What is the integral of sqrt(1+cos^2(x))sin(2x)cos(2x) ?

The integral of sqrt(1+cos^2(x))sin(2x)cos(2x) is -1/6 (1+cos^2(2x))^{3/2}+C

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integral of sqrt(1+cos^2(x))sin(2x)cos(2x)

\int 1 + cos^{2} (x) sin (2 x) cos (2 x) d x

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

Solution steps

\int 1 + cos^{2} (x) sin (2 x) cos (2 x) d x

Apply u-substitution

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

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

Apply u-substitution

: \int v^{2} d v

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

Apply the Power Rule

: \frac{v ^{3}}{3}

= - \frac{1}{2} \cdot \frac{v ^{3}}{3}

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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