What is the integral of (cos^5(x))/(sin^2(x)) ?

The integral of (cos^5(x))/(sin^2(x)) is -csc(x)-2sin(x)+1/3 sin^3(x)+C

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integral of (cos^5(x))/(sin^2(x))

\int \frac{cos ^{5} ( x )}{sin ^{2} ( x )} d x

- csc (x) - 2 sin (x) + \frac{1}{3} sin^{3} (x) + C

Solution steps

\int \frac{cos ^{5} ( x )}{sin ^{2} ( x )} d x

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

\frac{( 1 - u ^{2} ) ^{2}}{u ^{2}} : \frac{1}{u ^{2}} - 2 + u^{2}

= \int \frac{1}{u ^{2}} - 2 + 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 \frac{1}{u ^{2}} d u - \int 2 d u + \int u^{2} d u

\int \frac{1}{u ^{2}} d u = - \frac{1}{u}

\int 2 d u = 2 u

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

= - \frac{1}{u} - 2 u + \frac{u ^{3}}{3}

Substitute back

u = sin (x)

= - \frac{1}{sin ( x )} - 2 sin (x) + \frac{sin ^{3} ( x )}{3}

Simplify

- \frac{1}{sin ( x )} - 2 sin (x) + \frac{sin ^{3} ( x )}{3} : - csc (x) - 2 sin (x) + \frac{1}{3} sin^{3} (x)

= - csc (x) - 2 sin (x) + \frac{1}{3} sin^{3} (x)

Add a constant to the solution

= - csc (x) - 2 sin (x) + \frac{1}{3} sin^{3} (x) + C

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