Intigration of dx divided by 1+sinx+cosx

Question

\(∫\frac{1}{1\ +\ sinx\ +\ cosx}dx\)

Answer 1

I = \(∫\frac{1}{1\ +\ sinx\ +\ cosx}dx\)

\(∫\frac{1}{1+\frac{2tan\frac{x}{2}}{1+tan^2\frac{x}{2}}+\ \frac{1-\tan^2\frac{x}{2}}{1+tan^2\frac{x}{2}}}dx\)

= \(∫\frac{1+tan^2\frac{x}{2}}{1+tan^2\frac{x}{2}+2tan\frac{x}{2}+1-tan^2\frac{x}{2}}dx\)

= \(∫\frac{sec^2\frac{x}{2}dx}{2+2tan\frac{x}{2}}\)

Tan x/2 = t

⇒ 1/2 · sec² x/2 = dt/dx

⇒ dt = 1/2 · sec² x/2 dx

I = \(\frac{1}{2}∫\frac{sec^2\frac{x}{2}\ dx}{1+tan\frac{x}{2}}\)

\(∫\frac{dt}{1+t}=ln|1 + t|+c \)

\(∫\frac{1}{1+sinx+cosx}dx\)

= ln|1 + tan x/2| + c

\(∫\frac{1}{asinx\ +\ bcosx}dx\)