What is the integral of (sin^2(2x)cot(2x))/(sec(2x)) ?

The integral of (sin^2(2x)cot(2x))/(sec(2x)) is -1/6 cos^3(2x)+C

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integral of (sin^2(2x)cot(2x))/(sec(2x))

\int \frac{sin ^{2} ( 2 x ) cot ( 2 x )}{sec ( 2 x )} d x

- \frac{1}{6} cos^{3} (2 x) + C

Solution steps

\int \frac{sin ^{2} ( 2 x ) cot ( 2 x )}{sec ( 2 x )} d x

Rewrite using trig identities

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

Apply u-substitution

= \int cos^{2} (u) sin (u) \frac{1}{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 cos^{2} (u) sin (u) d u

Apply u-substitution

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

Take the constant out:

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

= \frac{1}{2} (- \int v^{2} d v)

Apply the Power Rule

= \frac{1}{2} (- \frac{v ^{3}}{3})

Substitute back

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

Simplify

\frac{1}{2} (- \frac{cos ^{3} ( 2 x )}{3}) : - \frac{1}{6} cos^{3} (2 x)

= - \frac{1}{6} cos^{3} (2 x)

Add a constant to the solution

= - \frac{1}{6} cos^{3} (2 x) + C

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What is the integral of (sin^2(2x)cot(2x))/(sec(2x)) ?
The integral of (sin^2(2x)cot(2x))/(sec(2x)) is -1/6 cos^3(2x)+C