Question

`int_0^1 x (tan^-1 x)^2 "d"x` का मान ज्ञात कीजिए।

Answer 1

I = `int_0^1 x (tan^-1 x)^2 "d"x` 

समाकलन द्वारा हमें प्राप्त होता है:

I = `x^2 [(tan^-1 x)^2]_0^1 - 1/2 int_0^1 x^2 * 2 (tan^-1 x)/(1 + x^2)  "d"x`

= `pi^2/32 - int_0^1 x^2/(1 + x^2) * tan^-1 x"d"x`

= `pi^2/32 - "I"_1`, जहाँ I₁ = `int_0^1 x^2/(1 + x^2) tan^-1 x"d"x` है।

अब I₁ = `int_0^1 (x^2 + 1 - 1)/(1 + x^2) tan^-1x "d"x`

= `int_0^1 tan^-1 x"d"x - int_0^1 1/(1 + x^2) tan^-1 x"d"x`

= `"I"_2 - 1/2 ((tan^-1x)^2)_0^1`

= `"I"_2 - pi^2/32`

यहाँ I₂ = `int_0^1 tan^-1 x"d"x = (x tan^-1x)_0^1 - int_0^1 x/(1 + x^2)  "d"x`

= `pi/4 - 1/2(log|1 + x^2|)_0^1`

= `pi/4 - 1/2 log2`

इस प्रकार,  I₁ = `pi/4 - 1/2 log 2 - pi^2/32`

अत:, I = `pi^2/32 - pi/4 + 1/2 log2 + pi^2/32`

= `pi^2/16 - pi/4 + 1/2 log2`

= `(pi^2 - 4pi)/16 + log sqrt(2)`