What is the integral of sin^2(x)cos^7(x) ?

The integral of sin^2(x)cos^7(x) is (sin^3(x))/3-(3sin^5(x))/5+(3sin^7(x))/7-(sin^9(x))/9+C

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integral of sin^2(x)cos^7(x)

\int sin^{2} (x) cos^{7} (x) d x

\frac{sin ^{3} ( x )}{3} - \frac{3 sin ^{5} ( x )}{5} + \frac{3 sin ^{7} ( x )}{7} - \frac{sin ^{9} ( x )}{9} + C

Solution steps

\int sin^{2} (x) cos^{7} (x) d x

Rewrite using trig identities

= \int (1 - sin^{2} (x))^{3} cos (x) sin^{2} (x) d x

Apply u-substitution

: \int u^{2} (1 - u^{2})^{3} d u

= \int u^{2} (1 - u^{2})^{3} d u

Expand

u^{2} (1 - u^{2})^{3} : u^{2} - 3 u^{4} + 3 u^{6} - u^{8}

= \int u^{2} - 3 u^{4} + 3 u^{6} - u^{8} 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 u^{2} d u - \int 3 u^{4} d u + \int 3 u^{6} d u - \int u^{8} d u

\int u^{2} d u = \frac{u ^{3}}{3}

\int 3 u^{4} d u = \frac{3 u ^{5}}{5}

\int 3 u^{6} d u = \frac{3 u ^{7}}{7}

\int u^{8} d u = \frac{u ^{9}}{9}

= \frac{u ^{3}}{3} - \frac{3 u ^{5}}{5} + \frac{3 u ^{7}}{7} - \frac{u ^{9}}{9}

Substitute back

u = sin (x)

= \frac{sin ^{3} ( x )}{3} - \frac{3 sin ^{5} ( x )}{5} + \frac{3 sin ^{7} ( x )}{7} - \frac{sin ^{9} ( x )}{9}

Add a constant to the solution

= \frac{sin ^{3} ( x )}{3} - \frac{3 sin ^{5} ( x )}{5} + \frac{3 sin ^{7} ( x )}{7} - \frac{sin ^{9} ( x )}{9} + C

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What is the integral of sin^2(x)cos^7(x) ?
The integral of sin^2(x)cos^7(x) is (sin^3(x))/3-(3sin^5(x))/5+(3sin^7(x))/7-(sin^9(x))/9+C