#### Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise 18.27 Question 5

Answer: $I=\frac{e^{2 x}}{8}[\sin 2 x-\cos 2 x]+c$

Hint: $[\sin 2 x-\cos 2 x]$

Given: $\int e^{2 x} \sin x \cos x d x$

Solution: $I=\frac{1}{2} \int e^{2 x} \sin x \cos x d x$

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

\begin{aligned} &=\frac{1}{4} \sin 2 x e^{2 x}-\frac{1}{2} \int e^{2 x} \cos 2 x d x \\ &=\frac{1}{4} \sin 2 x e^{2 x}-\frac{1}{2}\left[\cos 2 x \frac{e^{2 x}}{2}+\int 2 \sin 2 x \frac{e^{2 x}}{2} d x\right] \\ &=\frac{1}{4} e^{2 x} \sin 2 x-\frac{1}{4} e^{2 x} \cos 2 x-I+c \end{aligned}

\begin{aligned} &I+I=\frac{1}{4} e^{2 x} \sin 2 x-\frac{1}{4} e^{2 x} \cos 2 x+c \\ &2 I=\frac{1}{4} e^{2 x} \sin 2 x-\frac{1}{4} e^{2 x} \cos 2 x+c \\ &I=\frac{1}{2}\left[\frac{1}{4} e^{2 x} \sin 2 x-\frac{1}{4} e^{2 x} \cos 2 x\right]+c \\ &I=\frac{e^{2 x}}{8}[\sin 2 x-\cos 2 x]+c \end{aligned}