What is the integral of (cos^4(x))/(sin^6(x)) ?

The integral of (cos^4(x))/(sin^6(x)) is -(cot^5(x))/5+C

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integral of (cos^4(x))/(sin^6(x))

\int \frac{cos ^{4} ( x )}{sin ^{6} ( x )} d x

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

Solution steps

\int \frac{cos ^{4} ( x )}{sin ^{6} ( x )} d x

Rewrite using trig identities

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

Apply u-substitution

: \int - u^{4} d u

= \int - u^{4} d u

Take the constant out:

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

= - \int u^{4} d u

Apply the Power Rule

: \frac{u ^{5}}{5}

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

Substitute back

u = cot (x)

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

Add a constant to the solution

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

\int - 7 csc^{2} (x) d x

y^{^{''}} + y = k \cdot (e^{(- a \cdot t)} - e^{(- b \cdot t)})

\int x^{- 3} ln (x) d x

\frac{\partial}{\partial s} (e^{s + t})

\frac{\partial}{\partial x} (5 \cdot (y - 3)^{3} \cdot (x - 6))

What is the integral of (cos^4(x))/(sin^6(x)) ?
The integral of (cos^4(x))/(sin^6(x)) is -(cot^5(x))/5+C