#### Provide Solution For R.D.Sharma Maths Class 12 Chapter 19 definite Integrals Exercise 19.1 Question 53 Maths Textbook Solution.

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

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

Given: $\int_{0}^{\pi} e^{2 x} \sin \left(\frac{\pi}{4}+x\right) d x$

Solution: Let$I=\int_{0}^{\pi} e^{2 x} \sin \left(\frac{\pi}{4}+x\right) d x$                                                ...............................(1)

Apply integration by parts method

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

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

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

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

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

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

$\left[\because \frac{d}{d x} \cos a x=-\sin a x . a, \int e^{a x} d x=\frac{e^{a x}}{a}, \sin (\pi+x)=-\sin x\right] \\$

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

$\inline \left [ \because \sin \frac{\pi }{4}=\frac{1}{\sqrt{2}} \right ]$

\inline \begin{aligned} &\Rightarrow I=\frac{1}{2} \cdot \frac{-1}{\sqrt{2}}\left[e^{2 \pi}+1\right]-\frac{1}{4}\left[-\cos \frac{\pi}{4} e^{2 \pi}-\cos \frac{\pi}{4}\right]-\frac{1}{4} I \quad[\because \cos (\pi+x)=-\cos x] \\ &\Rightarrow I+\frac{1}{4} I=\frac{-1}{2 \sqrt{2}}\left[e^{2 \pi}+1\right]+\frac{1}{4}\left[\frac{1}{\sqrt{2}} e^{2 \pi}+\frac{1}{\sqrt{2}}\right] \quad\left[\because \cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}\right] \\ &\Rightarrow \frac{4 I+I}{4}=\frac{-1}{2 \sqrt{2}}\left[e^{2 \pi}+1\right]+\frac{1}{4} \frac{1}{\sqrt{2}}\left(e^{2 \pi}+1\right) \\ &\Rightarrow \frac{5 I}{4}=\left[e^{2 \pi}+1\right]\left(\frac{1}{4 \sqrt{2}}-\frac{1}{2 \sqrt{2}}\right)=\left[e^{2 \pi}+1\right]\left(\frac{1-2}{4 \sqrt{2}}\right) \\ &\Rightarrow I=\frac{4}{5}\left(\frac{-1}{4 \sqrt{2}}\right)\left(e^{2 \pi}+1\right)=\frac{-1}{5 \sqrt{2}}\left(e^{2 \pi}+1\right) \end{aligned}