What is the integral of (cos^2(x))/(sin^6(x)) ?

The integral of (cos^2(x))/(sin^6(x)) is -1/5 cot^5(x)-1/3 cot^3(x)+C

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integral of (cos^2(x))/(sin^6(x))

\int \frac{cos ^{2} ( x )}{sin ^{6} ( x )} d x

- \frac{1}{5} cot^{5} (x) - \frac{1}{3} cot^{3} (x) + C

Solution steps

\int \frac{cos ^{2} ( x )}{sin ^{6} ( x )} d x

Rewrite using trig identities

= \int cot^{2} (x) csc^{4} (x) d x

Apply Trigonometric Substitution

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

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Expand

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

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

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

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

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

Substitute back

u = cot (x)

= - (\frac{cot ^{5} ( x )}{5} + \frac{cot ^{3} ( x )}{3})

Simplify

- (\frac{cot ^{5} ( x )}{5} + \frac{cot ^{3} ( x )}{3}) : - \frac{1}{5} cot^{5} (x) - \frac{1}{3} cot^{3} (x)

= - \frac{1}{5} cot^{5} (x) - \frac{1}{3} cot^{3} (x)

Add a constant to the solution

= - \frac{1}{5} cot^{5} (x) - \frac{1}{3} cot^{3} (x) + C

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What is the integral of (cos^2(x))/(sin^6(x)) ?
The integral of (cos^2(x))/(sin^6(x)) is -1/5 cot^5(x)-1/3 cot^3(x)+C