Using the identity $\sin x\cos y= frac12[\sin(x+y)+\sin(x-y)]$, the v…

Question

Using the identity $\sin x\cos y=	frac12[\sin(x+y)+\sin(x-y)]$, the value of $\int \sin 2x\cos 3x\,dx$ is

Accepted Answer

B. $-\dfrac{1}{10}\cos 5x+\dfrac{1}{2}\cos x+C$

1. $\sin 2x\cos 3x=	frac12[\sin 5x+\sin(-x)]=	frac12[\sin 5x-\sin x]$.
2. $\int \sin 5x\,dx=-	frac15\cos 5x$ and $\int \sin x\,dx=-\cos x$.
3. So integral $=	frac12\left(-	frac15\cos 5x+\cos xight)$.
4. $=-\dfrac{1}{10}\cos 5x+\dfrac{1}{2}\cos x+C$. Option A flips the $\cos x$ sign.
_Source: NCERT Class 12 Mathematics Ch 7 "Integrals", p.305_

Answer

$-\dfrac{1}{10}\cos 5x-\dfrac{1}{2}\cos x+C$

Answer

$\dfrac{1}{10}\cos 5x+\dfrac{1}{2}\cos x+C$

Answer

$-\dfrac{1}{5}\cos 5x+\cos x+C$

Using the identity $\sin x\cos y=\tfrac12[\sin(x+y)+\sin(x-y)]$, the value of $\int \sin 2x\cos 3x\,dx$ is