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Evaluate: $$\int^{\frac{\pi}{2}}_0 \frac{\cos^2x\,dx}{\cos^2x+4\sin^2x}$$

First approach :

$$\int^{\frac{\pi}{2}}_0 \frac{\cos^2x\,dx}{\cos^2x+4(1-\cos^2x)}$$

$$=\int^{\frac{\pi}{2}}_0 \frac{\cos^2xdx}{4 - 3\cos^2x}$$

$$=\int^{\frac{\pi}{2}}_0 \frac{1}{3}\{\frac{4-3\cos^2x-4}{4 - 3\cos^2x}\}\,dx$$

$$=\int^{\frac{\pi}{2}}_0 \frac{1}{3}\{ 1- \frac{4}{4 - 3\cos^2x}\}\,dx$$

$$=\int^{\frac{\pi}{2}}_0 \frac{1}{3} 1\,dx- \int^{\frac{\pi}{2}}_0 \frac{1}{3} \frac{4\sec^2x}{4\sec^2x - 3}\,dx$$

$$=\int^{\frac{\pi}{2}}_0 \frac{1}{3} 1\,dx- \int^{\frac{\pi}{2}}_0 \frac{1}{3} \frac{4 \sec^2x}{4(1+\tan^2x) - 3}\,dx$$

Now I can easily put $\tan x = t$ and I get $\sec^2x \,dx =dt$

Second approach :

$$\int^{\frac{\pi}{2}}_0 \frac{\cos^2x\,dx}{\cos^2x+4\sin^2x}$$

Dividing numerator and denominator by $\cos^2x$ we get :

$$=\int^{\frac{\pi}{2}}_0 \frac{dx}{1 +4\tan^2x}$$

$$=\int^{\frac{\pi}{2}}_0 \frac{dx}{(4)\{\frac{1}{4} +\tan^2x\}}$$

$$=\int^{\frac{\pi}{2}}_0 \frac{dx}{(4)\{\{\frac{1}{2}\}^2 +(\tan x)^2\}}$$

Can we apply this formula of integral here :

$$\int \frac{1}{a^2+x^2}dx = \frac{1}{a}\tan^{-1}\frac{x}{a}$$

I tried but its not working here, I think doing some manipulation we can implement this here.. Please suggest thanks...

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  • $\begingroup$ No. that formula won't help in this situation because you have a function in place of $x$ and the $dx$ doesn't correspond to $\tan x$ as it is supposed to in the formula. $\endgroup$ – Paras Khosla Jan 22 '19 at 8:35
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Use the fact that

$$3 \int_0^{\pi/2} dx \frac{\cos^2{x}}{\cos^2{x}+4 \sin^2{x}} = 3 \int_0^{\pi/2} dx \frac{\cos^2{x}}{4-3 \cos^2{x}} = 4 \int_0^{\pi/2} \frac{dx}{4-3 \cos^2{x}} - \int_0^{\pi/2} dx$$

So consider the first integral on the RHS:

$$4 \int_0^{\pi/2} \frac{dx}{4-3 \cos^2{x}} = 2 \int_0^{2 \pi} \frac{dy}{5-3 \cos{y}}$$

We may evaluate this latter integral using residue theory. Let $z=e^{i y}$, $dy = -i dz/z$ so that

$$2 \int_0^{2 \pi} \frac{dy}{5-3 \cos{y}} = -i 4 \oint_{|z|=1} \frac{dz}{10z-3(z^2+1)}$$

The denominator has zeroes at $z=1/3$ and $z=3$; of these, only $z=1/3$ lies within the unit circle. Thus the integral is

$$(i 2 \pi) \frac{-i 4}{10-6 (1/3)} = \pi$$

Then

$$\int_0^{\pi/2} dx \frac{\cos^2{x}}{\cos^2{x}+4 \sin^2{x}} = \frac{\pi-(\pi/2)}{3} = \frac{\pi}{6}$$

ADDENDUM

Alternatively, you may substitute $t=\tan{(y/2)}$, $dy = 2 dt/(1+t^2)$, $\cos{y} = (1-t^2)/(1+t^2)$, and then

$$2 \int_0^{2 \pi} \frac{dy}{5-3 \cos{y}} = \int_0^{\infty} \frac{dt}{(1/4)+t^2} = \frac{\pi}{2} 2 = \pi$$

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You multiply and divide by $\sec^{4}(x)$ and see if it's working. The denominator becomes $$\sec^{2}(x) + 4 \tan^{2}(x) \sec^{2}(x)$$ Now use $1+\tan^{2}(x)=\sec^{2}(x)$.

So

  • In numerator you have $\sec^{2}(x)$

  • In denominator you have $\sec^{2}(x) \cdot \left(1+4\tan^{2}(x)\right) = (1+\tan^{2}(x))\cdot(1+4\tan^{2}(x))$.

  • Now put $t=\tan(x)$ and your integral becomes $$\int_{0}^{\infty} \frac{dt}{(1+t^{2}) \cdot (1+4t^{2})} = \int_{0}^{\infty} \left\{ \frac{1}{1+t^2} -\frac{4}{1+4t^2}\right\} \ dt$$


Oh. I missed a trick here. Note that you had $$\int_{0}^{\pi/2} \frac{dx}{1+4\tan^{2}(x)}$$ Now put $t = \tan(x)$. Then you have $dt = \sec^{2}(x) \ dx $ and so $dx = \frac{dt}{(1+t^{2})}$.

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