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

The integral of cos(x)sin^3(2x) is 8(-(cos^5(x))/5+(cos^7(x))/7)+C

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

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

8 (- \frac{cos ^{5} ( x )}{5} + \frac{cos ^{7} ( x )}{7}) + C

Solution steps

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

Rewrite using trig identities

= \int 8 cos^{4} (x) sin^{3} (x) d x

Take the constant out:

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

= 8 \cdot \int cos^{4} (x) sin^{3} (x) d x

Rewrite using trig identities

= 8 \cdot \int (1 - cos^{2} (x)) sin (x) cos^{4} (x) d x

Apply u-substitution

: \int - u^{4} (1 - u^{2}) d u

= 8 \cdot \int - u^{4} (1 - u^{2}) d u

Expand

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

= 8 \cdot \int - u^{4} + u^{6} 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^{4} d u + \int u^{6} d u)

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

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

= 8 (- \frac{u ^{5}}{5} + \frac{u ^{7}}{7})

Substitute back

u = cos (x)

= 8 (- \frac{cos ^{5} ( x )}{5} + \frac{cos ^{7} ( x )}{7})

Add a constant to the solution

= 8 (- \frac{cos ^{5} ( x )}{5} + \frac{cos ^{7} ( x )}{7}) + C

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What is the integral of cos(x)sin^3(2x) ?
The integral of cos(x)sin^3(2x) is 8(-(cos^5(x))/5+(cos^7(x))/7)+C