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Find the integral integral xsqrt(1-x2)\,dx and select the correct option.

Mathematics

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Answer

B: −13(1−x2)3/2+c\text{B: } -\frac{1}{3}(1-x^2)^{3/2} + c

Step-by-step solution

5 steps

To find the integral ∫x1−x2 dx\int x\sqrt{1-x^2}\,dx, we will use a substitution method.

  1. Set up the substitution.

    Let u=1−x2u = 1-x^2.

  2. Find the differential dudu.

    Differentiate uu with respect to xx:

    dudx=−2x\frac{du}{dx} = -2x

    Rearrange to find x dxx\,dx:

    du=−2x dxdu = -2x\,dx x dx=−12 dux\,dx = -\frac{1}{2}\,du
  3. Substitute into the integral.

    Replace 1−x21-x^2 with uu and x dxx\,dx with −12 du-\frac{1}{2}\,du:

    ∫x1−x2 dx=∫u(−12) du\int x\sqrt{1-x^2}\,dx = \int \sqrt{u} \left(-\frac{1}{2}\right)\,du =−12∫u1/2 du= -\frac{1}{2} \int u^{1/2}\,du
  4. Integrate with respect to uu.

    Use the power rule for integration, ∫un du=un+1n+1+C\int u^n\,du = \frac{u^{n+1}}{n+1} + C:

    −12∫u1/2 du=−12(u1/2+11/2+1)+C-\frac{1}{2} \int u^{1/2}\,du = -\frac{1}{2} \left( \frac{u^{1/2+1}}{1/2+1} \right) + C =−12(u3/23/2)+C= -\frac{1}{2} \left( \frac{u^{3/2}}{3/2} \right) + C =−12(23u3/2)+C= -\frac{1}{2} \left( \frac{2}{3}u^{3/2} \right) + C =−13u3/2+C= -\frac{1}{3}u^{3/2} + C
  5. Substitute back xx for uu.

    Replace uu with 1−x21-x^2:

    −13(1−x2)3/2+C-\frac{1}{3}(1-x^2)^{3/2} + C

    Comparing this result with the given options, it matches option B.

    The final answer is B: −13(1−x2)3/2+c\text{B: } -\frac{1}{3}(1-x^2)^{3/2} + c.

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