Evaluate: ^ /2 _ 0 ^ 2 x dx 1 + 3 ^ 2 x - Vidyadip

Question

Evaluate:
$\int\limits^{\pi/2}_{0} \frac{\cos^{2} \text{x dx}}{1 + 3\sin^{2}\text{x}}$

✓

Answer

$\text{I} = \int\limits^\frac{\pi}{2}_{0}\frac{\text{dx}}{1 + 4 \tan^{2}\text{x}} = \int\limits^\frac{\pi}{2}_0 \frac{\sec^{2}\text{x}}{( 1 + \tan^{2} \text{x})(1 + 4\tan^{2}\text{x})} \text{dx}$$\text{Put} \tan \text{x = t}$
$\text{I} = \int\limits^{\infty}_{0}\frac{\text{dt}}{(1 + \text{t}^{2)} (1 + 4 \text{t}^{2})} = -\frac{1}{3} \int\limits^\infty_{0} \frac{\text{dt}}{1 + \text{t}^{2}} + \frac{4}{3}\int\limits^\infty_{0} \frac{\text{dt}}{1 + (\text{2 t)}^{2}}$
$= - \frac{1}{3} \tan^{-1}\text{t}\bigg]^{\infty}_{0} + \frac{4}{3 \times 2} \tan^{-1}\text{(2 t})\bigg]^{\infty}_{0}$
$= -\frac{1}{3}\bigg(\frac{\pi}{2}\bigg) + \frac{2}{3}\bigg(\frac{\pi}{2}\bigg) = \frac{\pi}{6}$

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