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  • Need solution for RD Sharma Maths Class 12 Chapter Definite Integrals exercise 19.2 question 18.

Need solution for RD Sharma Maths Class 12 Chapter Definite Integrals exercise 19.2 question 18.

Answers (1)

Answer: \frac{\pi}{8}

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

Given: \int_{0}^{\frac{\pi}{2}}\frac{\sin x\cos x}{1+\sin^4 x}dx

Solution:I=\int_{0}^{\frac{\pi}{2}}\frac{\sin x\cos x}{1+\sin^4 x}dx

Put \sin2x=t

2\sin x \cos x=dt

\sin x \cos xdx=\frac{dt}{2}

When x=0  then t=0  and when x=\frac{\pi}{2}  then t=1

\begin{aligned} &I=\int_{0}^{1} \frac{1}{1+t^{2}} \frac{d t}{2} \\ &=\frac{1}{2} \int_{0}^{1} \frac{1}{1+t^{2}} d t \\ &=\frac{1}{2}\left[\tan ^{-1}(t)\right]_{0}^{1} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\int \frac{1}{a^{2}+x^{2}} d x=\frac{1}{a} \tan ^{-1} \frac{x}{a}\right] \end{aligned}

\begin{aligned} &=\frac{1}{2}\left[\tan ^{-1} 1-\tan ^{-1} 0\right]\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\tan ^{-1} 1=\frac{\pi}{4}, \tan ^{-1} 0=0\right] \\ &=\frac{1}{2}\left[\frac{\pi}{4}-0\right] \\ &=\frac{1}{2} \cdot \frac{\pi}{4} \\ &=\frac{\pi}{8} \end{aligned}

 

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