#### Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise 18.27 Question 8

Solution:

We need

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

Now perform the integration by parts

\begin{aligned} &u=1-\cos 2 x \Rightarrow u^{\prime}=2 \sin 2 x \\ &v^{\prime}=e^{x} \Rightarrow v=e^{x} \end{aligned}

\begin{aligned} &\text { Therefore, }\\ &I=\frac{1}{2}\left(e^{x}(1-\cos 2 x)-\int e^{x} 2 \sin 2 x d x\right)\\ &=\frac{e^{x}}{2}(1-\cos 2 x)-\int e^{x} 2 \sin 2 x d x \end{aligned}

Again integration by parts

\begin{aligned} &u=\sin 2 x \Rightarrow u^{\prime}=2 \cos 2 x \\ &v^{\prime}=e^{x} \Rightarrow v=e^{x} \\ &I=\frac{1}{2} \int e^{x} d x-\frac{1}{2} \int e^{x} \cos 2 x d x \\ &=\frac{e^{x}}{2}(1-\cos 2 x)-e^{x} \sin 2 x+2 \int e^{x} \cos 2 x d x \end{aligned}

\begin{aligned} &\text { Therefore, }\\ &\frac{5}{2} \int e^{x} \cos 2 x d x=\frac{e^{x}}{2}-\frac{e^{x}}{2}+\frac{1}{2} e^{x} \cos 2 x+e^{x} \sin 2 x\\ &\int e^{x} \cos 2 x d x=\int\left(e^{x}\right)\left(1-2 \sin ^{2} x\right) d x=e^{x}-2 \int e^{x} \sin ^{2} d x \end{aligned}

\begin{aligned} &\text { So, }\\ &e^{x}-2 \int e^{x} \sin ^{2} x d x=\frac{2}{5}\left(e^{x}\right)\left(\frac{1}{2} \cos 2 x+\sin 2 x\right)\\ &\int e^{x} \sin ^{2} x d x=\frac{e^{x}}{2}-\frac{1}{10} e^{x} \cos 2 x-\frac{1}{5} e^{x} \sin 2 x+c\\ &I=\frac{e^{x}}{2}-\frac{1}{10} e^{x} \cos 2 x-\frac{1}{5} e^{x} \sin 2 x+c \end{aligned}