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

The integral of sin^{-4}(x/2)cos^3(x/2) is 2(-1/3 csc^3(x/2)+csc(x/2))+C

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integral of sin^{-4}(x/2)cos^3(x/2)

\int sin^{- 4} (\frac{x}{2}) cos^{3} (\frac{x}{2}) d x

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

Solution steps

\int sin^{- 4} (\frac{x}{2}) cos^{3} (\frac{x}{2}) d x

Apply u-substitution

: \int \frac{2 cos ^{3} ( u )}{sin ^{4} ( u )} d u

= \int \frac{2 cos ^{3} ( u )}{sin ^{4} ( u )} d u

Take the constant out:

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

= 2 \cdot \int \frac{cos ^{3} ( u )}{sin ^{4} ( u )} d u

Rewrite using trig identities

= 2 \cdot \int \frac{( 1 - sin ^{2} ( u ) ) cos ( u )}{sin ^{4} ( u )} d u

Apply u-substitution

: \int \frac{1 - v ^{2}}{v ^{4}} d v

= 2 \cdot \int \frac{1 - v ^{2}}{v ^{4}} d v

Expand

\frac{1 - v ^{2}}{v ^{4}} : \frac{1}{v ^{4}} - \frac{1}{v ^{2}}

= 2 \cdot \int \frac{1}{v ^{4}} - \frac{1}{v ^{2}} 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

= 2 (\int \frac{1}{v ^{4}} d v - \int \frac{1}{v ^{2}} d v)

\int \frac{1}{v ^{4}} d v = - \frac{1}{3 v ^{3}}

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

= 2 (- \frac{1}{3 v ^{3}} - (- \frac{1}{v}))

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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What is the integral of sin^{-4}(x/2)cos^3(x/2) ?
The integral of sin^{-4}(x/2)cos^3(x/2) is 2(-1/3 csc^3(x/2)+csc(x/2))+C