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

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

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

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

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

Solution steps

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

Rewrite using trig identities

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

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 (y)

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

Add a constant to the solution

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

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