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  • Need solution for RD Sharma maths class 12 chapter 19 Definite Integrals exercise Fill in the blanks question 11

Need solution for RD Sharma maths class 12 chapter 19 Definite Integrals exercise Fill in the blanks question 11

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Answer: \frac{\pi }{8}

Hint: Use \int \frac{1}{1+x^{2}} d x

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

Solution:  

\mathrm{I}=\int_{0}^{\pi / 2} \frac{\sin x \cos x}{1+\sin ^{4} x} d x

Put

    \begin{aligned} &\sin ^{2} x=t \\\\ &2 \sin x \cos x \; d x=d t \\\\ &\sin x \cos x \; d x=\frac{d t}{2} \end{aligned}

When \mathrm{x}=0, \mathrm{t}=0

When \mathrm{x}=\frac{\pi}{2}, \mathrm{t}=1

I=\int_{0}^{\pi / 2} \frac{\sin x \cos x}{1+\sin ^{4} x} d x

    \begin{aligned} &=\int_{0}^{1} \frac{1}{1+t^{2}} \cdot \frac{d t}{2} \\\\ &=\frac{1}{2}\left[\tan ^{-1} t\right]_{0}^{1} \end{aligned}

    \begin{aligned} &=\frac{1}{2}\left[\tan ^{-1}(1)-\tan ^{-1}(0)\right] \\\\ &=\frac{1}{2}\left[\tan ^{-1}(1)-\tan ^{-1}(0)\right] \end{aligned}

    \begin{aligned} &=\frac{1}{2}\left[\tan ^{-1}\left(\tan \frac{\pi}{4}\right)-\tan ^{-1}(\tan 0)\right] \\\\ &=\frac{1}{2}\left(\frac{\pi}{4}\right)-\frac{1}{2}(0) \\\\ &=\frac{\pi}{8} \end{aligned}

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