What is the integral of $\int {sin}^{3} x {cos}^{2} x d x$ $int sin^3xcos^2x dx$ from $0$ $0$ to $\frac{π}{2}$ $pi/2$ ?

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What is the integral of $\int {sin}^{3} x {cos}^{2} x d x$ $int sin^3xcos^2x dx$ from $0$ $0$ to $\frac{π}{2}$ $pi/2$ ?

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Answer 1 · 2018-05-06T20:17:10

$\frac{2}{15}$ $2/15$

Explanation:

Rewrite the integrand as $\int_{0}^{\frac{π}{2}} {sin}^{2} x {cos}^{2} x sin x d x$ $int_0^(pi/2)sin^2xcos^2xsinxdx$

Recalling the identity ${sin}^{2} x = 1 - {cos}^{2} x,$ $sin^2x=1-cos^2x,$ we get

$\int_{0}^{\frac{π}{2}} (1 - {cos}^{2} x) {cos}^{2} x sin x d x$ $int_0^(pi/2)(1-cos^2x)cos^2xsinxdx$

We can now solve this with a simple substitution.

$u = cos x$ $u=cosx$
$d u = - sin x d x$ $du=-sinxdx$

Calculate the new bounds:

Upper: $u = cos (\frac{π}{2}) = 0$ $u=cos(pi/2)=0$
Lower: $u = cos 0 = 1$ $u=cos0=1$

$- \int_{1}^{0} u^{2} (1 - u^{2}) d u = - \int_{1}^{0} (u^{2} - u^{4}) d u$ $-int_1^0u^2(1-u^2)du=-int_1^0(u^2-u^4)du$

$= - (\frac{1}{3} u^{3} - \frac{1}{5} u^{5}) {∣ ∣ ∣}_{1}^{0}$ $=-(1/3u^3-1/5u^5)|_1^0$
$- (- \frac{1}{3} + \frac{1}{5}) = \frac{2}{15}$ $-(-1/3+1/5)=2/15$

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What is the integral of $\int {sin}^{3} x {cos}^{2} x d x$ $int sin^3xcos^2x dx$ from $0$ $0$ to $\frac{π}{2}$ $pi/2$ ?

1 Answer

Explanation:

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What is the integral of $\int {sin}^{3} x {cos}^{2} x d x$ $int sin^3xcos^2x dx$ from $0$ $0$ to $\frac{π}{2}$ $pi/2$ ?