What is the integral of cot^4(1/2 x) ?

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

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integral of cot^4(1/2 x)

\int cot^{4} (\frac{1}{2} x) d x

2 (cot (\frac{x}{2}) - \frac{1}{3} cot^{3} (\frac{x}{2}) + arctan (tan (\frac{x}{2}))) + C

Solution steps

\int cot^{4} (\frac{1}{2} x) d x

Rewrite using trig identities

= \int \frac{1}{tan ^{4} ( \frac{x}{2} )} d x

Use the basic trigonometric identity:

\frac{1}{tan ( x )} = cot (x)

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

Apply u-substitution

: \int \frac{2}{u ^{4} ( 1 + u ^{2} )} d u

= \int \frac{2}{u ^{4} ( 1 + u ^{2} )} d u

Take the constant out:

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

= 2 \cdot \int \frac{1}{u ^{4} ( 1 + u ^{2} )} d u

Take the partial fraction of

\frac{1}{u ^{4} ( 1 + u ^{2} )} : - \frac{1}{u ^{2}} + \frac{1}{u ^{4}} + \frac{1}{u ^{2} + 1}

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

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

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

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

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

= 2 (- (- \frac{1}{u}) - \frac{1}{3 u ^{3}} + arctan (u))

Substitute back

u = tan (\frac{x}{2})

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

Simplify

2 (- (- \frac{1}{tan ( \frac{x}{2} )}) - \frac{1}{3 tan ^{3} ( \frac{x}{2} )} + arctan (tan (\frac{x}{2}))) : 2 (cot (\frac{x}{2}) - \frac{1}{3} cot^{3} (\frac{x}{2}) + arctan (tan (\frac{x}{2})))

= 2 (cot (\frac{x}{2}) - \frac{1}{3} cot^{3} (\frac{x}{2}) + arctan (tan (\frac{x}{2})))

Add a constant to the solution

= 2 (cot (\frac{x}{2}) - \frac{1}{3} cot^{3} (\frac{x}{2}) + arctan (tan (\frac{x}{2}))) + C

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