What is the integral of tan^5(7x) ?

The integral of tan^5(7x) is 1/7 ln|sec(7x)|-(sec^2(7x))/7+(sec^4(7x))/(28)+C

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integral of tan^5(7x)

\int tan^{5} (7 x) d x

\frac{1}{7} ln ∣ sec (7 x) ∣ - \frac{sec ^{2} ( 7 x )}{7} + \frac{sec ^{4} ( 7 x )}{28} + C

Solution steps

\int tan^{5} (7 x) d x

Rewrite using trig identities

= \int (- 1 + sec^{2} (7 x))^{2} tan (7 x) d x

Apply u-substitution

= \int \frac{( - 1 + u ^{2} ) ^{2}}{7 u} d u

Expand

\frac{( - 1 + u ^{2} ) ^{2}}{7 u} : \frac{1}{7 u} - \frac{2 u}{7} + \frac{u ^{3}}{7}

= \int \frac{1}{7 u} - \frac{2 u}{7} + \frac{u ^{3}}{7} 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}{7 u} d u - \int \frac{2 u}{7} d u + \int \frac{u ^{3}}{7} d u

\int \frac{1}{7 u} d u = \frac{1}{7} ln ∣ u ∣

\int \frac{2 u}{7} d u = \frac{u ^{2}}{7}

\int \frac{u ^{3}}{7} d u = \frac{u ^{4}}{28}

= \frac{1}{7} ln ∣ u ∣ - \frac{u ^{2}}{7} + \frac{u ^{4}}{28}

Substitute back

u = sec (7 x)

= \frac{1}{7} ln ∣ sec (7 x) ∣ - \frac{sec ^{2} ( 7 x )}{7} + \frac{sec ^{4} ( 7 x )}{28}

Add a constant to the solution

= \frac{1}{7} ln ∣ sec (7 x) ∣ - \frac{sec ^{2} ( 7 x )}{7} + \frac{sec ^{4} ( 7 x )}{28} + C

What is the integral of tan^5(7x) ?
The integral of tan^5(7x) is 1/7 ln|sec(7x)|-(sec^2(7x))/7+(sec^4(7x))/(28)+C