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Explain solution RD Sharma class 12 Chapter 19 Definite Integrals Exercise Revision Exercise question 53

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Answer:  \frac{2}{\sqrt{2}} \log (\sqrt{2}+1)

Hint: To solve this equation we use   \int_{0}^{a} f\left ( x \right )  formula

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


 \begin{aligned} &I=\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{2}\left(\frac{\pi}{2}-x\right)}{\sin \left(\frac{\pi}{2}-x\right)+\cos \left(\frac{\pi}{2}-x\right)} d x \\ & \end{aligned}

=\int_{0}^{\frac{\pi}{2}} \frac{\cos ^{2} x}{\cos x+\sin x} d x

\begin{aligned} &\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{2} x}{\cos x+\sin x} d x \\ & \end{aligned}

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

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

=\frac{1}{\sqrt{2}} \int_{0}^{\frac{\pi}{2}} \frac{1}{\cos \left(x-\frac{\pi}{4}\right)} d x \\

=\frac{1}{\sqrt{2}} \int_{0}^{\frac{\pi}{2}} \sec \left(x-\frac{\pi}{4}\right) d x

\begin{aligned} &=\frac{1}{2}\left\{\log \sec \left|x-\frac{\pi}{9}\right|+\tan \left|x-\frac{\pi}{9}\right|\right\}_{0}^{\frac{\pi}{2}} \\ & \end{aligned}

=\frac{1}{\sqrt{2}} \log \left|\sec \frac{\pi}{4}+\tan \frac{\pi}{4}\right|-\log \left|\sec \left(-\frac{\pi}{4}\right)+\tan \left(-\frac{\pi}{4}\right)\right| \\

=\frac{1}{\sqrt{2}}\{\log (\sqrt{2}+1)-\log (\sqrt{2}-1)\}

\begin{aligned} &=\frac{1}{\sqrt{2}} \log \left(\frac{\sqrt{2}+1}{\sqrt{2}-1}\right) \\ & \end{aligned}

=\frac{1}{\sqrt{2}} \log \left(\frac{(\sqrt{2}+1)^{2}}{2-1}\right) \\

=\frac{2}{\sqrt{2}} \log (\sqrt{2}+1)

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