INTEGRALS — Maths STD 12 Science — Question

3. $2$

Solution:

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

$=\int\limits^\pi_0\frac{1}{1+\sin\text{x}}\times\frac{1-\sin\text{x}}{1-\sin\text{x}}\text{dx}$

$= \int\limits^\pi_0\frac{1-\sin\text{x}}{1-\sin^2\text{x}}\text{dx}$

$= \int\limits^\pi_0\frac{1-\sin\text{x}}{\cos^2\text{x}}\text{dx}$

$=\int\limits^\pi_0(\sec^2\text{x}-\sec\text{x}\tan\text{x})\text{dx}$

$=\big[\tan\text{x}-\sec\text{x}\big]^\pi_0$

$=0+1-0+1$

$=2$