What is the integral of tan^5(x)*sec^3(x) ?

The integral of tan^5(x)*sec^3(x) is (sec^7(x))/7-(2sec^5(x))/5+(sec^3(x))/3+C

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integral of tan^5(x)*sec^3(x)

\int tan^{5} (x) \cdot sec^{3} (x) d x

\frac{sec ^{7} ( x )}{7} - \frac{2 sec ^{5} ( x )}{5} + \frac{sec ^{3} ( x )}{3} + C

Solution steps

\int tan^{5} (x) sec^{3} (x) d x

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

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

= \int u^{6} - 2 u^{4} + u^{2} 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^{6} d u - \int 2 u^{4} d u + \int u^{2} d u

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

\int 2 u^{4} d u = \frac{2 u ^{5}}{5}

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

= \frac{u ^{7}}{7} - \frac{2 u ^{5}}{5} + \frac{u ^{3}}{3}

Substitute back

u = sec (x)

= \frac{sec ^{7} ( x )}{7} - \frac{2 sec ^{5} ( x )}{5} + \frac{sec ^{3} ( x )}{3}

Add a constant to the solution

= \frac{sec ^{7} ( x )}{7} - \frac{2 sec ^{5} ( x )}{5} + \frac{sec ^{3} ( x )}{3} + C

\frac{d y}{d x} = - 4^{- 2} - y^{- 1} + y^{- 1} + y

y^{^{''}} - \frac{3}{x} y^{^{'}} - \frac{12}{x ^{2}} y = 4 x^{2}

\int \frac{1}{x 36 x ^{2} - 1} d x

\int 6 x^{2} e^{6 x} d x

y = x^{3} + x^{2} - 6 x, y = 6 x

What is the integral of tan^5(x)*sec^3(x) ?
The integral of tan^5(x)*sec^3(x) is (sec^7(x))/7-(2sec^5(x))/5+(sec^3(x))/3+C