Question
Evaluate the following intregals:
$\int\frac{1}{1+3\sin^2\text{x}}\text{ dx}$
$\int\frac{1}{1+3\sin^2\text{x}}\text{ dx}$
✓
Answer
Let $\text{I}=\int\frac{1}{1+3\sin^2\text{x}}\text{ dx}$
Dividing numerator and denominator by $\cos^2\text{x}$
$\text{I}=\int\frac{\frac{1}{\cos^2\text{x}}}{\frac{1}{\cos^2\text{x}}+\frac{3\sin^2\text{x}}{\cos^2\text{x}}}\ \text{dx}$
$=\int\frac{\sec^2\text{x}}{\sec^2\text{x}+3\tan^2\text{x}}\ \text{dx}$
$=\int\frac{\sec^2\text{x}}{1+\tan^2\text{x}+3\tan^2\text{x}}\ \text{dx}$
$=\int\frac{\sec^2\text{x}}{1+4\tan^2\text{x}}\ \text{dx}$
$=\int\frac{\sec^2\text{x}}{1+(2\tan\text{x})}\ \text{dx}$
Let $2\tan\text{x}=\text{t}$
$2\sec^2\times\text{dx}=\text{dt}$
$\text{I}=\frac{1}{2}\int\frac{\text{dt}}{1+\text{t}^2}$
$=\frac{1}{2}\tan^{-1}\text{t}+\text{c}$
$\text{I}=\frac{1}{2}\tan^{-1}(2\tan\text{x})+\text{C}$
Dividing numerator and denominator by $\cos^2\text{x}$
$\text{I}=\int\frac{\frac{1}{\cos^2\text{x}}}{\frac{1}{\cos^2\text{x}}+\frac{3\sin^2\text{x}}{\cos^2\text{x}}}\ \text{dx}$
$=\int\frac{\sec^2\text{x}}{\sec^2\text{x}+3\tan^2\text{x}}\ \text{dx}$
$=\int\frac{\sec^2\text{x}}{1+\tan^2\text{x}+3\tan^2\text{x}}\ \text{dx}$
$=\int\frac{\sec^2\text{x}}{1+4\tan^2\text{x}}\ \text{dx}$
$=\int\frac{\sec^2\text{x}}{1+(2\tan\text{x})}\ \text{dx}$
Let $2\tan\text{x}=\text{t}$
$2\sec^2\times\text{dx}=\text{dt}$
$\text{I}=\frac{1}{2}\int\frac{\text{dt}}{1+\text{t}^2}$
$=\frac{1}{2}\tan^{-1}\text{t}+\text{c}$
$\text{I}=\frac{1}{2}\tan^{-1}(2\tan\text{x})+\text{C}$
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