What is the integral of 1/(csc^2(u)sec^5(u)) ?

The integral of 1/(csc^2(u)sec^5(u)) is (sin^3(u))/3-(2sin^5(u))/5+(sin^7(u))/7+C

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integral of 1/(csc^2(u)sec^5(u))

\int \frac{1}{csc ^{2} ( u ) sec ^{5} ( u )} d u

\frac{sin ^{3} ( u )}{3} - \frac{2 sin ^{5} ( u )}{5} + \frac{sin ^{7} ( u )}{7} + C

Solution steps

\int \frac{1}{csc ^{2} ( u ) sec ^{5} ( u )} d u

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

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

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

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

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

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

Substitute back

v = sin (u)

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

Add a constant to the solution

= \frac{sin ^{3} ( u )}{3} - \frac{2 sin ^{5} ( u )}{5} + \frac{sin ^{7} ( u )}{7} + C

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What is the integral of 1/(csc^2(u)sec^5(u)) ?
The integral of 1/(csc^2(u)sec^5(u)) is (sin^3(u))/3-(2sin^5(u))/5+(sin^7(u))/7+C