What is the integral of cos^3(y)sin^3(2y) ?

The integral of cos^3(y)sin^3(2y) is 8(-(cos^7(y))/7+(cos^9(y))/9)+C

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integral of cos^3(y)sin^3(2y)

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

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

Solution steps

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

Rewrite using trig identities

= \int 8 cos^{6} (y) sin^{3} (y) d y

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 8 \cdot \int cos^{6} (y) sin^{3} (y) d y

Rewrite using trig identities

= 8 \cdot \int (1 - cos^{2} (y)) sin (y) cos^{6} (y) d y

Apply u-substitution

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

= 8 \cdot \int - u^{6} (1 - u^{2}) d u

Expand

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

= 8 \cdot \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

= 8 (- \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}

= 8 (- \frac{u ^{7}}{7} + \frac{u ^{9}}{9})

Substitute back

u = cos (y)

= 8 (- \frac{cos ^{7} ( y )}{7} + \frac{cos ^{9} ( y )}{9})

Add a constant to the solution

= 8 (- \frac{cos ^{7} ( y )}{7} + \frac{cos ^{9} ( y )}{9}) + C

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What is the integral of cos^3(y)sin^3(2y) ?
The integral of cos^3(y)sin^3(2y) is 8(-(cos^7(y))/7+(cos^9(y))/9)+C