What is the integral of tan^4(2θ) ?

The integral of tan^4(2θ) is 1/6 tan^3(2θ)+θ-1/2 tan(2θ)+C

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integral of tan^4(2θ)

\int tan^{4} (2 θ) d θ

\frac{1}{6} tan^{3} (2 θ) + θ - \frac{1}{2} tan (2 θ) + C

Solution steps

\int tan^{4} (2 θ) d θ

Apply u-substitution

: \int tan^{4} (u) \frac{1}{2} d u

= \int tan^{4} (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 tan^{4} (u) d u

Apply Integral Reduction

= \frac{1}{2} (\frac{tan ^{3} ( u )}{3} - \int tan^{2} (u) d u)

\int tan^{2} (u) d u = - u + tan (u)

= \frac{1}{2} (\frac{tan ^{3} ( u )}{3} - (- u + tan (u)))

Substitute back

u = 2 θ

= \frac{1}{2} (\frac{tan ^{3} ( 2 θ )}{3} - (- 2 θ + tan (2 θ)))

Simplify

\frac{1}{2} (\frac{tan ^{3} ( 2 θ )}{3} - (- 2 θ + tan (2 θ))) : \frac{1}{6} tan^{3} (2 θ) + θ - \frac{1}{2} tan (2 θ)

= \frac{1}{6} tan^{3} (2 θ) + θ - \frac{1}{2} tan (2 θ)

Add a constant to the solution

= \frac{1}{6} tan^{3} (2 θ) + θ - \frac{1}{2} tan (2 θ) + C

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

\int (cos (x) + sin (2 x)) d x

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\int - 5 sin (x) + 3 cos (x) d x

\int \frac{1}{y + 2 y} d y

What is the integral of tan^4(2θ) ?
The integral of tan^4(2θ) is 1/6 tan^3(2θ)+θ-1/2 tan(2θ)+C