Question

${\displaystyle \int \frac{\cos x}{1-\sin x}\text{dx}=}$ ______.

विकल्प

• log |sec x + tan x| + c

• log |sec x(sex x + tan x)| + c

• log |sex - tan x| + c

• log |sec x(sec x - tan x)| + c

Answer 1

${\displaystyle \int \frac{\cos x}{1-\sin x}\text{dx}=}$ log |sec x(sex x + tan x)| + c.

Explanation:

Let I = ${\displaystyle \int \frac{\cos x}{1-\sin x}\text{dx}}$

Put 1 - sin x = t ⇒ cos x dx = - dt

${\displaystyle \therefore \text{I}=\int \frac{-\text{dt}}{\text{t}}=-\log \left|\text{t}\right|}$ + c

= - log |1 - sin x| + c

= log |(1 - sin x)^-1| + c

${\displaystyle =\log \left|\frac{1}{1-\sin x}\right|}$ + c

${\displaystyle =\log |\frac{1}{1-\sin x}\times \frac{1+\sin x}{1+\sin x}|}$+ c

${\displaystyle =\log \left|\frac{1+\sin x}{1-{\sin }^{2}x}\right|}$+ c

${\displaystyle =\log \left|\frac{1+\sin x}{{\cos }^{2}x}\right|}$ + c

${\displaystyle =\log |\frac{1}{{\cos }^{2}x}+\frac{\sin x}{{\cos }^{2}x}|}$ + c

= log |sec² x + sec x · tan x| + c

= log |sec x (sec x + tan x)| + c