Question

$\int \frac{1}{1+\tan x}dx=$

(a) log_e (x + sin x) + C

(b) log_e (sin x + cos x) + C

(c) $2{\sec }^2\frac{x}{2}+C$

(d) $\frac{1}{2}$ [x + log (sin x + cos x)] + C

Answer 1

Disclaimer : Generally here book is taking log_e x as log x . So we are writing ln x or log_e x instead log x only .
(d) $\frac{1}{2}$ [x + ln (sin x + cos x)] + C

$$\begin{gathered} \mathrm{Let}I=\int \frac{1}{1+\tan x}dx \\ =\int \frac{1}{1+{\displaystyle \frac{\sin x}{\cos x}}}dx \\ =\int \frac{\cos x}{\cos x+\sin x}dx \\ =\frac{1}{2}\int \frac{2\cos x}{\cos x+\sin x}dx \\ =\frac{1}{2}\int \left[\frac{\left(\cos x+\sin x\right)+\left(\cos x-\sin x\right)}{\left(\cos x+\sin x\right)}\right]dx \\ =\frac{1}{2}\int \left(\frac{\cos x+\sin x}{\cos x+\sin x}\right)dx+\frac{1}{2}\int \left(\frac{\cos x-\sin x}{\cos x+\sin x}\right)dx \\ =\frac{1}{2}\int dx+\frac{1}{2}\int \left(\frac{\cos x-\sin x}{\cos x+\sin x}\right)dx \\ \mathrm{Putting}\sin x+\cos x=t \\ \Rightarrow \left(\cos x-\sin x\right)dx=dt \\ \therefore I=\frac{1}{2}\int dx+\frac{1}{2}\int \frac{dt}{t} \\ =\frac{x}{2}+\frac{1}{2}\ln \left|t\right|+C \\ =\frac{x}{2}+\frac{1}{2}\ln \left|\cos x+\sin x\right|+C\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\left(\because \mathit{}t=\sin \mathit{}x+\cos \mathit{}x\right) \\ \mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{=}\frac{1}{2}\left[x+\ln \mathit{}\left(\sin \mathit{}x+\cos \mathit{}x\right)\right]+C\mathit{}\mathit{} \end{gathered}$$