allery$\int \sin^3 x\,dx$ 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: syllabus Mathematics Ch 7 "Integrals", p.306_
Keep practising Probability & Statistics
Browse all Probability & Statistics questions for AP Intermediate 1st Year, or explore more Mathematics topics.
Related questions
$\int x\cos x\,dx$ equalsThe integration-by-parts formula $\int u\,\dfrac{dv}{dx}\,dx$ equals$\int \dfrac{dx}{x(x^2+1)}$ equals$\int \dfrac{x\,dx}{(x-1)(x-2)}$ equalsIn decomposing $\dfrac{1}{(x+1)(x+2)}=\dfrac{A}{x+1}+\dfrac{B}{x+2}$, the constants are$\int \dfrac{dx}{(x+1)(x+2)}$ equals$\int \dfrac{dx}{x^2+2x+2}$ equals$\int \dfrac{dx}{x^2-6x+13}$ equals