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

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

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integral of (csc^4(x))/((1+cot(x))^2)

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

\frac{2}{cot ( x ) + 1} - cot (x) + 2 ln ∣ cot (x) + 1 ∣ + C

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Expand

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

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

= - (- \frac{1}{1 + u} + u - 2 ln ∣ u + 1 ∣ - \frac{1}{u + 1})

Substitute back

u = cot (x)

= - (- \frac{1}{1 + cot ( x )} + cot (x) - 2 ln ∣ cot (x) + 1 ∣ - \frac{1}{cot ( x ) + 1})

Simplify

- (- \frac{1}{1 + cot ( x )} + cot (x) - 2 ln ∣ cot (x) + 1 ∣ - \frac{1}{cot ( x ) + 1}) : \frac{2}{cot ( x ) + 1} - cot (x) + 2 ln ∣ cot (x) + 1 ∣

= \frac{2}{cot ( x ) + 1} - cot (x) + 2 ln ∣ cot (x) + 1 ∣

Add a constant to the solution

= \frac{2}{cot ( x ) + 1} - cot (x) + 2 ln ∣ cot (x) + 1 ∣ + C

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