#### Need solution for RD Sharma Maths Class 12 Chapter 19 Definite Integrals Excercise Revision Exercise Question 48

Answer:   $I=\frac{1}{\sqrt{2}} \ln |\sqrt{2}+1|$

Hint: To solve this question we use f(x) form

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

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

\begin{aligned} &\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x \\ & \end{aligned}

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

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

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

$2 I=\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{2} x+\cos ^{2} x}{\sin x+\cos x} d x$                                      #check the steps and contents again

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

$2 I=\frac{1}{\sqrt{2}} \int_{0}^{\frac{\pi}{2}} \frac{1}{\cos x \frac{1}{\sqrt{2}}+\sin x \frac{1}{\sqrt{2}}} d x\\$

$2 I=\frac{1}{\sqrt{2}} \int_{0}^{\frac{\pi}{2}} \frac{1}{\cos x \cos \frac{\pi}{4}+\sin x \sin \frac{\pi}{4}} d x$

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

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

\begin{aligned} &\left.2 I=\frac{1}{\sqrt{2}}\left[\ln \mid \sec \left(x-\frac{\pi}{4}\right)+\tan \left(x-\frac{\pi}{4}\right)\right]\right]_{0}^{\frac{\pi}{2}} \\ & \end{aligned}

$2 I=\frac{1}{\sqrt{2}}\left[\ln \mid \sec \left(\frac{\pi}{2}-\frac{\pi}{4}\right)+\tan \left(\frac{\pi}{2}-\frac{\pi}{4}\right)\right]-\ln \left[\sec \left(0-\frac{\pi}{4}\right)+\tan \left(0-\frac{\pi}{4}\right)\right]$

\begin{aligned} &2 I=\frac{1}{\sqrt{2}}\left[\ln \left|\frac{(\sqrt{2}+1)(\sqrt{2}+1)}{(\sqrt{2}-1) \sqrt{2}-1}\right|\right] \\ & \end{aligned}

$2 I=\frac{1}{\sqrt{2}}\left[\ln \left|\frac{(\sqrt{2}+1)^{2}}{(\sqrt{2})^{2}-(1)^{2}}\right|\right.$

\begin{aligned} &2 I=\frac{\sqrt{2} \cdot \sqrt{2}}{\sqrt{2}} \ln |\sqrt{2}+1| \\ & \end{aligned}

$I=\frac{1}{\sqrt{2}} \ln |\sqrt{2}+1|$