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Question -

$$\int_0^{\frac{\pi}{2}} \frac{dx}{3+2\sin x+\cos x}$$

I have tried this question and also tried to find online on google. But I only found questions that involves only 2 terms. So i dont have any other way except to ask it here.

Please help me.

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    After applying the method in one of those duplicates, plug in your endpoints to get $\arctan (2)-\pi/4$. – 2017-02-06

2 Answers 2

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HINT:

Use Weierstrass Substitution

For $\dfrac1{A+B\sin2y+C\cos2y}=\dfrac{\sec^2y}{A(1+\tan^2y)+2B\tan y+C(1-\tan^2y)}$

Set $\tan y=u$

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    @S.C.B., I've started with $\sin2y,\cos2y$ – 2017-02-06
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    But we dont have this method in our book. Other questions with 2 terms in denominator uses $sin x = \frac{2tan x}{1+tan^x}$ substitutions. Similarly for cos x. – 2017-02-06
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Calculate the indefinite integral:

$\int_0^{\frac{\pi}{2}} \frac{dx}{3+2\sin x+\cos x}$

Apply integration by substitution:

$u=\tan (\frac{x}{2})$

$\sin (x)=\frac{2u}{1+u^{2}}$, $\cos (x)=\frac{1-u^{2}}{1+u^{2}}$

$dx=\frac{2}{1+u^{2}}du$

$\int \dfrac{dx}{3+2\sin x+\cos x}=\int \dfrac{1}{3+2\cdot \frac{2u}{1+u^{2}}+\frac{1-u^{2}}{1+u^{2}}}\cdot \dfrac{2}{1+u^{2}}du=\int \dfrac{1}{u(u+2)+2}du=\int \dfrac{1}{(u+1)^{2}+1}du$

Apply integration by substitution:

$v=(u+1)$

$\frac{dv}{du}=1$, $dv=1du$, $du=1dv$

$\int \dfrac{1}{(u+1)^{2}+1}du=\int \dfrac{1}{v^{2}+1}\cdot 1dv=\int \dfrac{1}{v^{2}+1}dv=arc tan (v)$

Substitute in equation $v=(1+u)$, $u=tan(\frac{x}{2})$.

$\int \dfrac{1}{v^{2}+1}dv=arc tan (v)=arctan (tan(\frac{x}{2})+1) + C$

Calculate the limits:

$\int_0^{\frac{\pi}{2}} \frac{dx}{3+2\sin x+\cos x}=arctan (2)-\frac{\pi}{4}$