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

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

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integral of (csc(x)+cot(x))^2

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

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

Solution steps

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

Apply u-substitution

= \int \frac{2}{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

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

Take the partial fraction of

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

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

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

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

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

Substitute back

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

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

Use the basic trigonometric identity:

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

= 2 (- cot (\frac{x}{2}) - arctan (tan (\frac{x}{2})))

Add a constant to the solution

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

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