Question

$\int_{}^{}{\frac{dx}{\sin x + \cos x} =}$

• A. ${logtan}(\pi/8 + x/2) + c$
• B. ${logtan}(\pi/8 - x/2) + c$
• C. $\frac{1}{\sqrt{2}}{logtan}(\pi/8 + x/2) + c$
• D. None of these

Answer 1

Answer: (C)

$\frac{1}{\sqrt{2}}{logtan}(\pi/8 + x/2) + c$

Answer 2

$\int_{}^{}\frac{dx}{\frac{2\tan x/2}{1 + \tan^{2}x/2} + \frac{1 - \tan^{2}x/2}{1 + \tan^{2}x/2}}$

$\Rightarrow$ $\int_{}^{}{\frac{\sec^{2}x/2}{2\tan x/2 + 1 - \tan^{2}x/2}dx}$

Put $\tan x/2 = t$ $\Rightarrow \frac{1}{2}\sec^{2}x/2dx = dt$

$\therefore$ $I = 2\int_{}^{}\frac{dt}{2t + 1 - t^{2}} = 2\int_{}^{}\frac{dt}{2 - (t^{2} - 2t + 1)}$

$\Rightarrow I = 2\int_{}^{}\frac{dt}{(\sqrt{2})^{2} - (t - 1)^{2}} = \frac{2}{2\sqrt{2}}\log\left| \frac{\sqrt{2} + t - 1}{\sqrt{2} - t + 1} \right| + c$

$\Rightarrow$ $I = \frac{1}{\sqrt{2}}\log\frac{\lbrack(\sqrt{2} - 1) + \tan x/2\rbrack\lbrack\sqrt{2} - 1\rbrack}{\lbrack(\sqrt{2} + 1) - \tan x/2\rbrack\lbrack\sqrt{2} - 1\rbrack} + c$

$\Rightarrow$ $I = \frac{1}{\sqrt{2}}\log\frac{\tan\pi/8 + \tan x/2}{1 - (\sqrt{2} - 1)\tan x/2} + \frac{1}{\sqrt{2}}\log(\sqrt{2} - 1) + c$

$\Rightarrow$ $I = \frac{1}{\sqrt{2}}{logtan}(\pi/8 + x/2) + c_{1}$

where $c_{1} = \frac{1}{\sqrt{2}}\log(\sqrt{2} - 1) + c$

Trick : $I = \frac{1}{\sqrt{2}}\int_{}^{}\frac{dx}{\frac{1}{\sqrt{2}}\sin x + \frac{1}{\sqrt{2}}\cos x} = \frac{1}{\sqrt{2}}\int_{}^{}\frac{dx}{\sin(\pi/4 + x)}$ $= \frac{1}{\sqrt{2}}\int_{}^{}\text{cosec}(\pi/4 + x)dx$ $= \frac{1}{\sqrt{2}}{logtan}(\pi/8 + x/2) + c$.