What is the integral of cos^3(y)sin^7(y) ?

The integral of cos^3(y)sin^7(y) is -(cos^4(y))/4+(cos^6(y))/2-(3cos^8(y))/8+(cos^{10}(y))/(10)+C

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integral of cos^3(y)sin^7(y)

\int cos^{3} (y) sin^{7} (y) d y

- \frac{cos ^{4} ( y )}{4} + \frac{cos ^{6} ( y )}{2} - \frac{3 cos ^{8} ( y )}{8} + \frac{cos ^{10} ( y )}{10} + C

Solution steps

\int cos^{3} (y) sin^{7} (y) d y

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

- u^{3} (1 - u^{2})^{3} : - u^{3} + 3 u^{5} - 3 u^{7} + u^{9}

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

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

\int 3 u^{5} d u = \frac{u ^{6}}{2}

\int 3 u^{7} d u = \frac{3 u ^{8}}{8}

\int u^{9} d u = \frac{u ^{10}}{10}

= - \frac{u ^{4}}{4} + \frac{u ^{6}}{2} - \frac{3 u ^{8}}{8} + \frac{u ^{10}}{10}

Substitute back

u = cos (y)

= - \frac{cos ^{4} ( y )}{4} + \frac{cos ^{6} ( y )}{2} - \frac{3 cos ^{8} ( y )}{8} + \frac{cos ^{10} ( y )}{10}

Add a constant to the solution

= - \frac{cos ^{4} ( y )}{4} + \frac{cos ^{6} ( y )}{2} - \frac{3 cos ^{8} ( y )}{8} + \frac{cos ^{10} ( y )}{10} + C

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What is the integral of cos^3(y)sin^7(y) ?
The integral of cos^3(y)sin^7(y) is -(cos^4(y))/4+(cos^6(y))/2-(3cos^8(y))/8+(cos^{10}(y))/(10)+C