1 1- - dx= - Vidyadip

MCQ

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

A
$\log\big|1+\cot\frac{\pi}{2}\big|+\text{C}$
B
$\log\big|1-\tan\frac{\pi}{2}\big|+\text{C}$
✓
$\log\big|1-\cot\frac{\pi}{2}\big|+\text{C}$
D
$\log\big|1+\tan\frac{\pi}{2}\big|+\text{C}$

✓

Answer

Correct option: C.

$\log\big|1-\cot\frac{\pi}{2}\big|+\text{C}$

$\int\frac{\text{dx}}{1-\cos\text{x}-\sin\text{x}}$
Put $\text{t}=\tan\frac{\text{x}}{2}$
$\text{dx}=\frac{2\text{dt}}{1+\text{t}^2}$
$\cos\text{x}=\frac{1-\tan^2\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}=\frac{1-\text{t}^2}{1+\text{t}^2}$
$\sin\text{x}=\frac{2\tan\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}=\frac{2\text{t}}{1+\text{t}^2}$
Put in the question
$\text{I}=\int\frac{\frac{2\text{dt}}{1+\text{t}^2}}{1-\frac{1-\text{t}^2}{1+\text{t}^2}-\frac{2\text{t}}{1+\text{t}^2}}$
$\text{I}=\int\frac{\text{dt}}{\text{t}^2-1}$
$\text{I}=\int\frac{\text{dt}}{\text{t}^2-\text{t}+\frac{1}{4}-\frac{1}{4}}$
$\text{I}=\ln\Big|1-\cot\frac{\text{x}}{2}\Big|+\text{C}$

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