$\int \sin^3 x\,dx$ equals
A$-\cos x+\dfrac{1}{3}\cos^3 x+C$
B$\cos x-\dfrac{1}{3}\cos^3 x+C$
C$-\cos x-\dfrac{1}{3}\cos^3 x+C$
D$\dfrac{1}{4}\sin^4 x+C$
Answer & Solution
Correct answer: A. $-\cos x+\dfrac{1}{3}\cos^3 x+C$
1. Write $\sin^3 x=(1-\cos^2 x)\sin x$.
2. Put $t=\cos x$, $dt=-\sin x\,dx$.
3. Integral $=-\int(1-t^2)\,dt=-t+\dfrac{t^3}{3}$.
4. Back-substitute: $-\cos x+\dfrac{1}{3}\cos^3 x+C$. Option B flips both signs.
_Source: NCERT Class 12 Mathematics Ch 7 "Integrals", p.306_
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