Question

$\int e^x\sin 3x\cos 2xdx=$

• A. $\frac{e^x}{2}\left[\frac{1}{26}\right(\sin 5x-5\cos 5x)+\frac{1}{2}(\sin x-\cos x)]+c$
• B. $\frac{e^x}{2}\left[\frac{1}{26}\right(\sin 5x+5\cos 5x)+\frac{1}{2}(\sin x+\cos x\left)\right]+c$
• C. $\frac{e^x}{2}\left[\frac{1}{26}\right(\sin 5x-5\cos 5_X)+\frac{1}{2}(\sin x+\cos x\left)\right]+c$
• D. $\frac{e^x}{2}\left[\frac{1}{26}\right(\sin 5x-5\cos 5_X)-\frac{1}{2}(\sin x+\cos x\left)\right]+c$

Answer 1

The correct option is A $\frac{e^x}{2}\left[\frac{1}{26}\right(\sin 5x-5\cos 5x)+\frac{1}{2}(\sin x-\cos x)]+c$

$\sin 3x\cos 2x=\frac{1}{2}[\sin (3x+2x)+\sin (3x-2x)]=\frac{1}{2}(\sin 5x+\sin x)e^x\sin 3x\cos 2x=\frac{e^x}{2}(\sin 5x+\sin x)\int e^x(\sin 5x+\sin x)dx=\int e^x\sin 5xdx+\int e^x\sin xdx$

Let $\int e^x\sin 5xdx=I$

And $\int e^x\sin xdx=J$

$J=\frac{1}{2}\int e^x\sin xdx$

In integration by parts take $\sin x$ as a function of $1$ and $e^x$ as a function $2$

$\Rightarrow 2J=\int e^x\sin xdx{e}^x\sin x-\int e^x\cos xdx$

Again, on integration by parts.

$2J=e^x\sin x-[e^x\cos x+e^x\sin x]2J=\left(\sin x\right)e^x-\left(\cos x\right)e^x-2J4J=e^x(\sin x-\cos x)J=\frac{e^x}{4}(\sin x-\cos x)$

Similarly we can get that

$I=\frac{1}{52}{e}^x(\sin 5x-\cos 5x)$

Hence, $\int e^x\sin 3x\cos 2xdx$

$=\frac{1}{2}{e}^x\left[\frac{1}{26}\right(\sin 5x-\cos 5x)+\frac{1}{2}(\sin x-\cos x\left)\right]+C$