Find: d (4 + ^ 2 ) (2- ^ 2 ) d - Vidyadip

Question

$\text{Find}:\int \frac{\sin \theta\ \text{d}\theta}{(4 + \cos^{2} \theta) (2- \sin^{2} \theta)} \text{d} \theta$

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Answer

$\text{I} = \int \frac{\sin \theta\ \text{d}\theta}{(4 + \cos^{2} \theta) (2 - \sin^{2} \theta)} = \int\frac{\sin \theta\ \text{d}\theta}{(4 + \cos^{2} \theta) (1 + \cos^{2}\theta)}$$= -\int \frac{\text{dt}}{(4 + \text{t}^{2}) (1 + \text{t}^{2})}, \text{ where } \cos \theta = \text{t}$
$= \int \frac{\frac{1}{3}}{4 + \text{t}^{2}} \text{dt} - \int \frac{\frac{1}{3}}{1 + \text{t}^{2}} \text{dt}$
$= \frac{1}{6}\tan^{-1} \frac{\text{t}}{2} - \frac{1}{3} \tan^{-1} \text{t} + \text{c}$
$ = \frac{1}{6}\tan^{-1} \big(\frac{\cos \theta}{2}\big) - \frac{1}{3} \tan^{-1} \cos \theta + \text{c}$

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