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Let \int \frac{x^{1/2}}{\sqrt{1-x^{3}}}dx=\frac{2}{3}gof(x)+C\: \: \: then

  • Option 1)

    f(x) = \sqrt{x} & g(x) = sin-1x

  • Option 2)

    f (x) = x3/2 & g(x) = sin-1x

  • Option 3)

    f(x) = x2/3 & g(x) = cos-1x

  • Option 4)

    g(x) = sin-1x and & f(x) = x2

 

Answers (1)

best_answer

As we learnt

Case for special type of indefinite integration -

\int x^{m}(a+bx^{n})^{p}dx

When P is an integer if P> 0 then apply expanded form

P< 0 then we put x=t^{k}

- wherein

Where k is the common denominator of m and n

 

 

Putting x3/2 = t, we have \sqrt{x} dx = 2dt/3

                              So \int \frac{x^{1/2}}{\sqrt{1-x^{3}}}dx=\frac{2}{3}\int \frac{dt}{\sqrt{1-t^{2}}}=\frac{2}{3}\sin^{-1}t+C=\frac{2}{3}\sin^{-1}x^{3/2}+C

                              Thus f(x) = x3/2 and g(x) = sin-1x.

 


Option 1)

f(x) = \sqrt{x} & g(x) = sin-1x

Option 2)

f (x) = x3/2 & g(x) = sin-1x

Option 3)

f(x) = x2/3 & g(x) = cos-1x

Option 4)

g(x) = sin-1x and & f(x) = x2

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Aadil

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