How do you integrate $\int {sin}^{3} x$ $int sin^3x$ by integration by parts method?

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Answer 1 · 2016-08-02T15:39:55

$= \frac{1}{3} {cos}^{3} - cos x + C$ $=1/3 cos^3 - cos x + C$

$int \ sin^3 x \ dx$

$= int \d/dx (-cos x) sin^2 x \ dx$

which by IBP

$=-cos x sin^2x + int \cos x d/dx( sin^2 x) \ dx$

$=-cos x sin^2 x + 2 int \cos^2 x sin x \ dx$

$=-cos x sin^2 x + 2 int \d/dx ( - 1/3 cos^3 x) \ dx$

$=-cos x sin^2 x - 2/3 cos^3 x + C$

$=-cos x (1 - cos^2 x) - 2/3 cos^3 x + C$

$=-cos x + cos^3 x - 2/3 cos^3 x + C$

$=1/3 cos^3 - cos x + C$

How do you integrate int sin^3x∫sin3xint sin^3x by integration by parts method?