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प्रश्न
Integrate the following with respect to x:
`tan^-1 ((8x)/(1 - 16x^2))`
उत्तर
`int tan^-1 ((8x)/(1 - 16x^2)) "d"x = int tan^-1 ((2 xx 4x)/(1 - (4x)^2)) "d"x`
Put 4x = tan θ
4 dx = sec2θ dθ
`int tan^-1 ((8x)/(1 - 16x^2)) "d"x = inttan^-1 ((2 tan theta)/(1 - tan^2theta)) (sec^2theta)/4 "d"theta`
= `1/4 int tan^-1 (tan 2theta) sec^2 theta "d"theta`
= `1/4 int 2theta sin^2theta "d"theta`
= `1/2 int theta sec^2theta "d"theta` .........(1)
Consider `int theta sec^2theta "d"theta`
u = θ
u' = 1
u" = 0
dv = `sin^2theta "d"theta`
⇒ v = `int tan theta "d"theta`
⇒ v = tan θ
v1 = `int "v" "d"theta`
= `int tan theta "d"theta`
= `log |sec theta|`
v2 = `int "v"_1 "d"theta`
= `int log |sec theta| "d"theta`
`int "u" "dv"` = uv – u'v1 + u"v2 – .............
`int theta sec^2theta "d"theta = theta tan theta - 1 xx log |sec theta| + 0 xx int log |sec theta| "d"theta + "c"`
`int theta sec^2theta "d"theta = theta tan theta - log |sec theta| + "c"`
Substituting in eqquation (1) we get
`int tan^-1 ((8x)/(1 - 16x^2)) "d"x = 1/2 [theta tan theta - log |sec theta|] + "c"` ........(2)
4x = tan θ
⇒ θ = `tan^-1 (4x)`
sec θ = `sqrt(1 + tan^2theta)`
= `sqrt(1 + (4x)^2`
= `sqrt(1 + 16x^2)`
Substituting in equation (2) we get
`int tan^-1 ((8x)/(1 - 16x^2)) "d"x = 1/2 [tan^-1 (4x) xx 4x - log |sqrt(1 + 16x^2)|] + "c"`
= `1/2 [4x tan^-1 (4x) - log |sqrt(1 + 16x^2)|] + "c"`
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