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Evaluate : $ \int\large\frac{dx}{2\sin^2x+5\cos^2x} $

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  • $\int \large\frac{dx}{x^2+a^2}=\frac{1}{a}$$\tan^{-1}\big(\large\frac{x}{a}\big)$
  • $\large\frac{d}{dx}$$(\tan x)=\sec^2x$
Step 1:
Let $I=\int\large\frac{dx}{2\sin^2x+5\cos^2x}$
Divide the numerator and the denominator by $\cos^2x$
Now put $\tan x=t$
On differentiating w.r.t $x$ we get
Step 2:
On substituting we get,
$\;\;=\large\frac{1}{2}\int\large\frac{dt}{t^2+(\Large\frac{\sqrt 5}{\sqrt 2})^2}$
$\int \large\frac{dx}{x^2+a^2}=\frac{1}{a}$$\tan^{-1}\big(\large\frac{x}{a}\big)$
$\therefore I=\large\frac{1}{2}\times \frac{1}{\sqrt 5/\sqrt 2}$$\tan^{-1}\big(\large\frac{t}{\sqrt 5/\sqrt 2}\big)+c$
Step 3:
Substituting for $t$ we get,
$I=\large\frac{1}{2}\times \frac{\sqrt 2}{\sqrt 5}$$\tan^{-1}\big(\large\frac{\sqrt 2\tan x}{\sqrt 5}\big)+c$
$\;\;=\large \frac{1}{\sqrt {10}}$$\tan^{-1}\big(\large\frac{\sqrt 2\tan x}{\sqrt 5}\big)+c$
answered Nov 7, 2013 by sreemathi.v
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