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

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

Answers (1)

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)

Posted by

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