How do you find the antiderivative of $\int {cos}^{3} x {sin}^{2} x d x$ $int cos^3xsin^2xdx$ ?

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Answer 1 · 2017-01-18T03:57:26

$\frac{1}{3} {sin}^{3} x - \frac{1}{5} {sin}^{5} x + C$ $1/3sin^3x - 1/5sin^5x + C$

Factor:

$\int cos x ({cos}^{2} x) {sin}^{2} x d x$ $intcosx(cos^2x)sin^2xdx$

Rewrite using the pythagorean identity ${sin}^{2} θ + {cos}^{2} θ = 1$ $sin^2theta + cos^2theta = 1$ :

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

Expand:

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

Let $u = sin x$ $u = sinx$ . Then $d u = cos x d x = d x = \frac{d u}{cos x}$ $du = cosxdx = dx = (du)/cosx$ .

$\int cos x (u^{2} - u^{4}) \cdot \frac{d u}{cos x}$ $intcosx(u^2 - u^4) * (du)/cosx$

$\int u^{2} - u^{4} d u$ $intu^2 - u^4du$

This can be integrated as $\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + C$ $intx^ndx = x^(n + 1)/(n +1) + C$ , where $n \neq - 1$ $n != -1$

$\frac{1}{3} u^{3} - \frac{1}{5} u^{5} + C$ $1/3u^3 - 1/5u^5 + C$

Reverse the substitution:

$\frac{1}{3} {sin}^{3} x - \frac{1}{5} {sin}^{5} x + C$ $1/3sin^3x - 1/5sin^5x + C$

Hopefully this helps!

How do you find the antiderivative of ∫cos3xsin2xdxint cos^3xsin^2xdx?