Integrate (1 / 1 + cot x)

Cot x is an inverse trigonometric identity and can be expressed as cos x / sin x.

Answer:  ∫ (1 / 1 + cot x) dx is x/2 + 1/2 log(sin x + cos x) + c

Let's integrate.

Explanation:

⇒ ∫ 1 / 1 + cot x dx = ∫ 1 / 1 + cosx/sinx dx

⇒ ∫ 1 / (sin x + cos x) / sinx dx

Multiply both numerator and denominator by sin x

⇒ ∫ sinx / (sin x + cosx) dx.

Multiply and divide by 2

⇒ 1/2 ∫ 2 sin x / (sin x + cosx) dx

Add and subtract cos x in the numerator.

⇒ 1/2 ∫ sin x + sin x + cos x - cos x / (sin x + cosx) dx

⇒ 1/ 2 ∫ sin x + cos x + sin x - cos x / (sin x + cosx) dx

⇒ 1/ 2 ∫ [(sin x + cos x) / (sin x + cosx) + (sin x - cos x) / (sin x + cosx)] dx

⇒ 1/ 2  ∫ dx + 1/ 2 ∫ (sin x - cos x) / (sin x + cosx) dx

⇒ 1/2 x + 1/ 2 ∫ (sin x - cos x) / (sin x + cosx) dx

Let sin x + cos x = t

Differentiating w.r.t 'x'

cos x - sin x = dt/dx

⇒ 1/2 x - 1/2 ∫ [ dt / t ] dx

⇒ x/2 - 1/ 2 log[t] + c

⇒ x/2 - 1/2 log(sin x + cos x) + c

Thus, ∫ (1 / 1 + cot x) dx is x/2 + 1/2 log(sin x + cos x) + c