$\int \cos^2 x\,dx$ equals
A$\dfrac{\cos^3 x}{3}+C$
B$\dfrac{x}{2}-\dfrac{1}{4}\sin 2x+C$
C$\dfrac{x}{2}+\dfrac{1}{4}\sin 2x+C$
D$\dfrac{x}{2}+\dfrac{1}{2}\sin 2x+C$
Answer & Solution
Correct answer: C. $\dfrac{x}{2}+\dfrac{1}{4}\sin 2x+C$
1. Use $\cos^2 x=\dfrac{1+\cos 2x}{2}$.
2. $\int \cos^2 x\,dx=\dfrac{1}{2}\int(1+\cos 2x)\,dx$.
3. $=\dfrac{1}{2}\left(x+\dfrac{1}{2}\sin 2x\right)=\dfrac{x}{2}+\dfrac{1}{4}\sin 2x+C$.
4. Option B uses the $\sin^2$ identity instead.
_Source: NCERT Class 12 Mathematics Ch 7 "Integrals", p.305_
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