Question

Evaluate!!


Definite integrals..


Please Help!!!!

Answer 1

HELLO DEAR,

$\bold{I = \int\limits^1_0 {\frac{log(1 + x)}{1 + x^2}}\,dx}$

let x= tanA ${\Rightarrow }$ dx = sec²A*dA

Also, [x = 1 ${\Rightarrow }$ A = π/4] and [x = 0 ${\Rightarrow }$ A = 0 ]

$\bold{\therefore, I = \int\limits^{\pi/4}_0 {\frac{tan(1 + tanA)}{1 + tan^2A}*sec^2A}\,dA}$

$\bold{\Rightarrow I = \int\limits^{\pi/4}_0 {\frac{log(1 + tanA)}{sec^2A}*sec^2A}\,dA}$

$\bold{\Rightarrow I = \int\limits^{\pi/4}_0 {log(1 + tanA)}\,dA}$------------( 1 )

using property , $\bold{\int\limits^a_0 f(x) \,dx= \int\limits^a_0 f(a - x)\,dx}$

so, $\bold{\Rightarrow I = \int\limits^{\pi/4}_0 {log[1 + tan(\pi/4 - A)]}\,dA}$

we know, tan(A - B) = [tanA - tanB]/[1 + tanAtanB]

so, tan(π/4 - A) = (1 - tanA)/(1 + tanA)

[as tan(π/4) = 1]

$\bold{\Rightarrow I = \int\limits^{\pi/4}_0 {log[1 + (1 - tanA)/(1 + tanA)]}\,dA}$

$\bold{\Rightarrow I = \int\limits^{\pi/4}_0 {log\frac{2}{1 + tanA}}\,dA}$

$\bold{\Rightarrow I = \int\limits^{\pi/4}_0 [{log2 - log(1 + tanA)}]\,dA}$

$\bold{\Rightarrow I = \int\limits^{\pi/4}_0 {log2}\,dA - \int\limits^{\pi/4}_0{log(1 + tanA)}\,dA}$

$\bold{\Rightarrow 2I = \int\limits^{\pi/4}_0 {log2}\,dA}$------from( 1 )

$\bold{\Rightarrow 2I = log(2)(\frac{\pi}{4} - 0)}$

$\bold{\Rightarrow I = log(2)\frac{\pi}{8}}$

I HOPE ITS HELP YOU DEAR,

THANKS