#### Please Solve R.D.Sharma class 12 Chapter 19 Definite Integrals Exercise 19.1 Question 52 Maths textbook Solution.

Answer:$\frac{-3\sqrt{2}}{5}\left ( e^{2\pi}+1 \right )$

Hint: Use indefinite formula then put the limit to solve this integral

Given:

$\int_{0}^{2\pi }e^{x}\cos \left ( \frac{\pi }{4}+\frac{x}{2} \right )dx$

Solution:

$I=\int_{0}^{2\pi }e^{x}\cos \left ( \frac{\pi }{4}+\frac{x}{2} \right )dx$                                                            ....................(1)

Apply integration by parts method, then

$\int_{0}^{2 \pi} e^{x} \cos \left(\frac{\pi}{4}+\frac{x}{2}\right) d x=\left[\cos \left(\frac{\pi}{4}+\frac{x}{2}\right) \int e^{x} d x\right]_{0}^{2 \pi}-\int_{0}^{2 \pi}\left\{\frac{d}{d x} \cos \left(\frac{\pi}{4}+\frac{x}{2}\right) \int e^{x} d x\right\} d x$

$\Rightarrow I=\left[\cos \left(\frac{\pi}{4}+\frac{x}{2}\right)\right]_{0}^{2 \pi}-\int_{0}^{2 \pi}-\sin \left(\frac{\pi}{4}+\frac{x}{2}\right) \frac{1}{2} e^{x} d x \quad\left[\int e^{x} d x=e^{x} \frac{d \cos (a x)}{d x}=-\sin a x \cdot a\right]$

$\Rightarrow I=\left[e^{2 \pi} \cos \left(\frac{\pi}{4}+\frac{2 \pi}{2}\right)-e^{0} \cos \left(\frac{\pi}{4}+\frac{0}{2}\right)\right]+\frac{1}{2} \int_{0}^{2 \pi} \sin \left(\frac{\pi}{4}+\frac{x}{2}\right) e^{x} d x$

$\Rightarrow I=\left[e^{2 \pi} \cos \left(\frac{\pi}{4}+\pi\right)-\cos \frac{\pi}{4}\right]+\frac{1}{2}\left[\left\{\sin \left(\frac{\pi}{4}+\frac{x}{2}\right) \int e^{x} d x\right\}_{0}^{2 \pi}-\int_{0}^{2 \pi}\left\{\frac{d}{d x} \sin \left(\frac{\pi}{4}+\frac{x}{2}\right) \int e^{x} d x\right\} d x\right]$

$\Rightarrow I=\left[e^{2 \pi}\left(-\cos \frac{\pi}{4}\right)-\cos \left(\frac{\pi}{4}\right)\right]+\frac{1}{2}\left\{\left[\sin \left(\frac{\pi}{4}+\frac{x}{2}\right)\right]_{0}^{2 \pi}-\int_{0}^{2 \pi} \cos \left(\frac{\pi}{4}+\frac{x}{2}\right) \frac{1}{2} e^{x} d x\right\}$

(Again using integrating by parts method)

$\left[\frac{d}{d x} \sin a x=\cos a x \cdot a, \int e^{x}=\frac{e^{x}}{2 \pi}\right]$

$\inline \Rightarrow I=\left[e^{2 \pi}\left(\frac{-1}{\sqrt{2}}\right)-\frac{1}{\sqrt{2}}\right]+\frac{1}{2}\left[\sin \left(\frac{\pi}{4}+\frac{2 \pi}{2}\right) e^{2 \pi} \sin \left(\frac{\pi}{4}+\frac{0}{2}\right) e^{0}\right]-\frac{1}{4} \int_{0}^{2 \pi} \cos \left(\frac{\pi}{4}+\frac{x}{2}\right) e^{x} d x$

$\left [ \cos \frac{\pi }{4}=\frac{1}{\sqrt{2}} \right ]$

$\Rightarrow I=\left[\frac{-e^{2 \pi}}{\sqrt{2}}-\frac{1}{\sqrt{2}}\right]+\frac{1}{2}\left[\sin \left(\pi+\frac{\pi}{4}\right) e^{2 \pi}-\sin \frac{\pi}{4}\right]-\frac{1}{4} I$

$\Rightarrow I=\frac{-1}{\sqrt{2}}\left(e^{2 \pi}+1\right)+\frac{1}{2}\left(-\sin \frac{\pi}{4} e^{2 \pi}-\sin \frac{\pi}{4}\right)-\frac{1}{4} I$                                                $\left [ \sin \frac{\pi }{4}=\frac{1}{\sqrt{2}}\right ]$

$\Rightarrow I=\frac{-1}{\sqrt{2}}\left(e^{2 \pi}+1\right)-\frac{1}{2}\left[\frac{1}{\sqrt{2}} e^{2 \pi}+\frac{1}{\sqrt{2}}\right]-\frac{1}{4} I$

$\Rightarrow I+\frac{1}{4} I=\frac{-1}{\sqrt{2}}\left(e^{2 \pi}+1\right)-\frac{1}{2 \sqrt{2}} e^{2 \pi}-\frac{1}{2 \sqrt{2}}$

$\Rightarrow \frac{4 I+I}{4}=\frac{-1}{\sqrt{2}}\left(e^{2 \pi}+1\right)-\frac{1}{2 \sqrt{2}}\left(e^{2 \pi}+1\right)$

$\Rightarrow \frac{5 I}{4}=\left(e^{2 \pi}+1\right)\left[\frac{-1}{\sqrt{2}}-\frac{1}{2 \sqrt{2}}\right]=\left(e^{2 \pi}+1\right)\left[\frac{-2-1}{2 \sqrt{2}}\right]$

$\Rightarrow I=\frac{4}{5} \frac{-3}{2 \sqrt{2}}\left(e^{2 \pi}+1\right)=\frac{2 \times \sqrt{2} \times \sqrt{2}}{5} \frac{-3}{2 \sqrt{2}}\left(e^{2 \pi}+1\right)$

$\Rightarrow I=\frac{-3 \sqrt{2}}{5}\left(e^{2 \pi}+1\right)$