Question

Integration of 1/(a^2cos^2x+b^2sin^2x)^2

Answer 1

∫0π/2(a2cos2x+b2sin2x)−2dx∫0π/2(a2cos2⁡x+b2sin2⁡x)−2dx

with help from the partial derivatives with respect to a and b of

∫0π/2(a2cos2x+b2sin2x)−1dx∫0π/2(a2cos2⁡x+b2sin2⁡x)−1dx

I've calculated these derivatives

∫0π/2−2a∗cos2x(a2cos2x+b2sin2x)−2dx∫0π/2−2a∗cos2⁡x(a2cos2⁡x+b2sin2⁡x)−2dx

and

∫0π/2−2b∗sin2x(a2cos2x+b2sin2x)−2dx

Answer 2

Given :

Function f(x)

$f(x) = \frac{1}{({ a^2\sin^2x + b^2\cos^2 x})^{2} }$

To find :

Integration of f(x).

Solution :

It is given that function f(x) is :

$f(x) = \frac{1}{({ a^2\sin^2x + b^2\cos^2 x})^{2} }$

From denominator of function f(x),

taking a²cos²x common,

we get

$a^2\sin^2x + b^2\cos^2 x = \\ a^2\cos^2 x\left(\tan^2 x + \dfrac{b^2}{a^2}\right) \:$

So we will integrate as :

$I = \displaystyle \int \dfrac{dx}{(a^2\sin^2x + b^2\cos^2 x)^2} = \displaystyle \int \dfrac{dx}{a^4\cos^4 x \left(\tan^2 x + \dfrac{b^2}{a^2}\right)^2} \:$

To solve this equation, we have to choose some values so that we can make the above equation easy to integrate.

So,

$</p><p>\text{Let }c = \dfrac{b}{a}; \tan x = c\tan \theta \implies \sec^2 x \cdot dx = c\sec^2 \theta \cdot d\theta \:$

[tex]= \dfrac{1}{a^4c^3}\displaystyle \int (1+c^2\tan^2 \theta) \cdot \cos^2 \theta \cdot d\theta \\

= \dfrac{1}{a^4c^3}\left [\displaystyle \int \cos^2 \theta \cdot d\theta + \displaystyle \int c^2\sin^2 \theta \cdot d\theta \right] \\

= \dfrac{1}{2a^4c^3}\left [\displaystyle \int (1+\cos 2\theta) \cdot d\theta + \displaystyle \int c^2(1-\cos 2\theta)\cdot d\theta \right] \\

= \dfrac{1}{2a^4c^3}\left [\theta + \dfrac{\sin 2\theta}{2} + c^2\theta - \dfrac{c^2\sin 2\theta}{2} \right] + I_c \\

= \dfrac{1}{2a^4c^3}\left [\theta + \dfrac{\sin 2\theta}{2} + c^2\theta - \dfrac{c^2\sin 2\theta}{2} \right] + I_c \\

= \dfrac{(1+c^2)\theta}{2a^4c^3} - \dfrac{(c^2-1)\sin 2\theta}{4a^4c^3} + I_c \\

c\tan \theta = \tan x \implies \tan \theta = \dfrac{a\tan x}{b}

\implies \sin 2\theta = \dfrac{2\tan \theta}{1+\tan^2 \theta} = \dfrac{\frac{2a\tan x}{b}}{1+\frac{a^2\tan^2 x}{b^2}} = \dfrac{2ab\tan x}{b^2+a^2\tan^2 x} \\

I = \dfrac{\left(1+\frac{b^2}{a^2}\right)\tan^{-1}\left(\frac{a\tan x}{b} \right)}{2a^4\left(\frac{b}{a}\right)^3 } - \dfrac{(\frac{b^2}{a^2}-1)}{4a^4\left(\frac{b}{a}\right)^3}\cdot \dfrac{2ab\tan x}{b^2+a^2\tan^2 x} + I_c \\

[/tex]

Then

[tex]= \dfrac{(a^2+b^2)\tan^{-1}\left(\frac{a\tan x}{b} \right)}{2a^3b^3} - \dfrac{(b^2-a^2)\tan x}{2a^2b^2(b^2+a^2\tan^2 x)} + I_c \\

I = \dfrac{(a^2+b^2)\tan^{-1}\left(\frac{a\tan x}{b} \right)}{2a^3b^3} - \dfrac{(b^2-a^2)\tan x}{2a^2b^2(b^2+a^2\tan^2 x)} + I_c \\

\: [/tex]

[tex]= \dfrac{(a^2+b^2)\tan^{-1}\left(\frac{a\tan x}{b} \right)}{2a^3b^3} - \dfrac{(b^2-a^2)\sin 2x}{4a^2b^2(b^2\cos^2 x+a^2\sin^2 x)}

\text{Applying the limits from }0 \text{ to }\dfrac{\pi}{2}:

\: [/tex]

Then

[tex]I = \boxed{\dfrac{(a^2+b^2) \cdot \pi}{4a^3b^3} } \\

I = \displaystyle \int \dfrac{dx}{(a^2\sin^2x + b^2\cos^2 x)^2} = \displaystyle \int \dfrac{dx}{a^4\cos^4 x \left(\tan^2 x + \dfrac{b^2}{a^2}\right)^2} \\

\text{Let }c = \dfrac{b}{a}; \tan x = c\tan \theta \implies \sec^2 x \cdot dx = c\sec^2 \theta \cdot d\theta

I = \displaystyle \int \dfrac{\sec^2 x \cdot \sec^2 x \cdot dx }{a^4 (\tan^2 x + c^2)^2} \\

[/tex]

[tex]= \displaystyle \int \dfrac{(1+c^2\tan^2 \theta) \cdot c\sec^2 \theta \cdot d\theta }{a^4 c^4 \sec^4 \theta} \\

= \dfrac{1}{a^4c^3}\displaystyle \int (1+c^2\tan^2 \theta) \cdot \cos^2 \theta \cdot d\theta \\

\: [/tex]

[tex]= \dfrac{1}{a^4c^3}\left [\displaystyle \int \cos^2 \theta \cdot d\theta + \displaystyle \int c^2\sin^2 \theta \cdot d\theta \right] \\

= \dfrac{1}{2a^4c^3}\left [\displaystyle \int (1+\cos 2\theta) \cdot d\theta + \displaystyle \int c^2(1-\cos 2\theta)\cdot d\theta \right] \\

= \dfrac{1}{2a^4c^3}\left [\theta + \dfrac{\sin 2\theta}{2} + c^2\theta - \dfrac{c^2\sin 2\theta}{2} \right] + I_c \\

= \dfrac{1}{2a^4c^3}\left [\theta + \dfrac{\sin 2\theta}{2} + c^2\theta - \dfrac{c^2\sin 2\theta}{2} \right] + I_c \\

= \dfrac{(1+c^2)\theta}{2a^4c^3} - \dfrac{(c^2-1)\sin 2\theta}{4a^4c^3} + I_c \\

\: \: [/tex]

[tex]c\tan \theta = \tan x \implies \tan \theta = \dfrac{a\tan x}{b}

\implies \sin 2\theta = \dfrac{2\tan \theta}{1+\tan^2 \theta} = \dfrac{\frac{2a\tan x}{b}}{1+\frac{a^2\tan^2 x}{b^2}} = \dfrac{2ab\tan x}{b^2+a^2\tan^2 x} \\

\: [/tex]

[tex]I = \dfrac{\left(1+\frac{b^2}{a^2}\right)\tan^{-1}\left(\frac{a\tan x}{b} \right)}{2a^4\left(\frac{b}{a}\right)^3 } - \dfrac{(\frac{b^2}{a^2}-1)}{4a^4\left(\frac{b}{a}\right)^3}\cdot \dfrac{2ab\tan x}{b^2+a^2\tan^2 x} + I_c \\

= \dfrac{(a^2+b^2)\tan^{-1}\left(\frac{a\tan x}{b} \right)}{2a^3b^3} - \dfrac{(b^2-a^2)\tan x}{2a^2b^2(b^2+a^2\tan^2 x)} + I_c \\

I = \dfrac{(a^2+b^2)\tan^{-1}\left(\frac{a\tan x}{b} \right)}{2a^3b^3} - \dfrac{(b^2-a^2)\tan x}{2a^2b^2(b^2+a^2\tan^2 x)} + I_c \\

= \dfrac{(a^2+b^2)\tan^{-1}\left(\frac{a\tan x}{b} \right)}{2a^3b^3} - \dfrac{(b^2-a^2)\sin 2x}{4a^2b^2(b^2\cos^2 x+a^2\sin^2 x)}

\text{Applying the limits from }0 \text{ to }\dfrac{\pi}{2}: \\ \\

\: [/tex]

So the value of integrtaion will be :

$\bf \: I = \boxed{\dfrac{(a^2+b^2) \cdot \pi}{4a^3b^3} }$