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

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

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best_answer

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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