What is the integral of 1/((tan(x)+1)sin^2(x)) ?

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

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integral of 1/((tan(x)+1)sin^2(x))

\int \frac{1}{( tan ( x ) + 1 ) sin ^{2} ( x )} d x

- ln ∣ tan (x) ∣ - cot (x) + ln ∣ tan (x) + 1 ∣ + C

Solution steps

\int \frac{1}{( tan ( x ) + 1 ) sin ^{2} ( x )} d x

Apply Trigonometric Substitution

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

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

Take the partial fraction of

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

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

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

\int \frac{1}{u} d u = ln ∣ u ∣

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

\int \frac{1}{u + 1} d u = ln ∣ u + 1 ∣

= - ln ∣ u ∣ - \frac{1}{u} + ln ∣ u + 1 ∣

Substitute back

u = tan (x)

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

Use the basic trigonometric identity:

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

= - ln ∣ tan (x) ∣ - cot (x) + ln ∣ tan (x) + 1 ∣

Add a constant to the solution

= - ln ∣ tan (x) ∣ - cot (x) + ln ∣ tan (x) + 1 ∣ + C

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