^ _01 1+ dx equals : - Vidyadip

MCQ

$\int\limits^{\pi}_0\frac{1}{1+\sin\text{x}}\text{ dx}$ equals :

A
$0$
B
$\frac{1}{2}$
✓
$2$
D
$\frac{3}{2}$

✓

Answer

Correct option: C.

$2$

$\int\limits^{\pi}_0\frac{1}{1+\sin\text{x}}\text{ dx}$
$=\int\limits^\pi_0\frac{1}{1+\sin\text{x}}\times\frac{1-\sin\text{x}}{1-\sin\text{x}}\text{dx}$
$= \int\limits^\pi_0\frac{1-\sin\text{x}}{1-\sin^2\text{x}}\text{dx}$
$= \int\limits^\pi_0\frac{1-\sin\text{x}}{\cos^2\text{x}}\text{dx}$
$=\int\limits^\pi_0(\sec^2\text{x}-\sec\text{x}\tan\text{x})\text{dx}$
$=\big[\tan\text{x}-\sec\text{x}\big]^\pi_0$
$=0+1-0+1$
$=2$

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