What is the integral of (2cos(4x)-1)^{1/2}sin(4x) ?

The integral of (2cos(4x)-1)^{1/2}sin(4x) is -1/12 (2cos(4x)-1)^{3/2}+C

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integral of (2cos(4x)-1)^{1/2}sin(4x)

\int (2 cos (4 x) - 1)^{\frac{1}{2}} sin (4 x) d x

- \frac{1}{12} (2 cos (4 x) - 1)^{\frac{3}{2}} + C

Solution steps

\int (2 cos (4 x) - 1)^{\frac{1}{2}} sin (4 x) d x

Apply u-substitution

: \int - \frac{u ^{\frac{1}{2}}}{8} d u

= \int - \frac{u ^{\frac{1}{2}}}{8} d u

Take the constant out:

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

= - \frac{1}{8} \cdot \int u^{\frac{1}{2}} d u

Apply the Power Rule

: \frac{2}{3} u^{\frac{3}{2}}

= - \frac{1}{8} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 2 cos (4 x) - 1

= - \frac{1}{8} \cdot \frac{2}{3} (2 cos (4 x) - 1)^{\frac{3}{2}}

Simplify

- \frac{1}{8} \cdot \frac{2}{3} (2 cos (4 x) - 1)^{\frac{3}{2}} : - \frac{1}{12} (2 cos (4 x) - 1)^{\frac{3}{2}}

= - \frac{1}{12} (2 cos (4 x) - 1)^{\frac{3}{2}}

Add a constant to the solution

= - \frac{1}{12} (2 cos (4 x) - 1)^{\frac{3}{2}} + C

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What is the integral of (2cos(4x)-1)^{1/2}sin(4x) ?
The integral of (2cos(4x)-1)^{1/2}sin(4x) is -1/12 (2cos(4x)-1)^{3/2}+C