Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise 18.27 Question 2

Answer: $I=\frac{e^{ax}}{1+\frac{b^{2}}{a^{2}}}\left ( \sin \left ( bx+c \right ) -\frac{b}{a}\cos \left ( bx+c \right )\right )+c$

Hint: Using ILATE rule

Given: $\int e^{ax}\sin\left ( bx+c \right )dx$

Solution: $I=\int e^{ax}\sin\left ( bx+c \right )dx$

\begin{aligned} &I=\int e^{a x} \sin (b x+c) d x=\int \sin (b x+c) \cdot e^{a x} d x\\ &=\sin (b x+c) \int e^{a x} d x-\int \frac{d}{d x}(\sin (b x+c)) \int e^{a x} d x \int d x\\ &=\sin (b x+c) \frac{e^{a x}}{a}-\int \cos (b x+c) \frac{e^{a x}}{a} d x\\ &\text { Again using integration by parts } \end{aligned}

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