What is the integral of tan^5(2θ) ?

The integral of tan^5(2θ) is 1/2 (ln|sec(2θ)|-sec^2(2θ)+(sec^4(2θ))/4)+C

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integral of tan^5(2θ)

\int tan^{5} (2 θ) d θ

\frac{1}{2} (ln ∣ sec (2 θ) ∣ - sec^{2} (2 θ) + \frac{sec ^{4} ( 2 θ )}{4}) + C

Solution steps

\int tan^{5} (2 θ) d θ

Apply u-substitution

: \int tan^{5} (u) \frac{1}{2} d u

= \int tan^{5} (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^{5} (u) d u

Rewrite using trig identities

= \frac{1}{2} \cdot \int (- 1 + sec^{2} (u))^{2} tan (u) d u

Apply u-substitution

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

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

Expand

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

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int \frac{1}{v} d v = ln ∣ v ∣

\int 2 v d v = v^{2}

\int v^{3} d v = \frac{v ^{4}}{4}

= \frac{1}{2} (ln ∣ v ∣ - v^{2} + \frac{v ^{4}}{4})

Substitute back

= \frac{1}{2} (ln ∣ sec (2 θ) ∣ - sec^{2} (2 θ) + \frac{sec ^{4} ( 2 θ )}{4})

Add a constant to the solution

= \frac{1}{2} (ln ∣ sec (2 θ) ∣ - sec^{2} (2 θ) + \frac{sec ^{4} ( 2 θ )}{4}) + C

\int \frac{r ( r ^{2} - 1 )}{r ^{2} + 1} d r

\int \frac{csc ^{4} ( t )}{cot ^{2} ( t )} d t

\int cot^{5} (2 y) d y

\int \frac{1}{t ^{2} - t + 2} d t

\int \frac{( 1 + tan ( t ) ) ^{2}}{sec ^{2} ( t )} d t

What is the integral of tan^5(2θ) ?
The integral of tan^5(2θ) is 1/2 (ln|sec(2θ)|-sec^2(2θ)+(sec^4(2θ))/4)+C