Question

निम्नलिखित के मान निकालिए-

`int (sin^-1 x)/((1 - x)^(3/2)) "d"x`

Answer 1

मान लीजिए I = `int (sin^-1 x)/((1 - x)^(3/2)) "d"x`

x = sin θ रखिए

⇒ dx = cos θ dθ

I = `int (sin^-1(sin theta))/((1 - sin^2 theta)^(3/2)) * cos theta "d"theta`

= `int (theta * cos theta "d"theta)/((cos^2 theta)^(3/2))`

= `int (theta * cos theta)/(cos^3 theta) "d"theta`

= `int theta/(cos^2 theta) "d"theta`

= `int theta_"I" sec_"II"^2theta "d"theta`

=`theta * sec^2theta "d"theta - int ("D"(theta) * int sec^2theta "d"theta)"d"theta`  .....`["क्योंकि" int "u"_"I" * "v"_"II" "d"x = "u" * int "v" "d"x - int ("D"("u") int "v"  "dv")"dv" + "C"]`

= `theta * tan theta - int 1 * tan theta "d"theta`

= `theta * tan theta - log sec theta + "C"`

= `sin^-1x * x/sqrt(1 - x^2) - log|sqrt(1 - x^2)| + "C"`  ......`[("कब"  x = sin theta),("इसलिए" tan theta = x/sqrt(1 - x^2)  "और" sec theta = sqrt(1 - x^2))]`

अत:, I = `(x sin^-1x)/sqrt(1 - x^2) - log|sqrt(1 - x^2)| + "C"`