^ _01 1+ dx equals: 1. 0 2. 1 2 3. 2 4. 3 2

Question

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

$0$
$\frac{1}{2}$
$2$
$\frac{3}{2}$

✓

Answer

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

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