What is the integration of sin inverse x whole square ?

Solution :

We have, I = $$(sin^{-1}x)^2$$ dx

Let $$sin^{-1}x$$ = t, Then, x = sin t $$\implies$$ dx = cos t dt

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

I = $$\int$$ $$t^2$$ cos t dt

Applying integration by parts and,

Taking $$t^2$$ as first function and cos t as second function,

I = $$t^2$$ (sin t) – $$\int$$ (sin t) – $$\int$$ 2t sin t dt = $$t^2$$ sin t – 2 $$\int$$ t sin t dt

Again applying integration by parts and,

Taking t as first function and sin t as second function,

I = $$t^2$$ sin t – 2{ t ( -cos t) – $$\int$$ 1 $$\times$$ (-cos t) dt }

I = $$t^2$$ sin t – 2{ -t cos t + $$\int$$ cos t dt }

I = $$t^2$$ sin t – 2( -t cos t + sin t ) + C

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

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

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