Mixing
What is $inttan^2{x}dx$? what strategy can I use?
$begingroup$
I'm a bit stuck on how to find the inter
gral of
$$ int tan^2{x}dx$$
if I substitute, with $tan{x}$, there isn't a $sec^2{x}$ for me to substitute the dx out with. If I transform $tan^2{x}$ into $sec^2{x}-1$, then you subbing gets me nowhere again. Is there an identity I can use? If so, can someone show me the proof of it?
EDIT
I am told I can use this formula:
$$tan{frac{x}{2}} = sqrt{frac{1 - cos{x}}{1 + cos{x}}} $$
but I don't see how this helps me. Can someone show me the proof of how this was obtained via the double angle formula for sin and cosine?
calculus
edited Feb 1 at 14:32
Bernard
124k741117
asked Feb 1 at 14:02
Jwan622Jwan622
2,38711632
$endgroup$
add a comment |
$begingroup$
I'm a bit stuck on how to find the inter
gral of
$$ int tan^2{x}dx$$
if I substitute, with $tan{x}$, there isn't a $sec^2{x}$ for me to substitute the dx out with. If I transform $tan^2{x}$ into $sec^2{x}-1$, then you subbing gets me nowhere again. Is there an identity I can use? If so, can someone show me the proof of it?
EDIT
I am told I can use this formula:
$$tan{frac{x}{2}} = sqrt{frac{1 - cos{x}}{1 + cos{x}}} $$
but I don't see how this helps me. Can someone show me the proof of how this was obtained via the double angle formula for sin and cosine?
calculus
edited Feb 1 at 14:32
Bernard
124k741117
asked Feb 1 at 14:02
Jwan622Jwan622
2,38711632
$endgroup$
$begingroup$
Don't you mean $tan^2x = sec^2x - 1$? Type$tan x$to obtain $tan x$ and$sec x$to obtain $sec x$.
$endgroup$
– N. F. Taussig
Feb 1 at 14:06
2
$begingroup$
$sec^2(x)$ is the derivative of $tan(x)$.
$endgroup$
– Robert Z
Feb 1 at 14:06
$begingroup$
Using that formula seems harder than the original problem.
$endgroup$
– Randall
Feb 1 at 14:23
add a comment |
$begingroup$
I'm a bit stuck on how to find the inter
gral of
$$ int tan^2{x}dx$$
if I substitute, with $tan{x}$, there isn't a $sec^2{x}$ for me to substitute the dx out with. If I transform $tan^2{x}$ into $sec^2{x}-1$, then you subbing gets me nowhere again. Is there an identity I can use? If so, can someone show me the proof of it?
EDIT
I am told I can use this formula:
$$tan{frac{x}{2}} = sqrt{frac{1 - cos{x}}{1 + cos{x}}} $$
but I don't see how this helps me. Can someone show me the proof of how this was obtained via the double angle formula for sin and cosine?
calculus
edited Feb 1 at 14:32
Bernard
124k741117
asked Feb 1 at 14:02
Jwan622Jwan622
2,38711632
$endgroup$
I'm a bit stuck on how to find the inter
gral of
$$ int tan^2{x}dx$$
if I substitute, with $tan{x}$, there isn't a $sec^2{x}$ for me to substitute the dx out with. If I transform $tan^2{x}$ into $sec^2{x}-1$, then you subbing gets me nowhere again. Is there an identity I can use? If so, can someone show me the proof of it?
EDIT
I am told I can use this formula:
$$tan{frac{x}{2}} = sqrt{frac{1 - cos{x}}{1 + cos{x}}} $$
but I don't see how this helps me. Can someone show me the proof of how this was obtained via the double angle formula for sin and cosine?
calculus
calculus
edited Feb 1 at 14:32
Bernard
124k741117
asked Feb 1 at 14:02
Jwan622Jwan622
2,38711632
edited Feb 1 at 14:32
Bernard
124k741117
asked Feb 1 at 14:02
Jwan622Jwan622
2,38711632
edited Feb 1 at 14:32
Bernard
124k741117
edited Feb 1 at 14:32
Bernard
124k741117
edited Feb 1 at 14:32
Bernard
124k741117
124k741117
asked Feb 1 at 14:02
Jwan622Jwan622
2,38711632
asked Feb 1 at 14:02
Jwan622Jwan622
2,38711632
asked Feb 1 at 14:02
Jwan622Jwan622
2,38711632
2,38711632
$begingroup$
Don't you mean $tan^2x = sec^2x - 1$? Type$tan x$to obtain $tan x$ and$sec x$to obtain $sec x$.
$endgroup$
– N. F. Taussig
Feb 1 at 14:06
2
$begingroup$
$sec^2(x)$ is the derivative of $tan(x)$.
$endgroup$
– Robert Z
Feb 1 at 14:06
$begingroup$
Using that formula seems harder than the original problem.
$endgroup$
– Randall
Feb 1 at 14:23
add a comment |
$begingroup$
Don't you mean $tan^2x = sec^2x - 1$? Type$tan x$to obtain $tan x$ and$sec x$to obtain $sec x$.
$endgroup$
– N. F. Taussig
Feb 1 at 14:06
2
$begingroup$
$sec^2(x)$ is the derivative of $tan(x)$.
$endgroup$
– Robert Z
Feb 1 at 14:06
$begingroup$
Using that formula seems harder than the original problem.
$endgroup$
– Randall
Feb 1 at 14:23
$begingroup$
Don't you mean $tan^2x = sec^2x - 1$? Type
$tan x$ to obtain $tan x$ and $sec x$ to obtain $sec x$.$endgroup$
– N. F. Taussig
Feb 1 at 14:06
$begingroup$
Don't you mean $tan^2x = sec^2x - 1$? Type
$tan x$ to obtain $tan x$ and $sec x$ to obtain $sec x$.$endgroup$
– N. F. Taussig
Feb 1 at 14:06
2
2
$begingroup$
$sec^2(x)$ is the derivative of $tan(x)$.
$endgroup$
– Robert Z
Feb 1 at 14:06
$begingroup$
$sec^2(x)$ is the derivative of $tan(x)$.
$endgroup$
– Robert Z
Feb 1 at 14:06
$begingroup$
Using that formula seems harder than the original problem.
$endgroup$
– Randall
Feb 1 at 14:23
$begingroup$
Using that formula seems harder than the original problem.
$endgroup$
– Randall
Feb 1 at 14:23
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Don't forget that $tan^2{x}=sec^2-1$ and that $int sec^2x,dx=tan{x}+C$ because $frac{d}{dx}left(tan{x}right)=sec^2{x}$:
$$
begin{align}
int tan^2{x},dx
&=int (sec^2x-1),dx\
&=int sec^2x,dx-int,dx\
&=tan{x}-x+C.
end{align}
$$
If you want to do this integral using the half-angle formula for the tangent function, you're gong to have to use the so-called Weierstrass substitution.
$$
tan^2{frac{x}{2}}=frac{sin^2{frac{x}{2}}}{cos^2{frac{x}{2}}}=frac{frac{1-cos{x}}{2}}{frac{1+cos{x}}{2}}=frac{1-cos{x}}{1+cos{x}}.
$$
$$
tan^2{frac{(2x)}{2}}=tan^2{x}=frac{1-cos{(2x)}}{1+cos{(2x)}}.
$$
$$
begin{align}
inttan^2{x},dx
&=int frac{1-cos{(2x)}}{1+cos{(2x)}}\
&=intfrac{1}{1+cos{(2x)}},dx-intfrac{cos{(2x)}}{1+cos{(2x)}},dx\
&=frac{1}{2}intfrac{1}{1+cos{(2x)}}frac{d}{dx}left(2xright),dx-frac{1}{2}intfrac{cos{(2x)}}{1+cos{(2x)}}frac{d}{dx}left(2xright),dx (u=2x)\
&=frac{1}{2}intfrac{1}{1+cos{u}},du-frac{1}{2}intfrac{cos{u}}{1+cos{u}},du\
end{align}
$$
And now you apply the Weierstrass substitution formulas:
$$
cos{u}=frac{1-t^2}{1+t^2}, du=frac{2}{1+t^2}dt
$$
$$
begin{align}
=frac{1}{2}intfrac{1}{1+frac{1-t^2}{1+t^2}}frac{2}{1+t^2},dt-frac{1}{2}intfrac{frac{1-t^2}{1+t^2}}{1+frac{1-t^2}{1+t^2}}frac{2}{1+t^2},dt
end{align}
$$
At this point, what you've got are purely algebraic expressions under the integral signs. All you need to do is simplify them, take their integrals and do back-substitution.
PS: I hope I didn't make any mistake.
answered Feb 1 at 14:18
Michael RybkinMichael Rybkin
4,254422
$endgroup$
$begingroup$
Is there a way to do this with a half angle identity for $ tan^2x$
$endgroup$
– Jwan622
Feb 1 at 14:27
$begingroup$
Probably, there is, but it's not going to be an integral that's easy to integrate because you would end up with something like this: $$intfrac{1}{1+cos{2x}},dx-intfrac{cos{2x}}{1+cos{2x}},dx.$$ I think this one could be done using the Weierstrass substitution.
$endgroup$
– Michael Rybkin
Feb 1 at 14:43
add a comment |
$begingroup$
Hint
use$$(tan x)'=1+tan^2x$$
answered Feb 1 at 14:04
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
add a comment |
$begingroup$
Hint:
A useful but not so well known formula is
$$(tan x)'=tan^2x+1.$$
answered Feb 1 at 14:43
Yves DaoustYves Daoust
133k676231
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Don't forget that $tan^2{x}=sec^2-1$ and that $int sec^2x,dx=tan{x}+C$ because $frac{d}{dx}left(tan{x}right)=sec^2{x}$:
$$
begin{align}
int tan^2{x},dx
&=int (sec^2x-1),dx\
&=int sec^2x,dx-int,dx\
&=tan{x}-x+C.
end{align}
$$
If you want to do this integral using the half-angle formula for the tangent function, you're gong to have to use the so-called Weierstrass substitution.
$$
tan^2{frac{x}{2}}=frac{sin^2{frac{x}{2}}}{cos^2{frac{x}{2}}}=frac{frac{1-cos{x}}{2}}{frac{1+cos{x}}{2}}=frac{1-cos{x}}{1+cos{x}}.
$$
$$
tan^2{frac{(2x)}{2}}=tan^2{x}=frac{1-cos{(2x)}}{1+cos{(2x)}}.
$$
$$
begin{align}
inttan^2{x},dx
&=int frac{1-cos{(2x)}}{1+cos{(2x)}}\
&=intfrac{1}{1+cos{(2x)}},dx-intfrac{cos{(2x)}}{1+cos{(2x)}},dx\
&=frac{1}{2}intfrac{1}{1+cos{(2x)}}frac{d}{dx}left(2xright),dx-frac{1}{2}intfrac{cos{(2x)}}{1+cos{(2x)}}frac{d}{dx}left(2xright),dx (u=2x)\
&=frac{1}{2}intfrac{1}{1+cos{u}},du-frac{1}{2}intfrac{cos{u}}{1+cos{u}},du\
end{align}
$$
And now you apply the Weierstrass substitution formulas:
$$
cos{u}=frac{1-t^2}{1+t^2}, du=frac{2}{1+t^2}dt
$$
$$
begin{align}
=frac{1}{2}intfrac{1}{1+frac{1-t^2}{1+t^2}}frac{2}{1+t^2},dt-frac{1}{2}intfrac{frac{1-t^2}{1+t^2}}{1+frac{1-t^2}{1+t^2}}frac{2}{1+t^2},dt
end{align}
$$
At this point, what you've got are purely algebraic expressions under the integral signs. All you need to do is simplify them, take their integrals and do back-substitution.
PS: I hope I didn't make any mistake.
answered Feb 1 at 14:18
Michael RybkinMichael Rybkin
4,254422
$endgroup$
$begingroup$
Is there a way to do this with a half angle identity for $ tan^2x$
$endgroup$
– Jwan622
Feb 1 at 14:27
$begingroup$
Probably, there is, but it's not going to be an integral that's easy to integrate because you would end up with something like this: $$intfrac{1}{1+cos{2x}},dx-intfrac{cos{2x}}{1+cos{2x}},dx.$$ I think this one could be done using the Weierstrass substitution.
$endgroup$
– Michael Rybkin
Feb 1 at 14:43
add a comment |
$begingroup$
Don't forget that $tan^2{x}=sec^2-1$ and that $int sec^2x,dx=tan{x}+C$ because $frac{d}{dx}left(tan{x}right)=sec^2{x}$:
$$
begin{align}
int tan^2{x},dx
&=int (sec^2x-1),dx\
&=int sec^2x,dx-int,dx\
&=tan{x}-x+C.
end{align}
$$
If you want to do this integral using the half-angle formula for the tangent function, you're gong to have to use the so-called Weierstrass substitution.
$$
tan^2{frac{x}{2}}=frac{sin^2{frac{x}{2}}}{cos^2{frac{x}{2}}}=frac{frac{1-cos{x}}{2}}{frac{1+cos{x}}{2}}=frac{1-cos{x}}{1+cos{x}}.
$$
$$
tan^2{frac{(2x)}{2}}=tan^2{x}=frac{1-cos{(2x)}}{1+cos{(2x)}}.
$$
$$
begin{align}
inttan^2{x},dx
&=int frac{1-cos{(2x)}}{1+cos{(2x)}}\
&=intfrac{1}{1+cos{(2x)}},dx-intfrac{cos{(2x)}}{1+cos{(2x)}},dx\
&=frac{1}{2}intfrac{1}{1+cos{(2x)}}frac{d}{dx}left(2xright),dx-frac{1}{2}intfrac{cos{(2x)}}{1+cos{(2x)}}frac{d}{dx}left(2xright),dx (u=2x)\
&=frac{1}{2}intfrac{1}{1+cos{u}},du-frac{1}{2}intfrac{cos{u}}{1+cos{u}},du\
end{align}
$$
And now you apply the Weierstrass substitution formulas:
$$
cos{u}=frac{1-t^2}{1+t^2}, du=frac{2}{1+t^2}dt
$$
$$
begin{align}
=frac{1}{2}intfrac{1}{1+frac{1-t^2}{1+t^2}}frac{2}{1+t^2},dt-frac{1}{2}intfrac{frac{1-t^2}{1+t^2}}{1+frac{1-t^2}{1+t^2}}frac{2}{1+t^2},dt
end{align}
$$
At this point, what you've got are purely algebraic expressions under the integral signs. All you need to do is simplify them, take their integrals and do back-substitution.
PS: I hope I didn't make any mistake.
answered Feb 1 at 14:18
Michael RybkinMichael Rybkin
4,254422
$endgroup$
$begingroup$
Is there a way to do this with a half angle identity for $ tan^2x$
$endgroup$
– Jwan622
Feb 1 at 14:27
$begingroup$
Probably, there is, but it's not going to be an integral that's easy to integrate because you would end up with something like this: $$intfrac{1}{1+cos{2x}},dx-intfrac{cos{2x}}{1+cos{2x}},dx.$$ I think this one could be done using the Weierstrass substitution.
$endgroup$
– Michael Rybkin
Feb 1 at 14:43
add a comment |
$begingroup$
Don't forget that $tan^2{x}=sec^2-1$ and that $int sec^2x,dx=tan{x}+C$ because $frac{d}{dx}left(tan{x}right)=sec^2{x}$:
$$
begin{align}
int tan^2{x},dx
&=int (sec^2x-1),dx\
&=int sec^2x,dx-int,dx\
&=tan{x}-x+C.
end{align}
$$
If you want to do this integral using the half-angle formula for the tangent function, you're gong to have to use the so-called Weierstrass substitution.
$$
tan^2{frac{x}{2}}=frac{sin^2{frac{x}{2}}}{cos^2{frac{x}{2}}}=frac{frac{1-cos{x}}{2}}{frac{1+cos{x}}{2}}=frac{1-cos{x}}{1+cos{x}}.
$$
$$
tan^2{frac{(2x)}{2}}=tan^2{x}=frac{1-cos{(2x)}}{1+cos{(2x)}}.
$$
$$
begin{align}
inttan^2{x},dx
&=int frac{1-cos{(2x)}}{1+cos{(2x)}}\
&=intfrac{1}{1+cos{(2x)}},dx-intfrac{cos{(2x)}}{1+cos{(2x)}},dx\
&=frac{1}{2}intfrac{1}{1+cos{(2x)}}frac{d}{dx}left(2xright),dx-frac{1}{2}intfrac{cos{(2x)}}{1+cos{(2x)}}frac{d}{dx}left(2xright),dx (u=2x)\
&=frac{1}{2}intfrac{1}{1+cos{u}},du-frac{1}{2}intfrac{cos{u}}{1+cos{u}},du\
end{align}
$$
And now you apply the Weierstrass substitution formulas:
$$
cos{u}=frac{1-t^2}{1+t^2}, du=frac{2}{1+t^2}dt
$$
$$
begin{align}
=frac{1}{2}intfrac{1}{1+frac{1-t^2}{1+t^2}}frac{2}{1+t^2},dt-frac{1}{2}intfrac{frac{1-t^2}{1+t^2}}{1+frac{1-t^2}{1+t^2}}frac{2}{1+t^2},dt
end{align}
$$
At this point, what you've got are purely algebraic expressions under the integral signs. All you need to do is simplify them, take their integrals and do back-substitution.
PS: I hope I didn't make any mistake.
answered Feb 1 at 14:18
Michael RybkinMichael Rybkin
4,254422
$endgroup$
Don't forget that $tan^2{x}=sec^2-1$ and that $int sec^2x,dx=tan{x}+C$ because $frac{d}{dx}left(tan{x}right)=sec^2{x}$:
$$
begin{align}
int tan^2{x},dx
&=int (sec^2x-1),dx\
&=int sec^2x,dx-int,dx\
&=tan{x}-x+C.
end{align}
$$
If you want to do this integral using the half-angle formula for the tangent function, you're gong to have to use the so-called Weierstrass substitution.
$$
tan^2{frac{x}{2}}=frac{sin^2{frac{x}{2}}}{cos^2{frac{x}{2}}}=frac{frac{1-cos{x}}{2}}{frac{1+cos{x}}{2}}=frac{1-cos{x}}{1+cos{x}}.
$$
$$
tan^2{frac{(2x)}{2}}=tan^2{x}=frac{1-cos{(2x)}}{1+cos{(2x)}}.
$$
$$
begin{align}
inttan^2{x},dx
&=int frac{1-cos{(2x)}}{1+cos{(2x)}}\
&=intfrac{1}{1+cos{(2x)}},dx-intfrac{cos{(2x)}}{1+cos{(2x)}},dx\
&=frac{1}{2}intfrac{1}{1+cos{(2x)}}frac{d}{dx}left(2xright),dx-frac{1}{2}intfrac{cos{(2x)}}{1+cos{(2x)}}frac{d}{dx}left(2xright),dx (u=2x)\
&=frac{1}{2}intfrac{1}{1+cos{u}},du-frac{1}{2}intfrac{cos{u}}{1+cos{u}},du\
end{align}
$$
And now you apply the Weierstrass substitution formulas:
$$
cos{u}=frac{1-t^2}{1+t^2}, du=frac{2}{1+t^2}dt
$$
$$
begin{align}
=frac{1}{2}intfrac{1}{1+frac{1-t^2}{1+t^2}}frac{2}{1+t^2},dt-frac{1}{2}intfrac{frac{1-t^2}{1+t^2}}{1+frac{1-t^2}{1+t^2}}frac{2}{1+t^2},dt
end{align}
$$
At this point, what you've got are purely algebraic expressions under the integral signs. All you need to do is simplify them, take their integrals and do back-substitution.
PS: I hope I didn't make any mistake.
answered Feb 1 at 14:18
Michael RybkinMichael Rybkin
4,254422
edited Feb 1 at 15:11
answered Feb 1 at 14:18
Michael RybkinMichael Rybkin
4,254422
answered Feb 1 at 14:18
Michael RybkinMichael Rybkin
4,254422
answered Feb 1 at 14:18
Michael RybkinMichael Rybkin
4,254422
4,254422
$begingroup$
Is there a way to do this with a half angle identity for $ tan^2x$
$endgroup$
– Jwan622
Feb 1 at 14:27
$begingroup$
Probably, there is, but it's not going to be an integral that's easy to integrate because you would end up with something like this: $$intfrac{1}{1+cos{2x}},dx-intfrac{cos{2x}}{1+cos{2x}},dx.$$ I think this one could be done using the Weierstrass substitution.
$endgroup$
– Michael Rybkin
Feb 1 at 14:43
add a comment |
$begingroup$
Is there a way to do this with a half angle identity for $ tan^2x$
$endgroup$
– Jwan622
Feb 1 at 14:27
$begingroup$
Probably, there is, but it's not going to be an integral that's easy to integrate because you would end up with something like this: $$intfrac{1}{1+cos{2x}},dx-intfrac{cos{2x}}{1+cos{2x}},dx.$$ I think this one could be done using the Weierstrass substitution.
$endgroup$
– Michael Rybkin
Feb 1 at 14:43
$begingroup$
Is there a way to do this with a half angle identity for $ tan^2x$
$endgroup$
– Jwan622
Feb 1 at 14:27
$begingroup$
Is there a way to do this with a half angle identity for $ tan^2x$
$endgroup$
– Jwan622
Feb 1 at 14:27
$begingroup$
Probably, there is, but it's not going to be an integral that's easy to integrate because you would end up with something like this: $$intfrac{1}{1+cos{2x}},dx-intfrac{cos{2x}}{1+cos{2x}},dx.$$ I think this one could be done using the Weierstrass substitution.
$endgroup$
– Michael Rybkin
Feb 1 at 14:43
$begingroup$
Probably, there is, but it's not going to be an integral that's easy to integrate because you would end up with something like this: $$intfrac{1}{1+cos{2x}},dx-intfrac{cos{2x}}{1+cos{2x}},dx.$$ I think this one could be done using the Weierstrass substitution.
$endgroup$
– Michael Rybkin
Feb 1 at 14:43
add a comment |
$begingroup$
Hint
use$$(tan x)'=1+tan^2x$$
answered Feb 1 at 14:04
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
add a comment |
$begingroup$
Hint
use$$(tan x)'=1+tan^2x$$
answered Feb 1 at 14:04
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
add a comment |
$begingroup$
Hint
use$$(tan x)'=1+tan^2x$$
answered Feb 1 at 14:04
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
Hint
use$$(tan x)'=1+tan^2x$$
answered Feb 1 at 14:04
Mostafa AyazMostafa Ayaz
18.1k31040
answered Feb 1 at 14:04
Mostafa AyazMostafa Ayaz
18.1k31040
answered Feb 1 at 14:04
Mostafa AyazMostafa Ayaz
18.1k31040
answered Feb 1 at 14:04
Mostafa AyazMostafa Ayaz
18.1k31040
18.1k31040
add a comment |
add a comment |
$begingroup$
Hint:
A useful but not so well known formula is
$$(tan x)'=tan^2x+1.$$
answered Feb 1 at 14:43
Yves DaoustYves Daoust
133k676231
$endgroup$
add a comment |
$begingroup$
Hint:
A useful but not so well known formula is
$$(tan x)'=tan^2x+1.$$
answered Feb 1 at 14:43
Yves DaoustYves Daoust
133k676231
$endgroup$
add a comment |
$begingroup$
Hint:
A useful but not so well known formula is
$$(tan x)'=tan^2x+1.$$
answered Feb 1 at 14:43
Yves DaoustYves Daoust
133k676231
$endgroup$
Hint:
A useful but not so well known formula is
$$(tan x)'=tan^2x+1.$$
answered Feb 1 at 14:43
Yves DaoustYves Daoust
133k676231
answered Feb 1 at 14:43
Yves DaoustYves Daoust
133k676231
answered Feb 1 at 14:43
Yves DaoustYves Daoust
133k676231
answered Feb 1 at 14:43
Yves DaoustYves Daoust
133k676231
133k676231
add a comment |
add a comment |
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$begingroup$
Don't you mean $tan^2x = sec^2x - 1$? Type
$tan x$to obtain $tan x$ and$sec x$to obtain $sec x$.$endgroup$
– N. F. Taussig
Feb 1 at 14:06
2
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$sec^2(x)$ is the derivative of $tan(x)$.
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– Robert Z
Feb 1 at 14:06
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Using that formula seems harder than the original problem.
$endgroup$
– Randall
Feb 1 at 14:23