What is the integral from 0 to 1 of sin^2(x)cos^5(x) ?

The integral from 0 to 1 of sin^2(x)cos^5(x) is (sin^3(1))/3-(2sin^5(1))/5+(sin^7(1))/7

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integral from 0 to 1 of sin^2(x)cos^5(x)

\int_{0}^{1} sin^{2} (x) cos^{5} (x) d x

\frac{sin ^{3} ( 1 )}{3} - \frac{2 sin ^{5} ( 1 )}{5} + \frac{sin ^{7} ( 1 )}{7}

Decimal

0.07252 \dots

Solution steps

\int_{0}^{1} sin^{2} (x) cos^{5} (x) d x

Rewrite using trig identities

= \int_{0}^{1} (1 - sin^{2} (x))^{2} cos (x) sin^{2} (x) d x

Apply u-substitution

: \int_{0}^{s i n (1)} u^{2} (1 - u^{2})^{2} d u

= \int_{0}^{s i n (1)} u^{2} (1 - u^{2})^{2} d u

Expand

u^{2} (1 - u^{2})^{2} : u^{2} - 2 u^{4} + u^{6}

= \int_{0}^{s i n (1)} u^{2} - 2 u^{4} + u^{6} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{s i n (1)} u^{2} d u - \int_{0}^{s i n (1)} 2 u^{4} d u + \int_{0}^{s i n (1)} u^{6} d u

\int_{0}^{s i n (1)} u^{2} d u = \frac{sin ^{3} ( 1 )}{3}

\int_{0}^{s i n (1)} 2 u^{4} d u = \frac{2 sin ^{5} ( 1 )}{5}

\int_{0}^{s i n (1)} u^{6} d u = \frac{sin ^{7} ( 1 )}{7}

= \frac{sin ^{3} ( 1 )}{3} - \frac{2 sin ^{5} ( 1 )}{5} + \frac{sin ^{7} ( 1 )}{7}

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What is the integral from 0 to 1 of sin^2(x)cos^5(x) ?
The integral from 0 to 1 of sin^2(x)cos^5(x) is (sin^3(1))/3-(2sin^5(1))/5+(sin^7(1))/7