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  • Need solution for RD Sharma maths class 12 chapter 19 Definite Integrals exercise Fill in the blanks question 15

Need solution for RD Sharma maths class 12 chapter 19 Definite Integrals exercise Fill in the blanks question 15

Answers (1)

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Answer: \frac{\pi }{4}

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

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

Solution:  

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

Put 

        \begin{aligned} &\cos x=t \\\\ &-\sin x\; d x=d t \\\\ &\sin x \; d x=-d t \end{aligned}

When \mathrm{x}=0, \mathrm{t}=1

When \mathrm{x}=\frac{\pi}{2}, \mathrm{t}=0

\begin{aligned} \mathrm{I} &=-\int_{1}^{0} \frac{d t}{1+t^{2}} \\\\ &=\int_{0}^{1} \frac{d t}{1+t^{2}} \end{aligned}

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

    \begin{aligned} &=\left[\frac{\pi}{4}-0\right] \\\\ &=\frac{\pi}{4} \end{aligned}

 

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