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

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

Answers (1)

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|

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