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  • Need solution for RD Sharma Maths Class 12 Chapter Definite Integrals exercise 19.2 question 22.

Need solution for RD Sharma Maths Class 12 Chapter Definite Integrals exercise 19.2 question 22.
 

Answers (1)

Answer: \frac{\pi}{4}

Hint: We use indefinite integral formula then put limits to solve this integral.

Given: \int_{0}^{\pi}\frac{1}{3+2\sin x +\cos x}dx

Solution:I=\int_{0}^{\pi}\frac{1}{3+2\sin x +\cos x}dx

Put \sin x=\frac{2 \tan \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}, \cos x=\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}

 

I=\int_{0}^{\pi}\frac{1}{3+2\sin x +\cos x}dx

=\int_{0}^{\pi} \frac{1}{3+2 \frac{2 \tan \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}+\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}} d x

=\int_{0}^{\pi} \frac{1}{\frac{3\left(1+\tan ^{2} \frac{x}{2}\right)+4 \tan \frac{x}{2}+1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}} d x

\begin{aligned} &=\int_{0}^{\pi} \frac{1+\tan ^{2} \frac{x}{2}}{3+3 \tan ^{2} \frac{x}{2}+4 \tan \frac{x}{2}+1-\tan ^{2} \frac{x}{2}} d x \\ &=\int_{0}^{\pi} \frac{\sec ^{2} \frac{x}{2}}{4+2 \tan ^{2} \frac{x}{2}+4 \tan \frac{x}{2}} d x\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\sec ^{2} x=1+\tan ^{2} x\right] \end{aligned}

Put \tan \frac{x}{2}=t

\sec^2\frac{x}{2}.\frac{1}{2}dx=dt

\sec^2\frac{x}{2}dx=2dt

When x=0  then t=0  and when x =\pi   then t=\infty

\begin{aligned} &I=\int_{0}^{\infty} \frac{1}{4+2 t^{2}+4 t} 2 d t \\ &=2 \times \frac{1}{2} \int_{0}^{\infty} \frac{1}{2+t^{2}+2 t} d t \\ &=\int_{0}^{\infty} \frac{1}{t^{2}+2 t+1+1} d t \\ &=\int_{0}^{\infty} \frac{1}{\left(t^{2}+2 t+1\right)+1} d t \\ &=\int_{0}^{\infty} \frac{1}{(t+1)^{2}+1^{2}} d t \end{aligned}

\begin{aligned} &=\frac{1}{1}\left[\tan ^{-1}(t+1)\right]_{0}^{\infty} \\ &=\tan ^{-1}(\infty+1)-\tan ^{-1}(0+1) \\ &=\tan ^{-1} \infty-\tan ^{-1} 1 \\ &=\frac{\pi}{2}-\frac{\pi}{4} \; \; \; \; \; \; \; \; \; \; \quad\left[\tan ^{-1} \infty=\frac{\pi}{2}, \tan ^{-1} 1=\frac{\pi}{4}\right] \\ &=\frac{2 \pi-\pi}{4} \\ &=\frac{\pi}{4} \end{aligned}

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