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Worked solution

Integrate inverse sine using a known derivative

(i) Show that ddx1−x2=−x1−x2\dfrac d{dx}\sqrt{1-x^2}=\dfrac{-x}{\sqrt{1-x^2}}, ∣x∣<1|x|<1.

(ii) Hence find ∫sin⁡−1x dx\displaystyle\int\sin^{-1}x\,dx.

CJC · Discussion · Question 9

Answer

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

(i)

Chain rule gives 12(1−x2)−1/2(−2x)=−x/1−x2\dfrac12(1-x^2)^{-1/2}(-2x)=-x/\sqrt{1-x^2}, as required.

How did this part go?

(ii)

Use parts with u=arcsin x, dv/dx=1, du/dx=1/√(1−x²) and v=x. Then ∫sin⁡−1x dx=xsin⁡−1x−∫x/1−x2 dx\int\sin^{-1}x\,dx=x\sin^{-1}x-\int x/\sqrt{1-x^2}\,dx. Part (i) identifies the latter integral as −√(1−x²). Therefore the answer is xsin⁡−1x+1−x2+Cx\sin^{-1}x+\sqrt{1-x^2}+C, initially on −1<x<1.

How did this part go?

Common mistakes

Subtracting an integral whose antiderivative is negative produces the positive square-root term.

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