Definite Integral: $\int_{0}^{\frac{\pi}{2}}\frac{\cos x}{(a+b\cos x)^2}dx$

[juantheron]

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

 

[chisigma]

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

With the substitution sugggested in...

http://www.mathhelpboards.com/f10/defeinite-integral-2038/#post9364

... You arrive to an integral from 0 to 1 of a rational function...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

With the substitution sugggested in...

http://www.mathhelpboards.com/f10/defeinite-integral-2038/#post9364

... You arrive to an integral from 0 to 1 of a rational function...

Kind regards

$\chi$ $\sigma$

In order to simplify the task You can observe that, setting...

$\displaystyle f(b)= \int_{0}^{\frac{\pi}{2}} \frac{d x}{a + b\ \cos x}$ (1)

... is...

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

Applying to (1) the substitution suggested we arrive to the integral...

$\displaystyle f(b)= 2\ \int_{0}^{1} \frac{1 + t^{2}}{(a+b) + (a-b)\ t^{2}}\ dt$ (3)

... and after some computation [that I strongly reccomand to control (Thinking)...] You obtain...

$\displaystyle f(b)= \frac{2}{a+b} + 4 \frac{b}{a-b}\ \sqrt{\frac{a+b}{a-b}}\ tan^{-1} \sqrt{\frac{a-b}{a+b}}$ (4)

Deriving the (4) is left to You...

Kind regards

$\chi$ $\sigma$