What is the integral of sin^3(y)cos^6(y) ?

The integral of sin^3(y)cos^6(y) is -(cos^7(y))/7+(cos^9(y))/9+C

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integral of sin^3(y)cos^6(y)

\int sin^{3} (y) cos^{6} (y) d y

- \frac{cos ^{7} ( y )}{7} + \frac{cos ^{9} ( y )}{9} + C

Solution steps

\int sin^{3} (y) cos^{6} (y) d y

Rewrite using trig identities

= \int (1 - cos^{2} (y)) sin (y) cos^{6} (y) d y

Apply u-substitution

= \int - u^{6} (1 - u^{2}) d u

Expand

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

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

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

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

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

Substitute back

u = cos (y)

= - \frac{cos ^{7} ( y )}{7} + \frac{cos ^{9} ( y )}{9}

Add a constant to the solution

= - \frac{cos ^{7} ( y )}{7} + \frac{cos ^{9} ( y )}{9} + C

What is the integral of sin^3(y)cos^6(y) ?
The integral of sin^3(y)cos^6(y) is -(cos^7(y))/7+(cos^9(y))/9+C