What is the integral of sec^6(θ) ?

The integral of sec^6(θ) is tan(θ)+(2tan^3(θ))/3+(tan^5(θ))/5+C

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integral of sec^6(θ)

\int sec^{6} (θ) d θ

tan (θ) + \frac{2 tan ^{3} ( θ )}{3} + \frac{tan ^{5} ( θ )}{5} + C

Solution steps

\int sec^{6} (θ) d θ

Rewrite using trig identities

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

Apply u-substitution

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

Expand

(1 + u^{2})^{2} : 1 + 2 u^{2} + u^{4}

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

\int 1 d u = u

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

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

= u + \frac{2 u ^{3}}{3} + \frac{u ^{5}}{5}

Substitute back

u = tan (θ)

= tan (θ) + \frac{2 tan ^{3} ( θ )}{3} + \frac{tan ^{5} ( θ )}{5}

Add a constant to the solution

= tan (θ) + \frac{2 tan ^{3} ( θ )}{3} + \frac{tan ^{5} ( θ )}{5} + C

What is the integral of sec^6(θ) ?
The integral of sec^6(θ) is tan(θ)+(2tan^3(θ))/3+(tan^5(θ))/5+C