What is the integral of tan^3(x)sec^{11}(x) ?

The integral of tan^3(x)sec^{11}(x) is (sec^{13}(x))/(13)-(sec^{11}(x))/(11)+C

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integral of tan^3(x)sec^{11}(x)

\int tan^{3} (x) sec^{11} (x) d x

\frac{sec ^{13} ( x )}{13} - \frac{sec ^{11} ( x )}{11} + C

Solution steps

\int tan^{3} (x) sec^{11} (x) d x

Rewrite using trig identities

= \int (- 1 + sec^{2} (x)) tan (x) sec^{11} (x) d x

Apply u-substitution

: \int u^{10} (u^{2} - 1) d u

= \int u^{10} (u^{2} - 1) d u

Expand

u^{10} (u^{2} - 1) : u^{12} - u^{10}

= \int u^{12} - u^{10} 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^{12} d u - \int u^{10} d u

\int u^{12} d u = \frac{u ^{13}}{13}

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

= \frac{u ^{13}}{13} - \frac{u ^{11}}{11}

Substitute back

u = sec (x)

= \frac{sec ^{13} ( x )}{13} - \frac{sec ^{11} ( x )}{11}

Add a constant to the solution

= \frac{sec ^{13} ( x )}{13} - \frac{sec ^{11} ( x )}{11} + C

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What is the integral of tan^3(x)sec^{11}(x) ?
The integral of tan^3(x)sec^{11}(x) is (sec^{13}(x))/(13)-(sec^{11}(x))/(11)+C