Find `intsqrtx/sqrt(a^3-x^3)dx`

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#### Solution

`I=intsqrtx/sqrt(a^3-x^3)dx`

Let: `x^(3/2)=t`

`=>3/2x^(1/2)dx=dt`

`x^(1/2)dx=2/3dt`

Putting the values in *I*, we get

`I=intsqrtx/sqrt(a^3-x^3)dx`

`=2/3int1/(sqrt(a^3-t^2))dt`

Using the following formula of integration, we get

`intdx/sqrt(a^2-x^2)=sin^(-1)(x/a)`

`:.2/3int1/sqrt(a^3-t^2)dt=2/3sin^(-1)(t/(a^(3/2)))+C`

Again, putting the value of *t, *we get

`2/3int1/sqrt(a^3-t^2)dt=2/3sin^(-1)(t/a^(3/2))+C`

`=2/3sin^(-1)(x^(3/2)/a^(3/2))+C`

Here,* C* is constant of integration.

Is there an error in this question or solution?

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