What is the integral of (tan(x))^6 ?

The integral of (tan(x))^6 is (tan^5(x))/5-(tan^3(x))/3+tan(x)-arctan(tan(x))+C

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integral of (tan(x))^6

\int (tan (x))^{6} d x

\frac{tan ^{5} ( x )}{5} - \frac{tan ^{3} ( x )}{3} + tan (x) - arctan (tan (x)) + C

Solution steps

\int (tan (x))^{6} d x

Simplify

= \int tan^{6} (x) d x

Apply u-substitution

= \int \frac{u ^{6}}{1 + u ^{2}} d u

Long division

\frac{u ^{6}}{1 + u ^{2}} : u^{4} - u^{2} + 1 - \frac{1}{u ^{2} + 1}

= \int u^{4} - u^{2} + 1 - \frac{1}{u ^{2} + 1} 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 u^{4} d u - \int u^{2} d u + \int 1 d u - \int \frac{1}{u ^{2} + 1} d u

\int u^{4} d u = \frac{u ^{5}}{5}

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

\int 1 d u = u

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

= \frac{u ^{5}}{5} - \frac{u ^{3}}{3} + u - arctan (u)

Substitute back

u = tan (x)

= \frac{tan ^{5} ( x )}{5} - \frac{tan ^{3} ( x )}{3} + tan (x) - arctan (tan (x))

Add a constant to the solution

= \frac{tan ^{5} ( x )}{5} - \frac{tan ^{3} ( x )}{3} + tan (x) - arctan (tan (x)) + C

What is the integral of (tan(x))^6 ?
The integral of (tan(x))^6 is (tan^5(x))/5-(tan^3(x))/3+tan(x)-arctan(tan(x))+C