Question

Solve the Integral
$\int{(\sin^{-1}x)^{2}} \ dx$

Answer 1

GIVEN:

$\implies \tt \: \int \: ( \sin ^{ - 1} x) \: dx \:$

TO FIND:

$\tt \: Integrate ..\: \int \: (\sin^{ - 1} x) \: dx$

SOLUTION:

$\tt : \implies I = \int {({sin}^{ - 1} x)}^{2} x \: dx.$

Now, Substitute x = sin θ.

Differentiate with respect to ' x '

$\tt : \implies 1= cos (\theta) \frac{d \theta}{dx}$

$\tt : \implies dx= cos (\theta) d \theta$

$\tt \therefore I = \int { (\theta)}^{2} \times cos (\theta )\: d \theta$

Now, Integration by part.

$\tt : \implies \int (uv)dx = u \int v \: dx - \int\Bigg[\dfrac{du}{dx} \times \int v \: dx \Bigg]dx$

So, that.

$\tt : \implies I = {( \theta)}^{2} \int cos( \theta)d \theta - \int\Bigg[\dfrac{d { (\theta)}^{2} }{d \theta} \times \int cos \: d \theta\Bigg]d \theta$

$\tt : \implies I = {( \theta)}^{2} \times \sin( \theta )- \int((2 \theta)(sin \theta))d \theta$

$\tt : \implies I = {( \theta)}^{2} \times \sin( \theta )- 2 \int(\theta)(sin \theta)d \theta$

$\tt Applying again by part</p><p>\tt : \implies I = {( \theta)}^{2} \times \sin( \theta )- 2 \Bigg[] \: \int(\theta)(sin \theta)d \theta$

$\tt : \implies I = {( \theta)}^{2} \times \sin( \theta )- 2 \Bigg[ \theta \{sin (\theta) - \int \{ \frac{ d \theta}{d \theta} \times \int sin (\theta)d \theta \} \} ]$

$\tt : \implies I = {( \theta)}^{2} \times \sin( \theta )- 2 \Bigg[ \theta (cos (\theta)) - \int ( - cos (\theta))\Bigg]$

$\tt : \implies I = {( \theta)}^{2} \times \sin( \theta )- 2 \Bigg[ \theta (cos (\theta)) + \int ( cos (\theta))\Bigg]$

$\tt : \implies I = {( \theta)}^{2} \times \sin( \theta )- 2 \Bigg[ \theta ( - cos (\theta)) + sin (\theta)\Bigg] + c$

$\tt : \implies I = {( \theta)}^{2} \times \sin( \theta ) + 2\theta (cos (\theta)) - 2 sin (\theta) + c$

$\tt : \implies I = {( \theta)}^{2} \times \sin( \theta ) + 2\theta \sqrt{1 - {sin}^{2} ( \theta)} - 2 sin (\theta) + c$

Now, Replace ( θ )

$\tt {I = \{{ {sin}^{ - 1} x \}}^{2} (x) + 2\{{ {sin}^{ - 1} x \}} \times \sqrt{1 - x ^{2} } - 2x + c.}$

Answer 2

Answer

$\tt \int (sin^{-1}x)^2dx$      

Let x = sinθ                          

⇒ dx = cosθ dθ

$\implies \tt \int (sin^{-1}(sin\theta))^2cos\theta d\theta\\\\\implies \tt \int \theta^2cos\theta d \theta$

Now , Apply ILATE rule , we get ,

$\bf So,u= \theta ^2,dv=cos\theta d \theta\\\\\implies \bf du=2\theta d \theta , \int dv=\int cos\theta d \theta\\\\\implies \bf du=2\theta d\theta , v=sin\theta$

$\implies \tt \theta ^2.sin\theta-\int sin\theta .2\theta d \theta\\\\\implies \tt \theta^2.sin\theta - 2 \int \theta .sin \theta d\theta$

Now once again apply ILATE rule , we get ,

$\bf u =\theta , dv=sin\theta d \theta\\\\\implies \bf du=d \theta , v =-cos \theta$

$\implies \tt \theta^2.sin \theta - 2 [-\theta.cos \theta - \int -cos \theta d \theta]\\\\\implies \tt \theta^2.sin \theta +2 \theta cos \theta - 2sin \theta\\\\\implies \tt (sin^{-1}x)^2.x+2.sin^{-1}x.\sqrt{1-x^2}-2x+c$

So ,

$\bf {\blue{\underline{\int (sin^{-1}x)^2dx=(sin^{-1}x)^2.x+2.sin^{-1}x.\sqrt{1-x^2}-2x+c}}}$