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

  1. $0$

  2. $\frac{1}{2}$

  3. $2$

  4. $\frac{3}{2}$

Answer

  1. $2$

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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