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

The integral of tan^7(x) is -ln|sec(x)|+(3sec^2(x))/2-(3sec^4(x))/4+(sec^6(x))/6+C

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integral of tan^7(x)

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

- ln ∣ sec (x) ∣ + \frac{3 sec ^{2} ( x )}{2} - \frac{3 sec ^{4} ( x )}{4} + \frac{sec ^{6} ( x )}{6} + C

Solution steps

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

Rewrite using trig identities

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

Apply u-substitution

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

Expand

\frac{( - 1 + u ^{2} ) ^{3}}{u} : - \frac{1}{u} + 3 u - 3 u^{3} + u^{5}

= \int - \frac{1}{u} + 3 u - 3 u^{3} + u^{5} 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}{u} d u + \int 3 u d u - \int 3 u^{3} d u + \int u^{5} d u

\int \frac{1}{u} d u = ln ∣ u ∣

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

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

\int u^{5} d u = \frac{u ^{6}}{6}

= - ln ∣ u ∣ + \frac{3 u ^{2}}{2} - \frac{3 u ^{4}}{4} + \frac{u ^{6}}{6}

Substitute back

u = sec (x)

= - ln ∣ sec (x) ∣ + \frac{3 sec ^{2} ( x )}{2} - \frac{3 sec ^{4} ( x )}{4} + \frac{sec ^{6} ( x )}{6}

Add a constant to the solution

= - ln ∣ sec (x) ∣ + \frac{3 sec ^{2} ( x )}{2} - \frac{3 sec ^{4} ( x )}{4} + \frac{sec ^{6} ( x )}{6} + C

What is the integral of tan^7(x) ?
The integral of tan^7(x) is -ln|sec(x)|+(3sec^2(x))/2-(3sec^4(x))/4+(sec^6(x))/6+C