# What is the integration of x sin inverse x dx ?

## Solution :

We have, I = $$\int$$  $$x sin^{-1} x$$ dx

By using integration by parts formula,

I = $$sin^{-1} x$$ $$x^2\over 2$$ – $$\int$$ $$1\over \sqrt{1 – x^2}$$ $$\times$$ $$x^2\over 2$$ dx

I =  $$x^2\over 2$$ $$sin^{-1} x$$ + $$1\over 2$$ $$\int$$ $$-x^2\over \sqrt{1 – x^2}$$ dx

= $$x^2\over 2$$ $$sin^{-1} x$$ + $$1\over 2$$ $$\int$$ $$1 – x^2 – 1\over \sqrt{1 – x^2}$$  dx

= $$x^2\over 2$$ $$sin^{-1} x$$ + $$1\over 2$$ { $$\int$$ $$1 – x^2\over \sqrt{1 – x^2}$$ – $$\int$$ $$1\over \sqrt{1 -x^2}$$ } dx

$$\implies$$ I = $$x^2\over 2$$ $$sin^{-1} x$$ + $$1\over 2$$ { $$\int$$ $$\sqrt{1 – x^2}$$ – $$\int$$ $$1\over \sqrt{1 -x^2}$$ } dx

By using integration formula of $$\sqrt{a^2 – x^2}$$,

$$\implies$$ I = $$x^2\over 2$$ $$sin^{-1} x$$ + $$1\over 2$$ [{ $$1\over 2$$ $$x\sqrt{1 – x^2}$$ – $$1\over 2$$ $$sin^{-1} x$$ } – $$sin^{-1} x$$ ] + C

$$\implies$$ I = $$x^2\over 2$$ $$sin^{-1} x$$ +  $$1\over 4$$ $$x\sqrt{1 – x^2}$$ – $$1\over 4$$ $$sin^{-1} x$$ + C

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