Question

solve the integral

$\int \: ( \sin ^{ - 1} x) ^{2} dx$
solve only if you know the answer otherwise I will report​

Answer 1

TO FIND :–

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

SOLUTION :–

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

• Let's substitute $\: \: { \bold{x = \sin( \theta) }} \: \: -$

=> Differentiate with respect to 'x' –

$\\ \implies \: { \bold{1 = \cos( \theta) \dfrac{d \theta}{dx} }} \: \:$

$\\ \implies \: { \bold{dx = \cos( \theta) d \theta }} \: \:$

• So that –

$\\ \implies { \bold{I =\int \: ( \theta) ^{2} \cos( \theta) .d \theta }}$

• Now Using integration by parts –

$\\ \implies { \bold{ \int (u.v).dx = u \int \: v.dx - \int \left[ \dfrac{du}{dx} . \int \: v.dx \right].dx}}$

• So that –

$\\ \implies { \bold{I = \: ( \theta) ^{2} \int \cos( \theta) .d \theta - \int \left[ \frac{d( { \theta)}^{2} }{d \theta} . \int \cos( \theta) \right].d \theta}}$

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

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

• Applying again by parts –

$\\ \implies { \bold{I = \: ( \theta) ^{2} \sin( \theta) - 2 \left[ \theta. \int \sin( \theta) - \int \left \{ \dfrac{d \theta}{d \theta} \times \int \sin( \theta) .d \theta \right \} .d \theta \right] }}$

$\\ \implies { \bold{I = \: ( \theta) ^{2} \sin( \theta) - 2 \left[ \theta. ( - \cos( \theta)) - \int \left \{ - \cos( \theta) \right \} .d \theta \right] }}$

$\\ \implies { \bold{I = \: ( \theta) ^{2} \sin( \theta) - 2 \left[ \theta. ( - \cos( \theta)) + \int \left \{ \cos( \theta) \right \} .d \theta \right] }}$

$\\ \implies { \bold{I = \: ( \theta) ^{2} \sin( \theta) - 2 \left[ \theta. ( - \cos( \theta)) + \sin( \theta) \right] + c}}$

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

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

$\\ \: \: { \huge{.}} \: \: { \bold{Now \: \: replace \: \: \theta \: - }}$

$\\ \implies \large{ \boxed{ \bold{I = \: \{ {sin}^{ - 1} (x) \}^{2} (x) + 2 \{ {sin}^{ - 1} (x) \}. \sqrt{1 - {x}^{2} } - 2x + c }}}$

Answer 2

Step-by-step explanation:

TO FIND :–

\begin{gathered}\\ \implies { \bold{I =\int \: ( \sin ^{ - 1} x) ^{2}.dx}} \\\end{gathered}

⟹I=∫(sin

−1

x)

2

.dx

SOLUTION :–

\begin{gathered}\\ \implies { \bold{I =\int \: ( \sin ^{ - 1} x) ^{2}.dx}} \\\end{gathered}

⟹I=∫(sin

−1

x)

2

.dx

• Let's substitute \: \: { \bold{x = \sin( \theta) }} \: \: -x=sin(θ)−

=> Differentiate with respect to 'x' –

\begin{gathered}\\ \implies \: { \bold{1 = \cos( \theta) \dfrac{d \theta}{dx} }} \: \: \\\end{gathered}

⟹1=cos(θ)

dx

dθ

\begin{gathered}\\ \implies \: { \bold{dx = \cos( \theta) d \theta }} \: \: \\\end{gathered}

⟹dx=cos(θ)dθ

• So that –

\begin{gathered}\\ \implies { \bold{I =\int \: ( \theta) ^{2} \cos( \theta) .d \theta }} \\\end{gathered}

⟹I=∫(θ)

2

cos(θ).dθ

• Now Using integration by parts –

\begin{gathered}\\ \implies { \bold{ \int (u.v).dx = u \int \: v.dx - \int [ \dfrac{du}{dx} . \int \: v.dx ].dx}}\\\end{gathered}

⟹∫(u.v).dx=u∫v.dx−∫[

dx

du

.∫v.dx].dx

• So that –

\begin{gathered}\\ \implies { \bold{I = \: ( \theta) ^{2} \int \cos( \theta) .d \theta - \int [ \frac{d( { \theta)}^{2} }{d \theta} . \int \cos( \theta) ].d \theta}}\\\end{gathered}

⟹I=(θ)

2

∫cos(θ).dθ−∫[

dθ

d(θ)

2

.∫cos(θ)].dθ

\begin{gathered}\\ \implies { \bold{I = \: ( \theta) ^{2} \sin( \theta) - \int (2 \theta). ( \sin( \theta)) .d \theta}}\\\end{gathered}

⟹I=(θ)

2

sin(θ)−∫(2θ).(sin(θ)).dθ

\begin{gathered}\\ \implies { \bold{I = \: ( \theta) ^{2} \sin( \theta) - 2 \int (\theta). ( \sin \theta) .d \theta }}\\\end{gathered}

⟹I=(θ)

2

sin(θ)−2∫(θ).(sinθ).dθ

• Applying again by parts –

\begin{gathered}\\ \implies { \bold{I = \: ( \theta) ^{2} \sin( \theta) - 2 [ \theta. \int \sin( \theta) - \int \{ \dfrac{d \theta}{d \theta} \times \int \sin( \theta) .d \theta \} .d \theta ] }}\\\end{gathered}

⟹I=(θ)

2

sin(θ)−2[θ.∫sin(θ)−∫{

dθ

dθ

×∫sin(θ).dθ}.dθ]

\begin{gathered}\\ \implies { \bold{I = \: ( \theta) ^{2} \sin( \theta) - 2 [ \theta. ( - \cos( \theta)) - \int \{ - \cos( \theta) \} .d \theta ] }}\\\end{gathered}

⟹I=(θ)

2

sin(θ)−2[θ.(−cos(θ))−∫{−cos(θ)}.dθ]

\begin{gathered}\\ \implies { \bold{I = \: ( \theta) ^{2} \sin( \theta) - 2 [ \theta. ( - \cos( \theta)) + \int \{ \cos( \theta) \} .d \theta ] }}\\\end{gathered}

⟹I=(θ)

2

sin(θ)−2[θ.(−cos(θ))+∫{cos(θ)}.dθ]

\begin{gathered}\\ \implies { \bold{I = \: ( \theta) ^{2} \sin( \theta) - 2 [ \theta. ( - \cos( \theta)) + \sin( \theta) ] + c}}\\\end{gathered}

⟹I=(θ)

2

sin(θ)−2[θ.(−cos(θ))+sin(θ)]+c

\begin{gathered}\\ \implies { \bold{I = \: ( \theta) ^{2} \sin( \theta) + 2 (\theta). \cos( \theta) - 2 \sin( \theta) + c }}\\\end{gathered}

⟹I=(θ)

2

sin(θ)+2(θ).cos(θ)−2sin(θ)+c

\begin{gathered}\\ \implies { \bold{I = \: ( \theta) ^{2} \sin( \theta) + 2 (\theta). \sqrt{1 - { \sin}^{2} ( \theta)} - 2 \sin( \theta) + c }}\\\end{gathered}

⟹I=(θ)

2

sin(θ)+2(θ).

1−sin

2

(θ)

−2sin(θ)+c

\begin{gathered}\\ \: \: { \huge{.}} \: \: { \bold{Now \: \: replace \: \: \theta \: - }}\\\end{gathered}

.Nowreplaceθ−

\begin{gathered}\\ \implies \large{ \boxed{ \bold{I = \: \{ {sin}^{ - 1} (x) \}^{2} (x) + 2 \{ {sin}^{ - 1} (x) \}. \sqrt{1 - {x}^{2} } - 2x + c }}}\\\end{gathered}

⟹

I={sin

−1

(x)}

2

(x)+2{sin

−1

(x)}.

1−x

2

−2x+c

I hope it helps you...