Question

The integral ${\int }_{\frac{\pi }{4}}^{\frac{3\pi }{4}}\frac{dx}{1+\cos x}$

• A. $2$
• B. $4$
• C. $-1$
• D. $-2$

Answer 1

Answer: (A)

$2$

Multiply Numerator & denominator by $(1-cosx)$

$\int \frac{1-\cos x}{(1+\cos x)(1-\cos x)}dx=\int \frac{1-\cos x}{1-\cos x}dx$

$=\int \frac{1-\cos x}{{\sin }^{2}x}dx=\int \frac{1}{{\sin }^{2}x}dx-\int \frac{\cos x}{{\sin }^{2}x}dx$

$=-\cot x-\int \frac{1}{y^2}dy(y=\sin x,dy=\cos xdx)$

$=-\cot x+\frac{1}{y}={[-\cot x+\frac{1}{\sin x}+c]}_{\frac{\pi }{4}}^{{}^3\frac{\pi }{4}}$

$={[-\frac{1}{\tan x}+\frac{1}{\sin x}]}^{{}^3\frac{\pi }{4}}-{[-\frac{1}{\tan x}+\frac{1}{\sin x}]}^{{}^3\frac{\pi }{4}}$

$=(1+\sqrt{2})-(-1+\sqrt{2})=2$