Question
Evaluate the following intregals:
$\int\frac{1}{4\sin^2\text{x}+5\cos^2\text{x}}\text{ dx}$
$\int\frac{1}{4\sin^2\text{x}+5\cos^2\text{x}}\text{ dx}$
✓
Answer
Let $\text{I}=\int\frac{1}{4\sin^2\text{x}+5\cos^2\text{x}}\text{ dx}$
Dividing numerator and denominator by $\cos^2\text{x}$
$\text{I}=\int\frac{\frac{1}{\cos^2\text{x}}}{4\tan^2\text{x}+5}\ \text{dx}$
$=\int\frac{\sec^2\text{x}}{4\tan^2\text{x}+5}\ \text{dx}$
Let $\tan\text{x}=\text{t}$
$\sec^2\text{x}\ \text{dx}=\text{dt}$
$\text{I}=\int\frac{\text{dt}}{4+9(\text{t})^2}$
$=\int\frac{\text{dt}}{4\text{t}^2+5}$
Let $2\text{t}=\text{u}$
$2\text{dt}=\text{du}$
$\text{I}=\frac{1}{2}\int\frac{\text{du}}{(4)^2+(\sqrt{5})^2}$
$=\frac{1}{2}\times\frac{1}{\sqrt{5}}\times\tan^{-1}\Big(\frac{\text{u}}{\sqrt{5}}\Big)+\text{C}$
$=\frac{1}{2\sqrt{5}}\tan^{-1}\Big(\frac{2\text{t}}{\sqrt{5}}\Big)+\text{C}$
$\text{I}=\frac{1}{2\sqrt{5}}\tan^{-1}\Big(\frac{2\tan\text{x}}{\sqrt{5}}\Big)+\text{C}$
Dividing numerator and denominator by $\cos^2\text{x}$
$\text{I}=\int\frac{\frac{1}{\cos^2\text{x}}}{4\tan^2\text{x}+5}\ \text{dx}$
$=\int\frac{\sec^2\text{x}}{4\tan^2\text{x}+5}\ \text{dx}$
Let $\tan\text{x}=\text{t}$
$\sec^2\text{x}\ \text{dx}=\text{dt}$
$\text{I}=\int\frac{\text{dt}}{4+9(\text{t})^2}$
$=\int\frac{\text{dt}}{4\text{t}^2+5}$
Let $2\text{t}=\text{u}$
$2\text{dt}=\text{du}$
$\text{I}=\frac{1}{2}\int\frac{\text{du}}{(4)^2+(\sqrt{5})^2}$
$=\frac{1}{2}\times\frac{1}{\sqrt{5}}\times\tan^{-1}\Big(\frac{\text{u}}{\sqrt{5}}\Big)+\text{C}$
$=\frac{1}{2\sqrt{5}}\tan^{-1}\Big(\frac{2\text{t}}{\sqrt{5}}\Big)+\text{C}$
$\text{I}=\frac{1}{2\sqrt{5}}\tan^{-1}\Big(\frac{2\tan\text{x}}{\sqrt{5}}\Big)+\text{C}$
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