Question

$\int\limits^1_0 tan⁻¹x dx ,Evaluate the given integral expression:$

Answer 1

HELLO DEAR,


YOUR QUESTIONS IS-------------- $\sf{\int\limits^1_0 {tan^{-1}x \over (1 + x^2)}\,dx}$

let tan^{-1}x = t
=> dx/(1 + x²) = dt
Also, [x = 1 , t = π/4] and [x = 0 , t = 0]

so, I = $\sf{\int\limits^{\pi/4}_0 {t.dt}}$

I = $\sf{[t^2/2]^{\pi/4}_0}$

I = 1/2 * π²/16 - 0

I = π²/32


I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

we have to find the value of $\int\limits^1_0{tan^{-1}x}\,dx$

we have to use integration by part to get indefinite Integration of tan^-1x.
ILATE is the tricky words which helps to get first and 2nd function. here, tan^-1x is the first function and 1 is 2nd function.

we if $\int{f.g}\,dx$ is required to find
where, f is first function and g is 2nd function then , do $f.\int{g}\,dx-\int{f'(\int{g}\,dx)}\,dx$

so, $\int{tan^{-1}x}\,dx=x.tan^{-1}x-\int{\frac{1}{1+x^2}x}\,dx$

$x.tan^{-1}x-\frac{1}{2}\int{\frac{2x}{1+x^2}}\,dx$

$x.tan^{-1}x-\frac{1}{2}log(1+x^2)$

now take the limits ,
so, $[xtan^{-1}x - 1/2 log(1 + x^2)]^1_0$

= π/4 - 1/2 log2 or , π/4 - log√2

hence, answer is π/4 - log√2