Question

The derivative of log (sin(logx) (x > 0)​

Answer 1

Answer:

TO FIND :–

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

SOLUTION :–

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

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

=> Differentiate with respect to 'x' –

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

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

• So that –

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

• Now Using integration by parts –

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

• So that –

\begin{gathered}\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}\end{gathered}⟹I=(θ)2∫cos(θ).dθ−∫[dθd(θ)2.∫cos(θ)].dθ

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

\begin{gathered}\begin{gathered}\\ \implies { \bold{I = \: ( \theta) ^{2} \sin( \theta) - 2 \int (\theta). ( \sin \theta) .d \theta }}\\\end{gathered}\end{gathered}⟹I=(θ)2sin(θ)−2∫(θ).(sinθ).dθ

• Applying again by parts –

\begin{gathered}\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}\end{gathered}⟹I=(θ)2sin(θ)−2[θ.∫sin(θ)−∫{dθdθ×∫sin(θ).dθ}.dθ]

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

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

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

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

\begin{gathered}\begin{gathered}\\ \implies { \bold{I = \: ( \theta) ^{2} \sin( \theta) + 2 (\theta). \sqrt{1 - { \sin}^{2} ( \theta)} - 2 \sin( \theta) + c }}\\\end{gathered}\end{gathered}⟹I=(θ)2sin(θ)+2(θ).1−sin2(θ)−2sin(θ)+c

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

\begin{gathered}\begin{gathered}\\ \implies \large{ \boxed{ \bold{I = \: \{ {sin}^{ - 1} (x) \}^{2} (x) + 2 \{ {sin}^{ - 1} (x) \}. \sqrt{1 - {x}^{2} } - 2x + c }}}\\\end{gathered}\end{gathered}⟹I={sin−1(x)}2(x)+2{sin−1(x)}.1−x2−2x+c

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Answer 2

Answer:

TO FIND :–

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

SOLUTION :–

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

$⟹I=∫(sin−1x)2.dx$

• Let's substitute \: \: { \bold{x =

$=> Differentiate with respect to 'x' –$

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

$\end{gathered}⟹1=cos(θ)dxdθ</p><p>\begin{gathered}\begin{gathered}$ $\implies \: { \bold{dx = \cos( \theta) d \theta }} \: \: \\\end{gathered}$

$\end{gathered}⟹dx=cos(θ)dθ</p><p>• So that –$

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

$\end{gathered}⟹I=∫(θ)2cos(θ).dθ</p><p>• Now Using integration by parts –$

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

$⟹∫(u.v).dx=u∫v.dx−∫[dxdu.∫v.dx].dx$

• so that

$⟹I=(θ)2sin(θ)−∫(2θ).(sin(θ)).dθ$

$\sin( \theta) - 2 \int (\theta)$.

$⟹I=(θ)2sin(θ)−2∫(θ).(sinθ)$.dθ