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

The integral of tan^7(x)sec^5(x) is (sec^{11}(x)}{11}-\frac{sec^9(x))/3+(3sec^7(x))/7-(sec^5(x))/5+C

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integral of tan^7(x)sec^5(x)

\int tan^{7} (x) sec^{5} (x) d x

\frac{sec ^{11} ( x )}{11} - \frac{sec ^{9} ( x )}{3} + \frac{3 sec ^{7} ( x )}{7} - \frac{sec ^{5} ( x )}{5} + C

Solution steps

\int tan^{7} (x) sec^{5} (x) d x

Rewrite using trig identities

= \int (- 1 + sec^{2} (x))^{3} tan (x) sec^{5} (x) d x

Apply u-substitution

: \int u^{4} (u^{2} - 1)^{3} d u

= \int u^{4} (u^{2} - 1)^{3} d u

Expand

u^{4} (u^{2} - 1)^{3} : u^{10} - 3 u^{8} + 3 u^{6} - u^{4}

= \int u^{10} - 3 u^{8} + 3 u^{6} - 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 u^{10} d u - \int 3 u^{8} d u + \int 3 u^{6} d u - \int u^{4} d u

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

\int 3 u^{8} d u = \frac{u ^{9}}{3}

\int 3 u^{6} d u = \frac{3 u ^{7}}{7}

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

= \frac{u ^{11}}{11} - \frac{u ^{9}}{3} + \frac{3 u ^{7}}{7} - \frac{u ^{5}}{5}

Substitute back

u = sec (x)

= \frac{sec ^{11} ( x )}{11} - \frac{sec ^{9} ( x )}{3} + \frac{3 sec ^{7} ( x )}{7} - \frac{sec ^{5} ( x )}{5}

Add a constant to the solution

= \frac{sec ^{11} ( x )}{11} - \frac{sec ^{9} ( x )}{3} + \frac{3 sec ^{7} ( x )}{7} - \frac{sec ^{5} ( x )}{5} + C

y^{^{''}} - 3 y^{^{'}} + 2 y = 4

\frac{d}{d x} (\frac{4 ln ( x )}{x})

\frac{\partial}{\partial x} (y - \frac{1}{x})

y^{^{'}} - y = 4 x

\frac{\partial}{\partial x} (arctan (x t))

What is the integral of tan^7(x)sec^5(x) ?
The integral of tan^7(x)sec^5(x) is (sec^{11}(x)}{11}-\frac{sec^9(x))/3+(3sec^7(x))/7-(sec^5(x))/5+C