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Question #017cf

1 Answer

Nov 16, 2016

$\int {sin}^{3} x {cos}^{2} x d x = \frac{{cos}^{5} x}{5} - \frac{{cos}^{3} x}{3} + C$ $intsin^3xcos^2xdx=cos^5x/5-cos^3x/3 + C$

Explanation:

$\int {sin}^{3} x {cos}^{2} x d x = \int sin x {sin}^{2} x {cos}^{2} x d x$ $intsin^3xcos^2xdx = intsinxsin^2xcos^2xdx$

$= \int sin x (1 - {cos}^{2} x) {cos}^{2} x d x$ $=intsinx(1-cos^2x)cos^2xdx$

$= \int sin x ({cos}^{2} x - {cos}^{4} x) d x$ $=intsinx(cos^2x-cos^4x)dx$

Let $u = cos x \Rightarrow d u = - sin x d x$ $u = cosx => du = -sinxdx$

$\Rightarrow \int sin x ({cos}^{2} x - {cos}^{4} x) d x$ $=> intsinx(cos^2x-cos^4x)dx$

$= \int ({cos}^{4} x - {cos}^{2} x) (- sin x) d x$ $= int(cos^4x-cos^2x)(-sinx)dx$

$= \int (u^{4} - u^{2}) d u$ $=int(u^4-u^2)du$

$= \frac{u^{5}}{5} - \frac{u^{3}}{3} + C$ $=u^5/5-u^3/3 + C$

$= \frac{{cos}^{5} x}{5} - \frac{{cos}^{3} x}{3} + C$ $=cos^5x/5-cos^3x/3 + C$

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