#### Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise 18.27 Question 10

Answer: $I=\frac{e^{2x}}{4}+\frac{e^{2x}}{8}\left ( \sin 2x+\cos 2x \right )+c$

Hint: Using ILATE rule

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

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

\begin{aligned} &=\frac{1}{2} \int e^{2 x} d x+\frac{1}{2} \int e^{2 x} \cos 2 x d x \\ &I=\frac{e^{2 x}}{4}+\frac{1}{2} I_{1} \quad \rightarrow(1) \\ &I_{1}=\int e^{2 x} \cos 2 x d x \Rightarrow I=e^{2 x} \frac{\sin 2 x}{2}-\int 2 e^{2 x} \frac{\sin 2 x}{2} d x \\ &=e^{2 x} \frac{\sin 2 x}{2}-\left[e^{2 x}\left(\frac{-\cos 2 x}{2}\right)-\int 2 e^{2 x}\left(\frac{-\cos 2 x}{2}\right) d x\right] \\ &=e^{2 x} \frac{\sin 2 x}{2}+e^{2 x}\left(\frac{\cos 2 x}{2}\right)-\int e^{2 x} \cos 2 x d x \\ &I_{1}+I_{1}=\frac{e^{2 x}}{2}[\sin 2 x+\cos 2 x]+c \\ &I_{1}=\frac{e^{2 x}}{4}[\sin 2 x+\cos 2 x]+c \\ &I=\frac{e^{2 x}}{4}+\frac{e^{2 x}}{8}(\sin 2 x+\cos 2 x)+c[\text { from }(1)] \end{aligned}