# Differentiation under integral sign

#### Dhamnekar Winod

##### New member
Hello,
How to find $$\displaystyle \displaystyle\int_0^{\pi} \frac{dx}{(5+3*cos(x))^3}$$ and$$\displaystyle \displaystyle\int_0^{\pi}\frac{sin^2(x)}{(5+3*cos(x))^3}$$ by differentiating under the integral sign via Feynman's trick? I am given the general result that $$\displaystyle \displaystyle\int_0^{\pi}\frac{dx}{(a+b*cos(x))}=\frac{\pi}{\sqrt{(a^2-b^2)}}$$

Solution:-
If 0<a<b we have,

$$\displaystyle \frac12\displaystyle\int_0^{2\pi}\frac{d\theta}{(b+a*cos(\theta))^2}d\theta=\frac{\pi b}{(b^2-a^2)^{\frac32})} …(1)$$

By differentiating both sides of (1) with respect to b,

$$\displaystyle \displaystyle\int_0^{2\pi}\frac{d\theta}{(b+a*cos(\theta))^3}=\frac{\pi*(a^2+2*b^2)}{(b^2-a^2)^{\frac52}}…(2)$$

and by differentiating both sides of (1) with respect to a,

$$\displaystyle \displaystyle\int_0^{2\pi}\frac{cos(\theta)d\theta}{(b+a cos(\theta))^3}=\frac{3\pi a *b}{(b^2-a^2)^{\frac52}}…(3)$$

Now, how to proceed further, by using (2) and (3) above , to compute the required integrals?

If any member knows the correct answer, may reply with correct answer. I know the correct answer is $$\displaystyle \frac{59*\pi}{2048} and\frac{\pi}{128}$$ respectively................... edited

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#### Dhamnekar Winod

##### New member
Please read $$\displaystyle \frac{59*\pi}{2048}$$

#### Dhamnekar Winod

##### New member
Hello,
The value of first integral $$\displaystyle \displaystyle\int_0^\pi \frac{d\theta}{(5+3*cos(\theta))^3}=\frac{59*\pi}{2048}$$
The value of integral $$\displaystyle \displaystyle\int_0^\pi \frac{sin^2(\frac{\theta}{2})}{(5+3*cos(\theta))^3}d\theta=\frac{104*\pi}{4096}=\frac{13*\pi}{512}$$

Now how to compute second integral $$\displaystyle \displaystyle\int_0^\pi \frac{sin^2(\theta)}{(5+3*cos(\theta))^3}=?$$