How do you find the integral of $\int {cos}^{3} x {sin}^{2} x d x$ $int cos^3x sin^2x dx$ ?

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How do you find the integral of $\int {cos}^{3} x {sin}^{2} x d x$ $int cos^3x sin^2x dx$ ?

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Answer 1 · 2015-09-30T07:51:36

$\int {cos}^{3} x {sin}^{2} x d x = (\frac{1}{3}) {sin}^{3} x - (\frac{1}{5}) \cdot {sin}^{5} x + c$ $intcos^3xsin^2xdx=(1/3)sin^3x-(1/5)*sin^5x+c$

Explanation:

$\int {cos}^{3} x {sin}^{2} x d x$ $intcos^3xsin^2xdx$

$= \int {cos}^{2} x \cdot {sin}^{2} x \cdot (cos x) d x$ $=intcos^2x*sin^2x*(cosx)dx$

$= \int (1 - {sin}^{2}) {sin}^{2} x \cdot d (sin x)$ $=int (1-sin^2)sin^2x*d(sinx)$

Hints:

$\frac{d (sin x)}{d x} = cos x$ $(d(sinx))/dx=cosx$

$d (sin x) = cos x d x$ $d(sinx)=cosxdx$

Therefore,

$= \int ({sin}^{2} x - {sin}^{4} x) \cdot d (sin x)$ $=int(sin^2x-sin^4x)*d(sinx)$
$= = (\frac{1}{3}) {sin}^{3} x - (\frac{1}{5}) \cdot {sin}^{5} x + c$ $==(1/3)sin^3x-(1/5)*sin^5x+c$

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How do you find the integral of $\int {cos}^{3} x {sin}^{2} x d x$ $int cos^3x sin^2x dx$ ?

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How do you find the integral of $\int {cos}^{3} x {sin}^{2} x d x$ $int cos^3x sin^2x dx$ ?