What is the integral of tan^5(3θ) ?

The integral of tan^5(3θ) is 1/3 ln|sec(3θ)|-(sec^2(3θ))/3+(sec^4(3θ))/(12)+C

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integral of tan^5(3θ)

\int tan^{5} (3 θ) d θ

\frac{1}{3} ln ∣ sec (3 θ) ∣ - \frac{sec ^{2} ( 3 θ )}{3} + \frac{sec ^{4} ( 3 θ )}{12} + C

Solution steps

\int tan^{5} (3 θ) d θ

Rewrite using trig identities

= \int (- 1 + sec^{2} (3 θ))^{2} tan (3 θ) d θ

Apply u-substitution

: \int \frac{( - 1 + u ^{2} ) ^{2}}{3 u} d u

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

Expand

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

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

Apply the Sum Rule:

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

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

\int \frac{1}{3 u} d u = \frac{1}{3} ln ∣ u ∣

\int \frac{2 u}{3} d u = \frac{u ^{2}}{3}

\int \frac{u ^{3}}{3} d u = \frac{u ^{4}}{12}

= \frac{1}{3} ln ∣ u ∣ - \frac{u ^{2}}{3} + \frac{u ^{4}}{12}

Substitute back

u = sec (3 θ)

= \frac{1}{3} ln ∣ sec (3 θ) ∣ - \frac{sec ^{2} ( 3 θ )}{3} + \frac{sec ^{4} ( 3 θ )}{12}

Add a constant to the solution

= \frac{1}{3} ln ∣ sec (3 θ) ∣ - \frac{sec ^{2} ( 3 θ )}{3} + \frac{sec ^{4} ( 3 θ )}{12} + C

ln (\frac{x + 1}{x ^{2} + 4})

f (x) = 3 - 2 cos (x), at x = \frac{π}{2}

\int \frac{2}{x ^{2} + 1} d x

f (x) = \frac{x + 5}{x + 2}

\int \frac{1}{x \cdot x ^{2} - 1} d x

What is the integral of tan^5(3θ) ?
The integral of tan^5(3θ) is 1/3 ln|sec(3θ)|-(sec^2(3θ))/3+(sec^4(3θ))/(12)+C