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

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

Solution:

\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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