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

The integral of tan^5(θ)sec^7(θ) is (sec^{11}(θ)}{11}-\frac{2sec^9(θ))/9+(sec^7(θ))/7+C

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integral of tan^5(θ)sec^7(θ)

\int tan^{5} (θ) sec^{7} (θ) d θ

\frac{sec ^{11} ( θ )}{11} - \frac{2 sec ^{9} ( θ )}{9} + \frac{sec ^{7} ( θ )}{7} + C

Solution steps

\int tan^{5} (θ) sec^{7} (θ) d θ

Rewrite using trig identities

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

Apply u-substitution

= \int u^{6} (u^{2} - 1)^{2} d u

Expand

u^{6} (u^{2} - 1)^{2} : u^{10} - 2 u^{8} + u^{6}

= \int u^{10} - 2 u^{8} + u^{6} 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^{10} d u - \int 2 u^{8} d u + \int u^{6} d u

\int u^{10} d u = \frac{u ^{11}}{11}

\int 2 u^{8} d u = \frac{2 u ^{9}}{9}

\int u^{6} d u = \frac{u ^{7}}{7}

= \frac{u ^{11}}{11} - \frac{2 u ^{9}}{9} + \frac{u ^{7}}{7}

Substitute back

u = sec (θ)

= \frac{sec ^{11} ( θ )}{11} - \frac{2 sec ^{9} ( θ )}{9} + \frac{sec ^{7} ( θ )}{7}

Add a constant to the solution

= \frac{sec ^{11} ( θ )}{11} - \frac{2 sec ^{9} ( θ )}{9} + \frac{sec ^{7} ( θ )}{7} + C

What is the integral of tan^5(θ)sec^7(θ) ?
The integral of tan^5(θ)sec^7(θ) is (sec^{11}(θ)}{11}-\frac{2sec^9(θ))/9+(sec^7(θ))/7+C