Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise 18.27 Question 6

Answer: $I=\frac{4}{5}\left ( \frac{e^{2x\sin x}}{2}-\frac{e^{2x\cos x}}{4} \right )+c$

Hint: $\int f\left ( x \right )g\left ( x \right )dx=f\left ( x \right )\int g\left ( x \right )dx-\int \left ( {f}' \left ( x \right )\int g\left ( x \right )dx\right )dx$

Putting $f\left ( x \right )=e^{2x} \: and\: g\left ( x \right )=\sin x$

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

Solution: $I=\int e^{2x}\sin x dx$

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

\begin{aligned} &=\frac{e^{2 x} \sin x}{2}-\frac{1}{2} \frac{e^{2 x} \cos x}{2}-\frac{1}{4} I \\ &\Rightarrow 1+\frac{1}{4} I=\frac{e^{2 x} \sin x}{2}-\frac{1}{2} \frac{e^{2 x} \cos x}{2}+c \\ &\Rightarrow \frac{5}{4} I=\frac{e^{2 x} \sin x}{2}-\frac{1}{2} \frac{e^{2 x} \cos x}{2}+c \\ &I=\frac{4}{5}\left(\frac{e^{2 x} \sin x}{2}-\frac{e^{2 x} \cos x}{4}\right)+c \end{aligned}

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