#### Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise Very Short Answers Question 31

Answer: $\frac{e^{ax}}{a^{2}+b^{2}}\left ( a\sin bx-b\cos bx \right )+c$

Hints: You must know about the integral rule of trigonometric and exponential functions

Given: $\int e^{ax}\sin bxdx$

Solution:

$I=\int e^{ax}\sin bxdx$ and using integration by parts,

\begin{aligned} &\sin b x \int e^{a x} d x-\int b \cos b x \frac{e^{a x}}{a} d x \\ &=\frac{1}{a} \sin b x e^{a x}-\frac{b}{a} \int \cos b x e^{a x} d x \\ &=\frac{1}{a} \sin b x e^{a x}-\frac{b}{a} \cos b x \int\left[e^{a x} d x+\frac{b}{a}\left[\int \frac{d}{d x}(\cos b x)\right]-\int e^{a x} d x\right] d x \\ &I=\frac{1}{a} \sin b x e^{a x}-\frac{b}{a} \cos b x \cdot \frac{e^{a x}}{a}+\frac{b}{a} \int(-\sin b x) \cdot \frac{b e^{a x}}{a} d x \\ &I=\frac{1}{a} \sin b x e^{a x}-\frac{b}{a^{2}} \cos b x \cdot e^{a x}-\frac{b^{2}}{a^{2}} I \end{aligned}

\begin{aligned} &I\left(1+\frac{b^{2}}{a^{2}}\right)=\frac{1}{a} \sin b x \cdot e^{a x}-\frac{b}{a^{2}} \cos b x \cdot e^{a x} \\ &\Rightarrow \int e^{a x} \cdot \sin b x d x=\frac{e^{a x}}{a^{2}+b^{2}}(a \sin b x-b \cos b x)+c \end{aligned}