Question

If $\int \sqrt{x}(1-x^3)^{-1/2} dx = \frac{2}{3}g(f(x))+c$, then

• A. $f(x)=\sqrt{x}, g(x)=\sin^{-1} x$
• B. $f(x)=x^{3/2}, g(x)=\sin^{-1} x$
• C. $f(x)=x^{3/2}, g(x)=\cos^{-1} x$
• D. $f(x)=\sqrt{x}, g(x)=\cos^{-1} x$

Hint:

Integration by Substitution and Standard Forms.

• Recognize expressions like $\frac1\sqrt1-u^2$ that suggest $\sin^-1u$.
• Match integrals to forms like $\frac1\sqrta^2 - u^2$, $\frac1a^2 + u^2$.
• Substitution reduces composite expressions to simpler integrals.

Answer 1

The Correct Option is B

Let the integral be $I = \int \sqrt{x}(1-x^3)^{-1/2} dx = \int \frac{\sqrt{x}}{\sqrt{1-x^3}} dx$. Rewrite it as: \[ I = \int \frac{x^{1/2}}{\sqrt{1-(x^{3/2})^2}} dx \] Let $u = x^{3/2} \Rightarrow du = \frac{3}{2}\sqrt{x}dx \Rightarrow \sqrt{x}dx = \frac{2}{3}du$. Substituting: \[ I = \int \frac{1}{\sqrt{1-u^2}} \cdot \frac{2}{3} du = \frac{2}{3}\sin^{-1}(u) + c = \frac{2}{3}\sin^{-1}(x^{3/2}) + c \] Comparing with $\frac{2}{3}g(f(x)) + c$, we identify $f(x) = x^{3/2},\ g(x) = \sin^{-1}x$.

Answer 2

Step 1: Understand the given integral
We have:
\[ \int \sqrt{x} (1 - x^3)^{-1/2} \, dx = \frac{2}{3} g(f(x)) + c \]
We need to identify the functions \(f(x)\) and \(g(x)\).

Step 2: Use substitution
Let:
\[ t = x^{3/2} \]
Then differentiate \(t\) with respect to \(x\):
\[ \frac{dt}{dx} = \frac{3}{2} x^{1/2} \implies dt = \frac{3}{2} x^{1/2} dx \]
Rearranging:
\[ x^{1/2} dx = \frac{2}{3} dt \]

Step 3: Rewrite the integral in terms of \(t\)
Substitute \(x^{1/2} dx\) and \(1 - x^3 = 1 - t^2\):
\[ \int \sqrt{x} (1 - x^3)^{-1/2} dx = \int (1 - t^2)^{-1/2} \frac{2}{3} dt = \frac{2}{3} \int (1 - t^2)^{-1/2} dt \]

Step 4: Recognize the integral form
\[ \int (1 - t^2)^{-1/2} dt = \sin^{-1} t + c \]

Step 5: Write the solution back in terms of \(x\)
\[ \int \sqrt{x} (1 - x^3)^{-1/2} dx = \frac{2}{3} \sin^{-1} (x^{3/2}) + c \]
Thus:
\[ f(x) = x^{3/2}, \quad g(x) = \sin^{-1} x \]

Final answer:
\[ \boxed{f(x) = x^{3/2}, \quad g(x) = \sin^{-1} x} \]