Using the identity $\sin x\cos y=\tfrac12[\sin(x+y)+\sin(x-y)]$, the value of $\int \sin 2x\cos 3x\,dx$ is
A$-\dfrac{1}{10}\cos 5x-\dfrac{1}{2}\cos x+C$
B$-\dfrac{1}{10}\cos 5x+\dfrac{1}{2}\cos x+C$
C$\dfrac{1}{10}\cos 5x+\dfrac{1}{2}\cos x+C$
D$-\dfrac{1}{5}\cos 5x+\cos x+C$
Answer & Solution
Correct answer: B. $-\dfrac{1}{10}\cos 5x+\dfrac{1}{2}\cos x+C$
1. $\sin 2x\cos 3x=\tfrac12[\sin 5x+\sin(-x)]=\tfrac12[\sin 5x-\sin x]$.
2. $\int \sin 5x\,dx=-\tfrac15\cos 5x$ and $\int \sin x\,dx=-\cos x$.
3. So integral $=\tfrac12\left(-\tfrac15\cos 5x+\cos x\right)$.
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_
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