Question
$\int\limits^\frac{\pi}{2}_0\frac{\sin\text{x}}{\sin\text{x}+\cos\text{x}}\text{ dx}$ equals to:
-
$\pi$
-
$\frac{\pi}{2}$
-
$\frac{\pi}{3}$
-
$\frac{\pi}{4}$
$\pi$
$\frac{\pi}{2}$
$\frac{\pi}{3}$
$\frac{\pi}{4}$
Solution:
We have,
$\text{I}=\int\limits^\frac{\pi}{2}_0\frac{\sin\text{x}}{\sin\text{x}+\cos\text{x}}\text{dx}\ ...(\text{i})$
$\Rightarrow \text{I}=\int\limits^\frac{\pi}{2}_0\frac{\sin\big(\frac{\pi}{2}-\text{x}\big)}{\sin\big(\frac{\pi}{2}-\text{x}+\cos\big(\frac{\pi}{2}-\text{x}\big)}\text{dx}$
$\Rightarrow\text{I}=\int\limits^\frac{\pi}{2}_0\frac{\cos\text{x}}{\cos\text{x}+\sin\text{x}}\text{dx}$
$\therefore\ \text{I}=\int\limits^\frac{\pi}{2}_0\frac{\cos\text{x}}{\sin\text{x}+\cos\text{x}}\text{dx}\ ...(\text{ii})$
Adding (i) and (ii), we get
$2\text{I}=\int\limits^\frac{\pi}{2}_0\Big[\frac{\sin\text{x}}{\sin{\text{x}}+\cos\text{x}}+\frac{\cos\text{x}}{\cos\text{x}+\sin\text{x}}\Big]\text{dx}$
$=\int\limits^\frac{\pi}{2}_0\Big[\frac{\sin\text{x}+\cos\text{x}}{\sin\text{x}+\cos\text{x}}\Big]\text{dx}$
$=\int\limits^\frac{\pi}{2}_0\text{dx}$
$=\big[\text{x}\big]^\frac{\pi}{2}_0$
$=\frac{\pi}{2}$
Hence $\text{I}=\frac{\pi}{4}$
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