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प्रश्न
Evaluate the following definite integrals:
`int_0^1 (1 - x^2)/(1 + x^2)^2 "d"x`
उत्तर
Let I = `int_0^1 (1 - x^2)/(1 + x^2)^2 "d"x`
= `int_0^1[2/(1 + x^2)^2 - (1 + x^2)/(1 + x^2)^2]"d"x`
I = `int_0^1 [2/(1 + x^2)^2 - 1/((1 + x^2))]"d"x` ........(1)
I1 = `int_0^1 2/(1 + x^2)^2 "d"x`
Put x = tan θ
dx = sec2θ dθ
| x | 0 | 1 |
| θ | 0 | `pi/4` |
= `2 int_0^(pi/4) (sec^2theta "d"theta)/(1 + tan^2theta)^2`
= `2 int_0^(pi/4) (sec^2theta)/(sec^2theta)^2 "d"theta`
= `2 int_0^(pi/4) cos^2theta "d"theta`
= `2 int_0^(pi/4) ((1 + cos 2theta)/2) "d"theta`
= `(theta + (sin 2theta)/2)_0^(pi/4)`
= `pi/4 + 1/2`
I2 = `int_0^1 1/(1 + x^2) "d"x`
= `[tan^-1 x]_0^1`
= `pi/4`
(1) ⇒ I = `pi/4 + 1/2 - pi/4`
I = `1/2`
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