#### Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise 18.27 Question 7

Answer: $I=\frac{e^{2x}}{13}\left \{ 2\sin\left ( 3x+1 \right )-3\cos \left ( 3x+1 \right ) \right \}+c$

Hint: Consider $\sin \left ( 3x+1 \right )$ as function 1 and $e^{2x}$ as function 2

Given: $\int e^{2x}\sin \left ( 3x+1 \right )dx$

Solution: $I= \int e^{2x}\sin \left ( 3x+1 \right )dx$

$I=\int e^{2 x} \sin (3 x+1) d x\\\\ Now \; integrating \; by \; parts\; chosing \sin (3 x+1) as\: first\; function \; and\; e^{2 x} as\; second\; function\; we \; get\\\\ I=\sin (3 x+1) \int e^{2 x} d x-\int\left\{\frac{d}{d x} \sin (3 x+1) \int e^{2 x} d x\right\} d x\\\\ =\frac{e^{2 x}}{2} \sin (3 x+1)-\int \frac{3 e^{2 x}}{2} \cos (3 x+1) d x$

$Now\; again \; integrating \; by \; parts\: chosing \cos (3 x+1) as\; first \; function \; and \; e^{2 x} as\; second \; function\; we\; get\\\\ I=\frac{e^{2 x} \sin (3 x+1)}{2}-\frac{3}{2}\left[\cos (3 x+1) \int e^{2 x} d x-\int\left\{\frac{d}{d x} \cos (3 x+1) \int e^{2 x} d x\right\}\right] d x\\\\ =\frac{e^{2 x} \sin (3 x+1)}{2}-\frac{3}{2}\left[\cos (3 x+1) \frac{e^{2 x}}{2}-\int-3 \sin (3 x+1) \frac{e^{2 x}}{2} d x\right]+c_{1}\\\\ =\frac{e^{2 x} \sin (3 x+1)}{2}-\frac{3}{4} e^{2 x} \cos (3 x+1)-\frac{9}{4} \int e^{2 x} \sin (3 x+1) d x+c_{1}$

\begin{aligned} &\int e^{2 x} \sin (3 x+1) d x=I \\ &=\frac{e^{2 x} \sin (3 x+1)}{2}-\frac{3}{4} e^{2 x} \cos (3 x+1)-\frac{9}{4} I+c_{1} \\ &\Rightarrow \frac{13}{4} I=\frac{e^{2 x} \sin (3 x+1)}{2}-\frac{3 e^{2 x} \cos (3 x+1)}{4}+c_{1} \\ &\therefore I=\frac{e^{2 x}}{13}\{2 \sin (3 x+1)-3 \cos (3 x+1)\}+c \end{aligned}