# What is the integration of sin inverse root x ?

## Solution :

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

By Applying integration by parts,

Taking $$sin^{-1}\sqrt{x}$$ as first function and 1 as second function. Then

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

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

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

Put x = $$sin^2 t$$

dx = 2 sin t cos t dt

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

$$\implies$$ I = x$$sin^{-1}\sqrt{x}$$  – { $$1\over 2$$ $$\int$$ (1 – cos 2t) dt }

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

Now, sin 2t = 2 sin t cos t = $$2\sqrt{x} \sqrt{1 – x}$$ = $$2\sqrt{x – x^2}$$

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

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

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