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I'm trying to solve this integral - $\int_0^{\pi/2}{dx\over{4+9\cos^2x}}$

I've started by dividing numerator and denominator by $\cos^2x$

$\int_0^{\pi/2}{{dx\over{\cos^2x}}\over{{4+9\cos^2x}\over{\cos^2x}}}$

which gives me

$\int_0^{\pi/2}{{\sec^2x}{dx}\over{4{\sec^2x}+9}}$

I cant go any further. Can someone help / point me in the right direction please?

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    You don't solve integrals, you evaluate them.2012-10-04

4 Answers 4

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$\int_0^{\pi/2}{{\sec^2x}\;{dx}\over{4{\sec^2x}+9}}=\int_0^{\pi/2}{{\sec^2x}\;{dx}\over{4{(1+\tan^2x)}+9}}$

Now substitute $\tan x =t$ which gives $\sec^2x\; dx=dt$ and thus the integral becomes $\int_0^{\infty}{{dt}\over{4{(1+t^2)}+9}}=\int_0^{\infty}{{dt}\over{4{t^2}+13}}$ which is standard integral.

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    Does "\sec" work in $\TeX$?2012-10-04
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If you apply the substitution $u = \tan(x/2)$, some algebraic manipulation will give you a rational function in $u$ as an integrand. From there you can use partial fractions.

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Hint

Calculate

  1. $d(\tan{x})=$
  2. $1+\tan^2{x}=$
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You can work as following:

$\begin{align*} \int_0^{\pi/2}{dx\over{4+9cos^2x}} &=\int_0^{\pi/2}{{sec^2x}{dx}\over{4{sec^2x}+9}}\\ &=\int_0^{\pi/2}{{d\tan x}\over{4{\tan^2x}+13}} \end{align*}$

Set $y=\tan x$, you can get

$\begin{align*} \int_0^{\pi/2}{{d\tan x}\over{4{\tan^2x}+13}} &=\int_{0}^{+\infty}{{dy}\over{13+4y^{2}}}\\ &=\frac{1}{2\sqrt{13}}\arctan\frac{2}{\sqrt{13}}y|_{0}^{+\infty}\\ &=\frac{\pi}{4\sqrt{13}} \end{align*}$