What is the integral of (1-sin(2x))^{3/2}cos(2x) ?

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

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integral of (1-sin(2x))^{3/2}cos(2x)

\int (1 - sin (2 x))^{\frac{3}{2}} cos (2 x) d x

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

Solution steps

\int (1 - sin (2 x))^{\frac{3}{2}} cos (2 x) d x

Apply u-substitution

: \int - \frac{u ^{\frac{3}{2}}}{2} d u

= \int - \frac{u ^{\frac{3}{2}}}{2} d u

Take the constant out:

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

= - \frac{1}{2} \cdot \int u^{\frac{3}{2}} d u

Apply the Power Rule

: \frac{2}{5} u^{\frac{5}{2}}

= - \frac{1}{2} \cdot \frac{2}{5} u^{\frac{5}{2}}

Substitute back

u = 1 - sin (2 x)

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

Simplify

- \frac{1}{2} \cdot \frac{2}{5} (1 - sin (2 x))^{\frac{5}{2}} : - \frac{1}{5} (1 - sin (2 x))^{\frac{5}{2}}

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

Add a constant to the solution

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

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