+frac{1}{4}intfrac{sintheta}{cos^2theta}dtheta+intfrac{1}{16cos^2theta}dtheta$$
For the second integral, I subbed $u=costheta$ and $-du=sintheta dtheta$
$$frac{1}{4}int sec^2theta-1dtheta-frac{1}{4}intfrac{du}{u^2}+frac{1}{16}int sec^2theta dtheta$$
$$[frac{tantheta}{4}-frac{theta}{4}]+[frac{1}{4costheta}]+[frac{tantheta}{16}]+C$$
On a right triangle where $sintheta=frac{2x-1}{2}$ and $costheta=frac{sqrt{4-(2x-1)^2}}{2}$, I subbed these values back in to make the integral in terms of $x$:
$$frac{2x-1}{4sqrt{4-(2x-1)^2}}-frac{arcsin(frac{2x-1}{2})}{4}+frac{1}{4(frac{sqrt{4-(2x-1)^2}}{2})}+frac{2x-1}{16sqrt{4-(2x-1)^2}}+C$$
This does not seem to be the right answer when I compare it to a calculator answer, so I'm guessing I either completed the square wrong or made an algebraic mistake? I've done this problem three or four times now and am getting a little frustrated over what I'm guessing is a easy fix. If anyone can help, I'd really appreciate it.
calculus indefinite-integrals trigonometric-integrals
edited Jan 9 at 8:52
Mostafa Ayaz
15.4k3939
asked Jan 9 at 7:37
Ben DreslinskiBen Dreslinski
224
$endgroup$
add a comment |
$begingroup$
So I've looked up how to do this problem and when they complete the square it's using the $(frac{-b}{a})^2$ method where, in the denominator, they factor out the $4$, and then make:
$$x^2-x=x^2-x+{1over4}$$ and then
$$x^2-x+{1over4}=(x-frac{1}{2})^2$$
which I understand, but is not the first option that popped into my head.
What I don't understand is why all the tutorials I've looked up are not completing the square in the traditional method. Here's my work:
$intfrac{x^2}{sqrt{3+4x-4x^2}^3}dx$
denominator:
$$3+4x-4x^2$$
$$(-4x^2+4x-1)+4$$
$$-(4x^2-4x+1)+4$$
$$-(2x-1)^2+4$$
So the integral is now:
$$intfrac{x^2}{(4-u^2)^frac{3}{2}}$$
So I do the rest of the work by using trig substitution now:
$u=2x-1$
$x=frac{u+1}{2}$
Now I subbed in the trig identities:
$u=2sintheta$
$du=2costheta dtheta$
$$intfrac{(frac{u+1}{2})^22costheta}{sqrt{4-4sin^2theta}^3}dtheta$$
$$intfrac{frac{(2sintheta+1)^2}{4}2costheta}{sqrt{4cos^2theta}^3}dtheta$$
$$intfrac{frac{(2sintheta+1)^2}{4}2costheta}{(2costheta)^3}dtheta$$
$$intfrac{frac{(2sintheta+1)^2}{4}2costheta}{8cos^3theta}dtheta$$
$$intfrac{frac{(4sin^2theta+4sintheta+1)}{4}2costheta}{8cos^3theta}dtheta$$
$$intfrac{(frac{4sin^2theta}{4}+frac{4sintheta}{4}+frac{1}{4})2costheta}{8cos^3theta}dtheta$$
$$intfrac{(sin^2theta+sintheta+frac{1}{4})2costheta}{8cos^3theta}dtheta$$
$$intfrac{(sin^2theta+sintheta+frac{1}{4})}{8cos^3theta}*frac{2costheta}{1}dtheta$$
$$intfrac{(sin^2theta+sintheta+frac{1}{4})}{4cos^2theta}dtheta$$
$$intfrac{sin^2theta}{4cos^2theta}dtheta+intfrac{sintheta}{4cos^2theta}dtheta+intfrac{frac{1}{4}}{4cos^2theta}dtheta$$
$$frac{1}{4}int tan^2theta dtheta+frac{1}{4}intfrac{sintheta}{cos^2theta}dtheta+intfrac{1}{16cos^2theta}dtheta$$
For the second integral, I subbed $u=costheta$ and $-du=sintheta dtheta$
$$frac{1}{4}int sec^2theta-1dtheta-frac{1}{4}intfrac{du}{u^2}+frac{1}{16}int sec^2theta dtheta$$
$$[frac{tantheta}{4}-frac{theta}{4}]+[frac{1}{4costheta}]+[frac{tantheta}{16}]+C$$
On a right triangle where $sintheta=frac{2x-1}{2}$ and $costheta=frac{sqrt{4-(2x-1)^2}}{2}$, I subbed these values back in to make the integral in terms of $x$:
$$frac{2x-1}{4sqrt{4-(2x-1)^2}}-frac{arcsin(frac{2x-1}{2})}{4}+frac{1}{4(frac{sqrt{4-(2x-1)^2}}{2})}+frac{2x-1}{16sqrt{4-(2x-1)^2}}+C$$
This does not seem to be the right answer when I compare it to a calculator answer, so I'm guessing I either completed the square wrong or made an algebraic mistake? I've done this problem three or four times now and am getting a little frustrated over what I'm guessing is a easy fix. If anyone can help, I'd really appreciate it.
calculus indefinite-integrals trigonometric-integrals
edited Jan 9 at 8:52
Mostafa Ayaz
15.4k3939
asked Jan 9 at 7:37
Ben DreslinskiBen Dreslinski
224
$endgroup$
So I've looked up how to do this problem and when they complete the square it's using the $(frac{-b}{a})^2$ method where, in the denominator, they factor out the $4$, and then make:
$$x^2-x=x^2-x+{1over4}$$ and then
$$x^2-x+{1over4}=(x-frac{1}{2})^2$$
which I understand, but is not the first option that popped into my head.
What I don't understand is why all the tutorials I've looked up are not completing the square in the traditional method. Here's my work:
$intfrac{x^2}{sqrt{3+4x-4x^2}^3}dx$
denominator:
$$3+4x-4x^2$$
$$(-4x^2+4x-1)+4$$
$$-(4x^2-4x+1)+4$$
$$-(2x-1)^2+4$$
So the integral is now:
$$intfrac{x^2}{(4-u^2)^frac{3}{2}}$$
So I do the rest of the work by using trig substitution now:
$u=2x-1$
$x=frac{u+1}{2}$
Now I subbed in the trig identities:
$u=2sintheta$
$du=2costheta dtheta$
$$intfrac{(frac{u+1}{2})^22costheta}{sqrt{4-4sin^2theta}^3}dtheta$$
$$intfrac{frac{(2sintheta+1)^2}{4}2costheta}{sqrt{4cos^2theta}^3}dtheta$$
$$intfrac{frac{(2sintheta+1)^2}{4}2costheta}{(2costheta)^3}dtheta$$
$$intfrac{frac{(2sintheta+1)^2}{4}2costheta}{8cos^3theta}dtheta$$
$$intfrac{frac{(4sin^2theta+4sintheta+1)}{4}2costheta}{8cos^3theta}dtheta$$
$$intfrac{(frac{4sin^2theta}{4}+frac{4sintheta}{4}+frac{1}{4})2costheta}{8cos^3theta}dtheta$$
$$intfrac{(sin^2theta+sintheta+frac{1}{4})2costheta}{8cos^3theta}dtheta$$
$$intfrac{(sin^2theta+sintheta+frac{1}{4})}{8cos^3theta}*frac{2costheta}{1}dtheta$$
$$intfrac{(sin^2theta+sintheta+frac{1}{4})}{4cos^2theta}dtheta$$
$$intfrac{sin^2theta}{4cos^2theta}dtheta+intfrac{sintheta}{4cos^2theta}dtheta+intfrac{frac{1}{4}}{4cos^2theta}dtheta$$
$$frac{1}{4}int tan^2theta dtheta+frac{1}{4}intfrac{sintheta}{cos^2theta}dtheta+intfrac{1}{16cos^2theta}dtheta$$
For the second integral, I subbed $u=costheta$ and $-du=sintheta dtheta$
$$frac{1}{4}int sec^2theta-1dtheta-frac{1}{4}intfrac{du}{u^2}+frac{1}{16}int sec^2theta dtheta$$
$$[frac{tantheta}{4}-frac{theta}{4}]+[frac{1}{4costheta}]+[frac{tantheta}{16}]+C$$
On a right triangle where $sintheta=frac{2x-1}{2}$ and $costheta=frac{sqrt{4-(2x-1)^2}}{2}$, I subbed these values back in to make the integral in terms of $x$:
$$frac{2x-1}{4sqrt{4-(2x-1)^2}}-frac{arcsin(frac{2x-1}{2})}{4}+frac{1}{4(frac{sqrt{4-(2x-1)^2}}{2})}+frac{2x-1}{16sqrt{4-(2x-1)^2}}+C$$
This does not seem to be the right answer when I compare it to a calculator answer, so I'm guessing I either completed the square wrong or made an algebraic mistake? I've done this problem three or four times now and am getting a little frustrated over what I'm guessing is a easy fix. If anyone can help, I'd really appreciate it.
calculus indefinite-integrals trigonometric-integrals
calculus indefinite-integrals trigonometric-integrals
edited Jan 9 at 8:52
Mostafa Ayaz
15.4k3939
asked Jan 9 at 7:37
Ben DreslinskiBen Dreslinski
224
edited Jan 9 at 8:52
Mostafa Ayaz
15.4k3939
asked Jan 9 at 7:37
Ben DreslinskiBen Dreslinski
224
edited Jan 9 at 8:52
Mostafa Ayaz
15.4k3939
edited Jan 9 at 8:52
Mostafa Ayaz
15.4k3939
edited Jan 9 at 8:52
Mostafa Ayaz
15.4k3939
15.4k3939
asked Jan 9 at 7:37
Ben DreslinskiBen Dreslinski
224
asked Jan 9 at 7:37
Ben DreslinskiBen Dreslinski
224
asked Jan 9 at 7:37
Ben DreslinskiBen Dreslinski
224
224
add a comment |
add a comment |
2 Answers
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$begingroup$
Except some minor mistakes, your solution is correct but you led to wrong answer.
Mistake 1
In the first integral after substituting $u=sin theta$ you used $dx=du$ while $dx={1over 2}du$.
Mistake 2
From $2$nd to $3$rd integral, you used $|cos theta|=cos theta$ while you should have dealt with $|cos theta|$ till the result.
Fixing up
Fixing these two mistakes, the final integral to be solved would become$$intfrac{sin^2theta}{8costheta|costheta|}dtheta+intfrac{sintheta}{8costheta|costheta|}dtheta+intfrac{1}{32costheta|costheta|}dtheta\={4+5sintheta-4thetacosthetaover 32|costheta|}+C$$
answered Jan 9 at 8:55
Mostafa AyazMostafa Ayaz
15.4k3939
$endgroup$
$begingroup$
So my $du$ is incorrect because where $u=2x-1$, $du=2dx$, thus $dx=frac{1}{2}du$? I think I understand that, but doesn't $(4cos^2theta)^frac{3}{2}$ turn into $$sqrt{4cos^2theta}sqrt{4cos^2theta}sqrt{4cos^2theta}$$ which is equivalent to $$(2costheta)(2costheta)(2costheta)$$ so where do the absolute value signs come from? Unless that's just some rule I'm not aware of.
$endgroup$
– Ben Dreslinski
Jan 9 at 19:04
1
$begingroup$
Note that $$sqrt {x^2}=|x|ne x$$for $xin Bbb R$ for example $$sqrt{(-1)^2}=sqrt{1^2}=|-1|=|1|=1$$
$endgroup$
– Mostafa Ayaz
Jan 9 at 19:50
$begingroup$
So, just to make sure my algebra is right after re-doing this, the denominator is: $$sqrt{4cos^2theta}^3$$ $$(2|costheta|)^3$$ $$(2|costheta|)(2|costheta|)(2|costheta|)$$ So $(|costheta|)(|costheta|)=cos^2theta$ because the squared part makes the absolute values negligent, leaving: $$8cos^2theta|costheta|$$ And the $costheta$ up top makes: $$8costheta|costheta|$$ And that's how you got your denominator in your final answer in terms of $theta$?
$endgroup$
– Ben Dreslinski
Jan 9 at 21:35
$begingroup$
Yes you got it. That's accurate...
$endgroup$
– Mostafa Ayaz
Jan 10 at 10:11
add a comment |
$begingroup$
Find $I=intfrac{x^2}{(3+4x-4x^2)^{3/2}}dx$.
First 4 steps:
Change $x=t+frac12$
$$I=intfrac{t^2+t+frac14}{8(1-t^2)^{3/2}}dt\=frac18intfrac{t^2}{(1-t^2)^{3/2}}dt+
frac18intfrac{t}{(1-t^2)^{3/2}}dt+frac1{32}intfrac{1}{(1-t^2)^{3/2}}dt$$- $$intfrac{t}{(1-t^2)^{3/2}}dt=
frac{1}{sqrt{1-{{t}^{2}}}}$$ - $$intfrac{t^2}{(1-t^2)^{3/2}}dt=frac{t}{sqrt{1-{{t}^{2}}}}-arcsin(t)$$
- $$intfrac{1}{(1-t^2)^{3/2}}dt=frac{t}{sqrt{1-{{t}^{2}}}}$$
answered Jan 9 at 21:24
Aleksas DomarkasAleksas Domarkas
1,1946
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add a comment |
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2 Answers
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2 Answers
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$begingroup$
Except some minor mistakes, your solution is correct but you led to wrong answer.
Mistake 1
In the first integral after substituting $u=sin theta$ you used $dx=du$ while $dx={1over 2}du$.
Mistake 2
From $2$nd to $3$rd integral, you used $|cos theta|=cos theta$ while you should have dealt with $|cos theta|$ till the result.
Fixing up
Fixing these two mistakes, the final integral to be solved would become$$intfrac{sin^2theta}{8costheta|costheta|}dtheta+intfrac{sintheta}{8costheta|costheta|}dtheta+intfrac{1}{32costheta|costheta|}dtheta\={4+5sintheta-4thetacosthetaover 32|costheta|}+C$$
answered Jan 9 at 8:55
Mostafa AyazMostafa Ayaz
15.4k3939
$endgroup$
$begingroup$
So my $du$ is incorrect because where $u=2x-1$, $du=2dx$, thus $dx=frac{1}{2}du$? I think I understand that, but doesn't $(4cos^2theta)^frac{3}{2}$ turn into $$sqrt{4cos^2theta}sqrt{4cos^2theta}sqrt{4cos^2theta}$$ which is equivalent to $$(2costheta)(2costheta)(2costheta)$$ so where do the absolute value signs come from? Unless that's just some rule I'm not aware of.
$endgroup$
– Ben Dreslinski
Jan 9 at 19:04
1
$begingroup$
Note that $$sqrt {x^2}=|x|ne x$$for $xin Bbb R$ for example $$sqrt{(-1)^2}=sqrt{1^2}=|-1|=|1|=1$$
$endgroup$
– Mostafa Ayaz
Jan 9 at 19:50
$begingroup$
So, just to make sure my algebra is right after re-doing this, the denominator is: $$sqrt{4cos^2theta}^3$$ $$(2|costheta|)^3$$ $$(2|costheta|)(2|costheta|)(2|costheta|)$$ So $(|costheta|)(|costheta|)=cos^2theta$ because the squared part makes the absolute values negligent, leaving: $$8cos^2theta|costheta|$$ And the $costheta$ up top makes: $$8costheta|costheta|$$ And that's how you got your denominator in your final answer in terms of $theta$?
$endgroup$
– Ben Dreslinski
Jan 9 at 21:35
$begingroup$
Yes you got it. That's accurate...
$endgroup$
– Mostafa Ayaz
Jan 10 at 10:11
add a comment |
$begingroup$
Except some minor mistakes, your solution is correct but you led to wrong answer.
Mistake 1
In the first integral after substituting $u=sin theta$ you used $dx=du$ while $dx={1over 2}du$.
Mistake 2
From $2$nd to $3$rd integral, you used $|cos theta|=cos theta$ while you should have dealt with $|cos theta|$ till the result.
Fixing up
Fixing these two mistakes, the final integral to be solved would become$$intfrac{sin^2theta}{8costheta|costheta|}dtheta+intfrac{sintheta}{8costheta|costheta|}dtheta+intfrac{1}{32costheta|costheta|}dtheta\={4+5sintheta-4thetacosthetaover 32|costheta|}+C$$
answered Jan 9 at 8:55
Mostafa AyazMostafa Ayaz
15.4k3939
$endgroup$
$begingroup$
So my $du$ is incorrect because where $u=2x-1$, $du=2dx$, thus $dx=frac{1}{2}du$? I think I understand that, but doesn't $(4cos^2theta)^frac{3}{2}$ turn into $$sqrt{4cos^2theta}sqrt{4cos^2theta}sqrt{4cos^2theta}$$ which is equivalent to $$(2costheta)(2costheta)(2costheta)$$ so where do the absolute value signs come from? Unless that's just some rule I'm not aware of.
$endgroup$
– Ben Dreslinski
Jan 9 at 19:04
1
$begingroup$
Note that $$sqrt {x^2}=|x|ne x$$for $xin Bbb R$ for example $$sqrt{(-1)^2}=sqrt{1^2}=|-1|=|1|=1$$
$endgroup$
– Mostafa Ayaz
Jan 9 at 19:50
$begingroup$
So, just to make sure my algebra is right after re-doing this, the denominator is: $$sqrt{4cos^2theta}^3$$ $$(2|costheta|)^3$$ $$(2|costheta|)(2|costheta|)(2|costheta|)$$ So $(|costheta|)(|costheta|)=cos^2theta$ because the squared part makes the absolute values negligent, leaving: $$8cos^2theta|costheta|$$ And the $costheta$ up top makes: $$8costheta|costheta|$$ And that's how you got your denominator in your final answer in terms of $theta$?
$endgroup$
– Ben Dreslinski
Jan 9 at 21:35
$begingroup$
Yes you got it. That's accurate...
$endgroup$
– Mostafa Ayaz
Jan 10 at 10:11
add a comment |
$begingroup$
Except some minor mistakes, your solution is correct but you led to wrong answer.
Mistake 1
In the first integral after substituting $u=sin theta$ you used $dx=du$ while $dx={1over 2}du$.
Mistake 2
From $2$nd to $3$rd integral, you used $|cos theta|=cos theta$ while you should have dealt with $|cos theta|$ till the result.
Fixing up
Fixing these two mistakes, the final integral to be solved would become$$intfrac{sin^2theta}{8costheta|costheta|}dtheta+intfrac{sintheta}{8costheta|costheta|}dtheta+intfrac{1}{32costheta|costheta|}dtheta\={4+5sintheta-4thetacosthetaover 32|costheta|}+C$$
answered Jan 9 at 8:55
Mostafa AyazMostafa Ayaz
15.4k3939
$endgroup$
Except some minor mistakes, your solution is correct but you led to wrong answer.
Mistake 1
In the first integral after substituting $u=sin theta$ you used $dx=du$ while $dx={1over 2}du$.
Mistake 2
From $2$nd to $3$rd integral, you used $|cos theta|=cos theta$ while you should have dealt with $|cos theta|$ till the result.
Fixing up
Fixing these two mistakes, the final integral to be solved would become$$intfrac{sin^2theta}{8costheta|costheta|}dtheta+intfrac{sintheta}{8costheta|costheta|}dtheta+intfrac{1}{32costheta|costheta|}dtheta\={4+5sintheta-4thetacosthetaover 32|costheta|}+C$$
answered Jan 9 at 8:55
Mostafa AyazMostafa Ayaz
15.4k3939
answered Jan 9 at 8:55
Mostafa AyazMostafa Ayaz
15.4k3939
answered Jan 9 at 8:55
Mostafa AyazMostafa Ayaz
15.4k3939
answered Jan 9 at 8:55
Mostafa AyazMostafa Ayaz
15.4k3939
15.4k3939
$begingroup$
So my $du$ is incorrect because where $u=2x-1$, $du=2dx$, thus $dx=frac{1}{2}du$? I think I understand that, but doesn't $(4cos^2theta)^frac{3}{2}$ turn into $$sqrt{4cos^2theta}sqrt{4cos^2theta}sqrt{4cos^2theta}$$ which is equivalent to $$(2costheta)(2costheta)(2costheta)$$ so where do the absolute value signs come from? Unless that's just some rule I'm not aware of.
$endgroup$
– Ben Dreslinski
Jan 9 at 19:04
1
$begingroup$
Note that $$sqrt {x^2}=|x|ne x$$for $xin Bbb R$ for example $$sqrt{(-1)^2}=sqrt{1^2}=|-1|=|1|=1$$
$endgroup$
– Mostafa Ayaz
Jan 9 at 19:50
$begingroup$
So, just to make sure my algebra is right after re-doing this, the denominator is: $$sqrt{4cos^2theta}^3$$ $$(2|costheta|)^3$$ $$(2|costheta|)(2|costheta|)(2|costheta|)$$ So $(|costheta|)(|costheta|)=cos^2theta$ because the squared part makes the absolute values negligent, leaving: $$8cos^2theta|costheta|$$ And the $costheta$ up top makes: $$8costheta|costheta|$$ And that's how you got your denominator in your final answer in terms of $theta$?
$endgroup$
– Ben Dreslinski
Jan 9 at 21:35
$begingroup$
Yes you got it. That's accurate...
$endgroup$
– Mostafa Ayaz
Jan 10 at 10:11
add a comment |
$begingroup$
So my $du$ is incorrect because where $u=2x-1$, $du=2dx$, thus $dx=frac{1}{2}du$? I think I understand that, but doesn't $(4cos^2theta)^frac{3}{2}$ turn into $$sqrt{4cos^2theta}sqrt{4cos^2theta}sqrt{4cos^2theta}$$ which is equivalent to $$(2costheta)(2costheta)(2costheta)$$ so where do the absolute value signs come from? Unless that's just some rule I'm not aware of.
$endgroup$
– Ben Dreslinski
Jan 9 at 19:04
1
$begingroup$
Note that $$sqrt {x^2}=|x|ne x$$for $xin Bbb R$ for example $$sqrt{(-1)^2}=sqrt{1^2}=|-1|=|1|=1$$
$endgroup$
– Mostafa Ayaz
Jan 9 at 19:50
$begingroup$
So, just to make sure my algebra is right after re-doing this, the denominator is: $$sqrt{4cos^2theta}^3$$ $$(2|costheta|)^3$$ $$(2|costheta|)(2|costheta|)(2|costheta|)$$ So $(|costheta|)(|costheta|)=cos^2theta$ because the squared part makes the absolute values negligent, leaving: $$8cos^2theta|costheta|$$ And the $costheta$ up top makes: $$8costheta|costheta|$$ And that's how you got your denominator in your final answer in terms of $theta$?
$endgroup$
– Ben Dreslinski
Jan 9 at 21:35
$begingroup$
Yes you got it. That's accurate...
$endgroup$
– Mostafa Ayaz
Jan 10 at 10:11
$begingroup$
So my $du$ is incorrect because where $u=2x-1$, $du=2dx$, thus $dx=frac{1}{2}du$? I think I understand that, but doesn't $(4cos^2theta)^frac{3}{2}$ turn into $$sqrt{4cos^2theta}sqrt{4cos^2theta}sqrt{4cos^2theta}$$ which is equivalent to $$(2costheta)(2costheta)(2costheta)$$ so where do the absolute value signs come from? Unless that's just some rule I'm not aware of.
$endgroup$
– Ben Dreslinski
Jan 9 at 19:04
$begingroup$
So my $du$ is incorrect because where $u=2x-1$, $du=2dx$, thus $dx=frac{1}{2}du$? I think I understand that, but doesn't $(4cos^2theta)^frac{3}{2}$ turn into $$sqrt{4cos^2theta}sqrt{4cos^2theta}sqrt{4cos^2theta}$$ which is equivalent to $$(2costheta)(2costheta)(2costheta)$$ so where do the absolute value signs come from? Unless that's just some rule I'm not aware of.
$endgroup$
– Ben Dreslinski
Jan 9 at 19:04
1
1
$begingroup$
Note that $$sqrt {x^2}=|x|ne x$$for $xin Bbb R$ for example $$sqrt{(-1)^2}=sqrt{1^2}=|-1|=|1|=1$$
$endgroup$
– Mostafa Ayaz
Jan 9 at 19:50
$begingroup$
Note that $$sqrt {x^2}=|x|ne x$$for $xin Bbb R$ for example $$sqrt{(-1)^2}=sqrt{1^2}=|-1|=|1|=1$$
$endgroup$
– Mostafa Ayaz
Jan 9 at 19:50
$begingroup$
So, just to make sure my algebra is right after re-doing this, the denominator is: $$sqrt{4cos^2theta}^3$$ $$(2|costheta|)^3$$ $$(2|costheta|)(2|costheta|)(2|costheta|)$$ So $(|costheta|)(|costheta|)=cos^2theta$ because the squared part makes the absolute values negligent, leaving: $$8cos^2theta|costheta|$$ And the $costheta$ up top makes: $$8costheta|costheta|$$ And that's how you got your denominator in your final answer in terms of $theta$?
$endgroup$
– Ben Dreslinski
Jan 9 at 21:35
$begingroup$
So, just to make sure my algebra is right after re-doing this, the denominator is: $$sqrt{4cos^2theta}^3$$ $$(2|costheta|)^3$$ $$(2|costheta|)(2|costheta|)(2|costheta|)$$ So $(|costheta|)(|costheta|)=cos^2theta$ because the squared part makes the absolute values negligent, leaving: $$8cos^2theta|costheta|$$ And the $costheta$ up top makes: $$8costheta|costheta|$$ And that's how you got your denominator in your final answer in terms of $theta$?
$endgroup$
– Ben Dreslinski
Jan 9 at 21:35
$begingroup$
Yes you got it. That's accurate...
$endgroup$
– Mostafa Ayaz
Jan 10 at 10:11
$begingroup$
Yes you got it. That's accurate...
$endgroup$
– Mostafa Ayaz
Jan 10 at 10:11
add a comment |
$begingroup$
Find $I=intfrac{x^2}{(3+4x-4x^2)^{3/2}}dx$.
First 4 steps:
Change $x=t+frac12$
$$I=intfrac{t^2+t+frac14}{8(1-t^2)^{3/2}}dt\=frac18intfrac{t^2}{(1-t^2)^{3/2}}dt+
frac18intfrac{t}{(1-t^2)^{3/2}}dt+frac1{32}intfrac{1}{(1-t^2)^{3/2}}dt$$- $$intfrac{t}{(1-t^2)^{3/2}}dt=
frac{1}{sqrt{1-{{t}^{2}}}}$$ - $$intfrac{t^2}{(1-t^2)^{3/2}}dt=frac{t}{sqrt{1-{{t}^{2}}}}-arcsin(t)$$
- $$intfrac{1}{(1-t^2)^{3/2}}dt=frac{t}{sqrt{1-{{t}^{2}}}}$$
answered Jan 9 at 21:24
Aleksas DomarkasAleksas Domarkas
1,1946
$endgroup$
add a comment |
$begingroup$
Find $I=intfrac{x^2}{(3+4x-4x^2)^{3/2}}dx$.
First 4 steps:
Change $x=t+frac12$
$$I=intfrac{t^2+t+frac14}{8(1-t^2)^{3/2}}dt\=frac18intfrac{t^2}{(1-t^2)^{3/2}}dt+
frac18intfrac{t}{(1-t^2)^{3/2}}dt+frac1{32}intfrac{1}{(1-t^2)^{3/2}}dt$$- $$intfrac{t}{(1-t^2)^{3/2}}dt=
frac{1}{sqrt{1-{{t}^{2}}}}$$ - $$intfrac{t^2}{(1-t^2)^{3/2}}dt=frac{t}{sqrt{1-{{t}^{2}}}}-arcsin(t)$$
- $$intfrac{1}{(1-t^2)^{3/2}}dt=frac{t}{sqrt{1-{{t}^{2}}}}$$
answered Jan 9 at 21:24
Aleksas DomarkasAleksas Domarkas
1,1946
$endgroup$
add a comment |
$begingroup$
Find $I=intfrac{x^2}{(3+4x-4x^2)^{3/2}}dx$.
First 4 steps:
Change $x=t+frac12$
$$I=intfrac{t^2+t+frac14}{8(1-t^2)^{3/2}}dt\=frac18intfrac{t^2}{(1-t^2)^{3/2}}dt+
frac18intfrac{t}{(1-t^2)^{3/2}}dt+frac1{32}intfrac{1}{(1-t^2)^{3/2}}dt$$- $$intfrac{t}{(1-t^2)^{3/2}}dt=
frac{1}{sqrt{1-{{t}^{2}}}}$$ - $$intfrac{t^2}{(1-t^2)^{3/2}}dt=frac{t}{sqrt{1-{{t}^{2}}}}-arcsin(t)$$
- $$intfrac{1}{(1-t^2)^{3/2}}dt=frac{t}{sqrt{1-{{t}^{2}}}}$$
answered Jan 9 at 21:24
Aleksas DomarkasAleksas Domarkas
1,1946
$endgroup$
Find $I=intfrac{x^2}{(3+4x-4x^2)^{3/2}}dx$.
First 4 steps:
Change $x=t+frac12$
$$I=intfrac{t^2+t+frac14}{8(1-t^2)^{3/2}}dt\=frac18intfrac{t^2}{(1-t^2)^{3/2}}dt+
frac18intfrac{t}{(1-t^2)^{3/2}}dt+frac1{32}intfrac{1}{(1-t^2)^{3/2}}dt$$- $$intfrac{t}{(1-t^2)^{3/2}}dt=
frac{1}{sqrt{1-{{t}^{2}}}}$$ - $$intfrac{t^2}{(1-t^2)^{3/2}}dt=frac{t}{sqrt{1-{{t}^{2}}}}-arcsin(t)$$
- $$intfrac{1}{(1-t^2)^{3/2}}dt=frac{t}{sqrt{1-{{t}^{2}}}}$$
answered Jan 9 at 21:24
Aleksas DomarkasAleksas Domarkas
1,1946
answered Jan 9 at 21:24
Aleksas DomarkasAleksas Domarkas
1,1946
answered Jan 9 at 21:24
Aleksas DomarkasAleksas Domarkas
1,1946
answered Jan 9 at 21:24
Aleksas DomarkasAleksas Domarkas
1,1946
1,1946
add a comment |
add a comment |
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