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

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

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

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

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

Solution steps

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

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

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

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

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

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

= \frac{u ^{7}}{7} - \frac{u ^{5}}{5}

Substitute back

u = sec (x)

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (\frac{1 - ln ( x )}{x ^{2}})

\frac{\partial}{\partial x} (ln (9 x))

\frac{d}{d x} (\frac{4}{7 x ^{4}})

y^{^{'}} + sin (t) y = sin (t), y (0) = 7

y = x^{2} + 9 x, at x = - 8

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