What is the integral of sec^{14}(θ)tan(θ) ?

The integral of sec^{14}(θ)tan(θ) is

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integral of sec^{14}(θ)tan(θ)

\int sec^{14} (θ) tan (θ) d θ

\frac{tan ^{2} ( θ )}{2} + \frac{3 tan ^{4} ( θ )}{2} + \frac{5 tan ^{6} ( θ )}{2} + \frac{5 tan ^{8} ( θ )}{2} + \frac{3 tan ^{10} ( θ )}{2} + \frac{tan ^{12} ( θ )}{2} + \frac{tan ^{14} ( θ )}{14} + C

Solution steps

\int sec^{14} (θ) tan (θ) d θ

Rewrite using trig identities

= \int sec^{2} (θ) (1 + tan^{2} (θ))^{6} tan (θ) d θ

Apply u-substitution

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

Expand

u (1 + u^{2})^{6} : u + 6 u^{3} + 15 u^{5} + 20 u^{7} + 15 u^{9} + 6 u^{11} + u^{13}

= \int u + 6 u^{3} + 15 u^{5} + 20 u^{7} + 15 u^{9} + 6 u^{11} + u^{13} 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 d u + \int 6 u^{3} d u + \int 15 u^{5} d u + \int 20 u^{7} d u + \int 15 u^{9} d u + \int 6 u^{11} d u + \int u^{13} d u

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

\int 6 u^{3} d u = \frac{3 u ^{4}}{2}

\int 15 u^{5} d u = \frac{5 u ^{6}}{2}

\int 20 u^{7} d u = \frac{5 u ^{8}}{2}

\int 15 u^{9} d u = \frac{3 u ^{10}}{2}

\int 6 u^{11} d u = \frac{u ^{12}}{2}

\int u^{13} d u = \frac{u ^{14}}{14}

= \frac{u ^{2}}{2} + \frac{3 u ^{4}}{2} + \frac{5 u ^{6}}{2} + \frac{5 u ^{8}}{2} + \frac{3 u ^{10}}{2} + \frac{u ^{12}}{2} + \frac{u ^{14}}{14}

Substitute back

u = tan (θ)

= \frac{tan ^{2} ( θ )}{2} + \frac{3 tan ^{4} ( θ )}{2} + \frac{5 tan ^{6} ( θ )}{2} + \frac{5 tan ^{8} ( θ )}{2} + \frac{3 tan ^{10} ( θ )}{2} + \frac{tan ^{12} ( θ )}{2} + \frac{tan ^{14} ( θ )}{14}

Add a constant to the solution

= \frac{tan ^{2} ( θ )}{2} + \frac{3 tan ^{4} ( θ )}{2} + \frac{5 tan ^{6} ( θ )}{2} + \frac{5 tan ^{8} ( θ )}{2} + \frac{3 tan ^{10} ( θ )}{2} + \frac{tan ^{12} ( θ )}{2} + \frac{tan ^{14} ( θ )}{14} + C

What is the integral of sec^{14}(θ)tan(θ) ?
The integral of sec^{14}(θ)tan(θ) is