Mixing
Help on integrating a rational function of sine and cosine.
$begingroup$
I got stuck while trying to integrate this function, I've tried to make a new function so that I can solve the integral with a system of equations, but I couldn't get an answer.
Here's the function $h(x)=frac{8sin(x)+3cos(x)}{sin(x)+2cos(x)}$, and I am trying to calculate $int_{0}^{frac{pi}{2}} h(x) dx$
I'd appreciate any help on this.
calculus integration definite-integrals
edited Jan 31 at 21:35
John Wayland Bales
15.1k21238
asked Jan 31 at 21:32
Idriss MoIdriss Mo
158
$endgroup$
add a comment |
$begingroup$
I got stuck while trying to integrate this function, I've tried to make a new function so that I can solve the integral with a system of equations, but I couldn't get an answer.
Here's the function $h(x)=frac{8sin(x)+3cos(x)}{sin(x)+2cos(x)}$, and I am trying to calculate $int_{0}^{frac{pi}{2}} h(x) dx$
I'd appreciate any help on this.
calculus integration definite-integrals
edited Jan 31 at 21:35
John Wayland Bales
15.1k21238
asked Jan 31 at 21:32
Idriss MoIdriss Mo
158
$endgroup$
$begingroup$
try multiplying both the numerator by $sec^3(x)$
$endgroup$
– Aniruddh Venkatesan
Jan 31 at 21:36
1
$begingroup$
math.stackexchange.com/questions/29980/…
$endgroup$
– John Wayland Bales
Jan 31 at 21:38
add a comment |
$begingroup$
I got stuck while trying to integrate this function, I've tried to make a new function so that I can solve the integral with a system of equations, but I couldn't get an answer.
Here's the function $h(x)=frac{8sin(x)+3cos(x)}{sin(x)+2cos(x)}$, and I am trying to calculate $int_{0}^{frac{pi}{2}} h(x) dx$
I'd appreciate any help on this.
calculus integration definite-integrals
edited Jan 31 at 21:35
John Wayland Bales
15.1k21238
asked Jan 31 at 21:32
Idriss MoIdriss Mo
158
$endgroup$
I got stuck while trying to integrate this function, I've tried to make a new function so that I can solve the integral with a system of equations, but I couldn't get an answer.
Here's the function $h(x)=frac{8sin(x)+3cos(x)}{sin(x)+2cos(x)}$, and I am trying to calculate $int_{0}^{frac{pi}{2}} h(x) dx$
I'd appreciate any help on this.
calculus integration definite-integrals
calculus integration definite-integrals
edited Jan 31 at 21:35
John Wayland Bales
15.1k21238
asked Jan 31 at 21:32
Idriss MoIdriss Mo
158
edited Jan 31 at 21:35
John Wayland Bales
15.1k21238
asked Jan 31 at 21:32
Idriss MoIdriss Mo
158
edited Jan 31 at 21:35
John Wayland Bales
15.1k21238
edited Jan 31 at 21:35
John Wayland Bales
15.1k21238
edited Jan 31 at 21:35
John Wayland Bales
15.1k21238
15.1k21238
asked Jan 31 at 21:32
Idriss MoIdriss Mo
158
asked Jan 31 at 21:32
Idriss MoIdriss Mo
158
asked Jan 31 at 21:32
Idriss MoIdriss Mo
158
158
$begingroup$
try multiplying both the numerator by $sec^3(x)$
$endgroup$
– Aniruddh Venkatesan
Jan 31 at 21:36
1
$begingroup$
math.stackexchange.com/questions/29980/…
$endgroup$
– John Wayland Bales
Jan 31 at 21:38
add a comment |
$begingroup$
try multiplying both the numerator by $sec^3(x)$
$endgroup$
– Aniruddh Venkatesan
Jan 31 at 21:36
1
$begingroup$
math.stackexchange.com/questions/29980/…
$endgroup$
– John Wayland Bales
Jan 31 at 21:38
$begingroup$
try multiplying both the numerator by $sec^3(x)$
$endgroup$
– Aniruddh Venkatesan
Jan 31 at 21:36
$begingroup$
try multiplying both the numerator by $sec^3(x)$
$endgroup$
– Aniruddh Venkatesan
Jan 31 at 21:36
1
1
$begingroup$
math.stackexchange.com/questions/29980/…
$endgroup$
– John Wayland Bales
Jan 31 at 21:38
$begingroup$
math.stackexchange.com/questions/29980/…
$endgroup$
– John Wayland Bales
Jan 31 at 21:38
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
So, the idea is to split the integrand $frac{8sin x+3cos x}{sin x+2cos x}$ into two parts, both of which will be easy to integrate. One of those, clearly, should be a multiple of $1=frac{sin x+2cos x}{sin x+2cos x}$. The other? Well, it would be nice if we could just substitute $sin x+2cos x$. That works when the numerator is the derivative $cos x - 2sin x$, or a multiple of that.
So, there's our system:
$$8sin x+3cos x=a(sin x+2cos x)+b(-2sin x+cos x)$$
Equate coefficients for
$$a-2b = 8$$
$$2a+b = 3$$
Can you finish it from there?
answered Jan 31 at 21:46
jmerryjmerry
17k11633
$endgroup$
$begingroup$
Yes, that works for me.
$endgroup$
– Idriss Mo
Feb 1 at 6:44
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
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active
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votes
$begingroup$
So, the idea is to split the integrand $frac{8sin x+3cos x}{sin x+2cos x}$ into two parts, both of which will be easy to integrate. One of those, clearly, should be a multiple of $1=frac{sin x+2cos x}{sin x+2cos x}$. The other? Well, it would be nice if we could just substitute $sin x+2cos x$. That works when the numerator is the derivative $cos x - 2sin x$, or a multiple of that.
So, there's our system:
$$8sin x+3cos x=a(sin x+2cos x)+b(-2sin x+cos x)$$
Equate coefficients for
$$a-2b = 8$$
$$2a+b = 3$$
Can you finish it from there?
answered Jan 31 at 21:46
jmerryjmerry
17k11633
$endgroup$
$begingroup$
Yes, that works for me.
$endgroup$
– Idriss Mo
Feb 1 at 6:44
add a comment |
$begingroup$
So, the idea is to split the integrand $frac{8sin x+3cos x}{sin x+2cos x}$ into two parts, both of which will be easy to integrate. One of those, clearly, should be a multiple of $1=frac{sin x+2cos x}{sin x+2cos x}$. The other? Well, it would be nice if we could just substitute $sin x+2cos x$. That works when the numerator is the derivative $cos x - 2sin x$, or a multiple of that.
So, there's our system:
$$8sin x+3cos x=a(sin x+2cos x)+b(-2sin x+cos x)$$
Equate coefficients for
$$a-2b = 8$$
$$2a+b = 3$$
Can you finish it from there?
answered Jan 31 at 21:46
jmerryjmerry
17k11633
$endgroup$
$begingroup$
Yes, that works for me.
$endgroup$
– Idriss Mo
Feb 1 at 6:44
add a comment |
$begingroup$
So, the idea is to split the integrand $frac{8sin x+3cos x}{sin x+2cos x}$ into two parts, both of which will be easy to integrate. One of those, clearly, should be a multiple of $1=frac{sin x+2cos x}{sin x+2cos x}$. The other? Well, it would be nice if we could just substitute $sin x+2cos x$. That works when the numerator is the derivative $cos x - 2sin x$, or a multiple of that.
So, there's our system:
$$8sin x+3cos x=a(sin x+2cos x)+b(-2sin x+cos x)$$
Equate coefficients for
$$a-2b = 8$$
$$2a+b = 3$$
Can you finish it from there?
answered Jan 31 at 21:46
jmerryjmerry
17k11633
$endgroup$
So, the idea is to split the integrand $frac{8sin x+3cos x}{sin x+2cos x}$ into two parts, both of which will be easy to integrate. One of those, clearly, should be a multiple of $1=frac{sin x+2cos x}{sin x+2cos x}$. The other? Well, it would be nice if we could just substitute $sin x+2cos x$. That works when the numerator is the derivative $cos x - 2sin x$, or a multiple of that.
So, there's our system:
$$8sin x+3cos x=a(sin x+2cos x)+b(-2sin x+cos x)$$
Equate coefficients for
$$a-2b = 8$$
$$2a+b = 3$$
Can you finish it from there?
answered Jan 31 at 21:46
jmerryjmerry
17k11633
answered Jan 31 at 21:46
jmerryjmerry
17k11633
answered Jan 31 at 21:46
jmerryjmerry
17k11633
answered Jan 31 at 21:46
jmerryjmerry
17k11633
17k11633
$begingroup$
Yes, that works for me.
$endgroup$
– Idriss Mo
Feb 1 at 6:44
add a comment |
$begingroup$
Yes, that works for me.
$endgroup$
– Idriss Mo
Feb 1 at 6:44
$begingroup$
Yes, that works for me.
$endgroup$
– Idriss Mo
Feb 1 at 6:44
$begingroup$
Yes, that works for me.
$endgroup$
– Idriss Mo
Feb 1 at 6:44
add a comment |
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$begingroup$
try multiplying both the numerator by $sec^3(x)$
$endgroup$
– Aniruddh Venkatesan
Jan 31 at 21:36
1
$begingroup$
math.stackexchange.com/questions/29980/…
$endgroup$
– John Wayland Bales
Jan 31 at 21:38