मराठी

Evaluate the following: d∫sin-1x(1-x)32dx

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प्रश्न

Evaluate the following:

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

बेरीज
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उत्तर

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

Put x = sin θ

⇒ 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`  .....`[because 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"`  ......`[("When"  x = sin theta),(therefore tan theta = x/sqrt(1 - x^2) "and" sec theta = sqrt(1 - x^2))]`

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

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पाठ 7: Integrals - Exercise [पृष्ठ १६४]

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