What is the integral of cot^3(x/2)csc^4(x/2) ?

The integral of cot^3(x/2)csc^4(x/2) is 2(-1/6 csc^6(x/2)+1/4 csc^4(x/2))+C

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integral of cot^3(x/2)csc^4(x/2)

\int cot^{3} (\frac{x}{2}) csc^{4} (\frac{x}{2}) d x

2 (- \frac{1}{6} csc^{6} (\frac{x}{2}) + \frac{1}{4} csc^{4} (\frac{x}{2})) + C

Solution steps

\int cot^{3} (\frac{x}{2}) csc^{4} (\frac{x}{2}) d x

Apply u-substitution

: \int cot^{3} (u) csc^{4} (u) \cdot 2 d u

= \int cot^{3} (u) csc^{4} (u) \cdot 2 d u

Take the constant out:

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

= 2 \cdot \int cot^{3} (u) csc^{4} (u) d u

Rewrite using trig identities

= 2 \cdot \int \frac{( 1 - sin ^{2} ( u ) ) cos ( u )}{sin ^{7} ( u )} d u

Apply u-substitution

: \int \frac{1 - v ^{2}}{v ^{7}} d v

= 2 \cdot \int \frac{1 - v ^{2}}{v ^{7}} d v

Expand

\frac{1 - v ^{2}}{v ^{7}} : \frac{1}{v ^{7}} - \frac{1}{v ^{5}}

= 2 \cdot \int \frac{1}{v ^{7}} - \frac{1}{v ^{5}} 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 ^{7}} d v - \int \frac{1}{v ^{5}} d v)

\int \frac{1}{v ^{7}} d v = - \frac{1}{6 v ^{6}}

\int \frac{1}{v ^{5}} d v = - \frac{1}{4 v ^{4}}

= 2 (- \frac{1}{6 v ^{6}} - (- \frac{1}{4 v ^{4}}))

Substitute back

= 2 (- \frac{1}{6 sin ^{6} ( \frac{x}{2} )} - (- \frac{1}{4 sin ^{4} ( \frac{x}{2} )}))

Simplify

2 (- \frac{1}{6 sin ^{6} ( \frac{x}{2} )} - (- \frac{1}{4 sin ^{4} ( \frac{x}{2} )})) : 2 (- \frac{1}{6} csc^{6} (\frac{x}{2}) + \frac{1}{4} csc^{4} (\frac{x}{2}))

= 2 (- \frac{1}{6} csc^{6} (\frac{x}{2}) + \frac{1}{4} csc^{4} (\frac{x}{2}))

Add a constant to the solution

= 2 (- \frac{1}{6} csc^{6} (\frac{x}{2}) + \frac{1}{4} csc^{4} (\frac{x}{2})) + C

f (x) = 5 e^{x}, at x = 4

\int (sin (2 x))^{3} d x

\int (\frac{1}{x ^{3} + 1}) d x

x (2 x + 1)^{3}

\frac{d}{d x} (sinh^{2} (3 x) + arcsin (2 x))

What is the integral of cot^3(x/2)csc^4(x/2) ?
The integral of cot^3(x/2)csc^4(x/2) is 2(-1/6 csc^6(x/2)+1/4 csc^4(x/2))+C