Question

Compute the integral 

$F(x)=\int_0^\frac{\pi}{2}ln(cos^2(x)+t^2sin^2(x))dx , (t>0)$

Answer 1

Compute the integral.

$F(t)=\int_0^\frac{\pi}{2}ln(cos^2(x)+t^2sin^2(x))dx , (t>0)$

$=>F'(t)=\int_0^\frac{\pi}{2}\frac{2tsin^2(x)}{cos^2(x)+t^2sin^2(x)}dx =\int_0^\frac{\pi}{2}\frac{2t.tan^2x\frac{1}{cos^2x}}{1+t^2tan^2x\frac{1}{cos^2x}}dx$

Let $u=tanx=>du=\frac{1}{cos^2x}dx$ . then $x=0=>u->0, x=\frac{\pi}{2}=>u->\infty$

$=>F'(t)=\int_0^\infty \frac{2t.u^2}{(1+t^2u^2)(1+u^2)}du$

we have $\frac{2tu^2}{(1+t^2u^2)(1+u^2)}=(\frac{1}{1+(tu)^2}-\frac{1}{1+u^2}).\frac{2t}{1-t^2}$

$=>F'(t) =\frac{2t}{1-t^2}\int_0^\infty (\frac{1}{1+t^2u^2}-\frac{1}{1+u^2})du$

                  $= \frac{2t}{1-t^2}[\frac{1}{t}arctan(tu)-arctan(u)]\left.\right|_0^\infty$

                  $=\frac{2t}{1-t^2}[\frac{\pi}{2}(\frac{1}{t}-1)]=\frac{\pi}{1+t}$

$=>F(t)=\int\frac{\pi}{1+t}dt=\pi ln(1+t)+c$  , (*)

we have $F(1)=0$ but (*) $F(1)=\pi ln2+c$

$=>\pi ln2+c=0=>c=-\pi ln2$

Therefore, $F(t)=\pi ln(1+t)-\pi ln2=\pi ln(\frac{1+t}{2})$

 

Therefore, $F(t)=\pi ln(1+t)-\pi ln2=\pi ln(\frac{1+t}{2})$