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

The integral of sin^3(2x)cos^2(x) is -4/3 cos^6(x)+cos^8(x)+C

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

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

- \frac{4}{3} cos^{6} (x) + cos^{8} (x) + C

Solution steps

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

Rewrite using trig identities

= \int 8 cos^{5} (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^{5} (x) sin^{3} (x) d x

Rewrite using trig identities

= 8 \cdot \int (1 - cos^{2} (x)) sin (x) cos^{5} (x) d x

Apply u-substitution

: \int - u^{5} (1 - u^{2}) d u

= 8 \cdot \int - u^{5} (1 - u^{2}) d u

Expand

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

= 8 \cdot \int - u^{5} + u^{7} 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^{5} d u + \int u^{7} d u)

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

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

= 8 (- \frac{u ^{6}}{6} + \frac{u ^{8}}{8})

Substitute back

u = cos (x)

= 8 (- \frac{cos ^{6} ( x )}{6} + \frac{cos ^{8} ( x )}{8})

Simplify

8 (- \frac{cos ^{6} ( x )}{6} + \frac{cos ^{8} ( x )}{8}) : - \frac{4}{3} cos^{6} (x) + cos^{8} (x)

= - \frac{4}{3} cos^{6} (x) + cos^{8} (x)

Add a constant to the solution

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

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What is the integral of sin^3(2x)cos^2(x) ?
The integral of sin^3(2x)cos^2(x) is -4/3 cos^6(x)+cos^8(x)+C