What is the integral of tan^5(x)sec^{11}(x) ?

The integral of tan^5(x)sec^{11}(x) is (sec^{15}(x))/(15)-(2sec^{13}(x))/(13)+(sec^{11}(x))/(11)+C

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integral of tan^5(x)sec^{11}(x)

\int tan^{5} (x) sec^{11} (x) d x

\frac{sec ^{15} ( x )}{15} - \frac{2 sec ^{13} ( x )}{13} + \frac{sec ^{11} ( x )}{11} + C

Solution steps

\int tan^{5} (x) sec^{11} (x) d x

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

u^{10} (u^{2} - 1)^{2} : u^{14} - 2 u^{12} + u^{10}

= \int u^{14} - 2 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^{14} d u - \int 2 u^{12} d u + \int u^{10} d u

\int u^{14} d u = \frac{u ^{15}}{15}

\int 2 u^{12} d u = \frac{2 u ^{13}}{13}

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

= \frac{u ^{15}}{15} - \frac{2 u ^{13}}{13} + \frac{u ^{11}}{11}

Substitute back

u = sec (x)

= \frac{sec ^{15} ( x )}{15} - \frac{2 sec ^{13} ( x )}{13} + \frac{sec ^{11} ( x )}{11}

Add a constant to the solution

= \frac{sec ^{15} ( x )}{15} - \frac{2 sec ^{13} ( x )}{13} + \frac{sec ^{11} ( x )}{11} + C

x \to 0 lim (2 x \cdot sin (x))

y = x^{2} + x, at x = - 2

\int_{- \infty}^{0} \frac{1}{4 - 2 x} d x

\int_{0.5}^{t + 4} 3 d x

\int_{- 1}^{3} (4 - 2 x) d x

What is the integral of tan^5(x)sec^{11}(x) ?
The integral of tan^5(x)sec^{11}(x) is (sec^{15}(x))/(15)-(2sec^{13}(x))/(13)+(sec^{11}(x))/(11)+C