What is the integral of (4-3sin(2x))^4cos(2x) ?

The integral of (4-3sin(2x))^4cos(2x) is -1/30 (4-3sin(2x))^5+C

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integral of (4-3sin(2x))^4cos(2x)

\int (4 - 3 sin (2 x))^{4} cos (2 x) d x

- \frac{1}{30} (4 - 3 sin (2 x))^{5} + C

Solution steps

\int (4 - 3 sin (2 x))^{4} cos (2 x) d x

Apply u-substitution

= \int - \frac{u ^{4}}{6} d u

Take the constant out:

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

= - \frac{1}{6} \cdot \int u^{4} d u

Apply the Power Rule

= - \frac{1}{6} \cdot \frac{u ^{5}}{5}

Substitute back

u = 4 - 3 sin (2 x)

= - \frac{1}{6} \cdot \frac{( 4 - 3 sin ( 2 x ) ) ^{5}}{5}

Simplify

- \frac{1}{6} \cdot \frac{( 4 - 3 sin ( 2 x ) ) ^{5}}{5} : - \frac{1}{30} (4 - 3 sin (2 x))^{5}

= - \frac{1}{30} (4 - 3 sin (2 x))^{5}

Add a constant to the solution

= - \frac{1}{30} (4 - 3 sin (2 x))^{5} + C

What is the integral of (4-3sin(2x))^4cos(2x) ?
The integral of (4-3sin(2x))^4cos(2x) is -1/30 (4-3sin(2x))^5+C