What is the integral of sin^{-4}(x)cos^2(x) ?

The integral of sin^{-4}(x)cos^2(x) is -(cot^3(x))/3+C

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integral of sin^{-4}(x)cos^2(x)

\int sin^{- 4} (x) cos^{2} (x) d x

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

Solution steps

\int sin^{- 4} (x) cos^{2} (x) d x

Simplify

cos^{2} (x) : cos (x) cos (x)

= \int sin^{- 4} (x) cos (x) cos (x) d x

Simplify

sin^{- 4} (x) cos (x) cos (x) : csc^{4} (x) cos^{2} (x)

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

Rewrite using trig identities

= \int \frac{cot ^{2} ( x )}{sin ^{2} ( x )} d x

Apply u-substitution

: \int - u^{2} d u

= \int - u^{2} d u

Take the constant out:

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

= - \int u^{2} d u

Apply the Power Rule

: \frac{u ^{3}}{3}

= - \frac{u ^{3}}{3}

Substitute back

u = cot (x)

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

Add a constant to the solution

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

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What is the integral of sin^{-4}(x)cos^2(x) ?
The integral of sin^{-4}(x)cos^2(x) is -(cot^3(x))/3+C