Porisma

Worked solution

Use a shown antiderivative inside integration by parts

(i) Show that ∫x1−x2 dx=−1−x2+c\displaystyle\int\dfrac{x}{\sqrt{1-x^2}}\,dx=-\sqrt{1-x^2}+c, where cc is an arbitrary constant. marks

(ii) Hence find ∫(sin⁡−1x)x1−x2 dx\displaystyle\int(\sin^{-1}x)\dfrac{x}{\sqrt{1-x^2}}\,dx. marks

[2016/JJC/Promo/9(b)]

CJC · Review Problems · Question 5

Answer

(i) −1−x2+c-\sqrt{1-x^2}+c; (ii) −(sin⁡−1x)1−x2+x+C-(\sin^{-1}x)\sqrt{1-x^2}+x+C.

(i)

Write the integral as −12∫(−2x)(1−x2)−1/2dx=−1−x2+c-\frac12\int(-2x)(1-x^2)^{-1/2}dx=-\sqrt{1-x^2}+c, as required.

How did this part go?

(ii)

Integrate by parts with u=sin⁡−1xu=\sin^{-1}x and dv=x/1−x2 dxdv=x/\sqrt{1-x^2}\,dx. Part (i) gives v=−1−x2v=-\sqrt{1-x^2} and du=dx/1−x2du=dx/\sqrt{1-x^2}. Thus the result is −(sin⁡−1x)1−x2+∫1 dx=−(sin⁡−1x)1−x2+x+C-(\sin^{-1}x)\sqrt{1-x^2}+\int1\,dx=-(\sin^{-1}x)\sqrt{1-x^2}+x+C.

How did this part go?

Common mistakes

The minus sign in v changes the remaining parts integral into +∫1 dx.

Read the hint