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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. –  Stefan Smith Oct 4 '12 at 21:57
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4 Answers

up vote 0 down vote accepted

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

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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Thank you so much! –  Varchasvi Oct 4 '12 at 15:46
    
Does "\sec" work in $\TeX$? –  Stefan Smith Oct 4 '12 at 21:56
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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

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

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