Integral: How would you do it?

First write the integral as :

$\displaystyle I = \int_0^{\pi} \ln (9\sin^2 x +16) dx$

$\displaystyle = \frac{1}{2} \int_0^{2\pi} \ln (9\frac{1-\cos x}{2} +16) dx {\color{#3D99F6}{(x\to \frac{x}{2} )}}$

$\displaystyle = \frac{1}{2} \int_{-\pi}^{\pi} \ln (9\frac{1+\cos x}{2} +16) dx {\color{#3D99F6}{(x\to (\pi - x) )}}$

$\displaystyle = \int_{0}^{\pi} \ln (9\frac{1+\cos x}{2} +16) dx$

$\displaystyle = \int_{0}^{\pi} \ln (\frac{41}{2} + \frac{9}{2}\cos x ) dx$

Now using This formula :

$\displaystyle \int_{0}^{\pi} \ln (a+b\cos x ) dx = \pi \ln(\frac{a+\sqrt{a^2-b^2}}{2})$

we obtain :

$\displaystyle \boxed{I= 2\pi \ln (\frac{9}{2})}$

(1/ 2)(3/ 5)^2/ 2 + (1/ 2)(3/ 4)(3/ 5)^4/ 4 + (1/ 2)(3/ 4)(5/ 6)(3/ 5)^6/ 6 + ... = Ln (10/ 9) {You can check with computer.}

Ln (9 Sin^2 x + 16)

= Ln (25 - 9 Cos^2 x)

= 2 Ln 5 + Ln [1 + (3/ 5) Cos x] + Ln [1 - (3/ 5) Cos x]

= 2 Ln 5 - 2 [((3/ 5) Cos x)^2/ 2 + ((3/ 5) Cos x)^4/ 4 + ((3/ 5) Cos x)^6/ 6 + ...]

Integral = 2 Pi Ln 5 - 2 Pi Ln (10/ 9) = 2 Pi Ln (5 x 9/ 10) = 2 Pi Ln (9/ 2) = 9.4503970002856063847339045387971

Note:

Integrate Cos^N x d x from 0 to Pi

= Integrate 2 Cos^N x d x from 0 to Pi/ 2

Integrate Cos^N x d x from 0 to Pi/ 2 = (Pi/ 2)(1/ 2)(3/ 4)(5/ 6)... (N - 1)/ N for even N.

Therefore, 9^2 + 2^2 + 2^2 = 89

[This solution isn't the most correct way however introduce a value to a series.]