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Explain solution RD Sharma class 12 chapter 19 Definite Integrals exercise Very short answer type question 16 maths

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Answer: \frac{\pi }{4}

Hint: You must know the integration rules of trigonometric function with its limits

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

Solution:  \int_{0}^{\frac{\pi}{2}} \frac{\sin ^{n} x}{\sin ^{n} x+\cos ^{n} x} d x        ....................(i)

Property: \int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x

=\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{n}\left(\frac{\pi}{2}-x\right)}{\sin ^{n}\left(\frac{\pi}{2}-x\right)+\cos ^{n}\left(\frac{\pi}{2}-x\right)} d x

=\int_{0}^{\frac{\pi}{2}} \frac{\cos ^{n} x}{\sin ^{n} x+\cos ^{n} x} d x    ..................(ii)

Add (i) and (ii) 

\begin{aligned} &2 \mathrm{l}=\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{n} x+\cos ^{n} x}{\sin ^{n} x+\cos ^{n} x} d x \\\\ &2 \mathrm{I}=\int_{0}^{\frac{\pi}{2}} 1 \cdot d x \end{aligned}

\begin{aligned} &21=[x]_{0}^{\frac{\pi}{2}} \\\\ &2 I=\left[\frac{\pi}{2}-0\right] \\\\ &I=\frac{\pi}{4} \end{aligned}


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