Question

find value of t in
m1c1 (t1 - t) = m2 c2 (t - t2) pleasee​

Answer 1

Step-by-step explanation:

TO FIND :–

$\begin{gathered}\begin{gathered}\\ \implies { \bold{I =\int \: ( \sin ^{ - 1} x) ^{2}.dx}} \\\end{gathered}\end{gathered} </p><p>⟹I=∫(sin </p><p>−1</p><p> x) </p><p>2</p><p> .dx</p><p> </p><p> </p><p> </p><p> </p><p></p><p>SOLUTION :–</p><p></p><p>\begin{gathered}\begin{gathered}\\ \implies { \bold{I =\int \: ( \sin ^{ - 1} x) ^{2}.dx}} \\\end{gathered}\end{gathered} </p><p>⟹I=∫(sin </p><p>−1</p><p> x) </p><p>2</p><p> .dx</p><p> </p><p> </p><p> </p><p> </p><p></p><p>• Let's substitute \: \: { \bold{x = \sin( \theta) }} \: \: -x=sin(θ)−</p><p></p><p>=> Differentiate with respect to 'x' –</p><p></p><p>\begin{gathered}\begin{gathered}\\ \implies \: { \bold{1 = \cos( \theta) \dfrac{d \theta}{dx} }} \: \: \\\end{gathered}\end{gathered} </p><p>⟹1=cos(θ) </p><p>dx</p><p>dθ</p><p> </p><p> </p><p> </p><p> </p><p> </p><p> </p><p></p><p>\begin{gathered}\begin{gathered}\\ \implies \: { \bold{dx = \cos( \theta) d \theta }} \: \: \\\end{gathered}\end{gathered} </p><p>⟹dx=cos(θ)dθ</p><p> </p><p> </p><p> </p><p> </p><p></p><p>• So that –</p><p></p><p>\begin{gathered}\begin{gathered}\\ \implies { \bold{I =\int \: ( \theta) ^{2} \cos( \theta) .d \theta }} \\\end{gathered}\end{gathered} </p><p>⟹I=∫(θ) </p><p>2</p><p> cos(θ).dθ</p><p> </p><p> </p><p> </p><p> </p><p></p><p>• Now Using integration by parts –</p><p></p><p>\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} </p><p>⟹∫(u.v).dx=u∫v.dx−∫[ </p><p>dx</p><p>du</p><p> </p><p> .∫v.dx].dx</p><p> </p><p> </p><p> </p><p> </p><p></p><p>• So that –</p><p></p><p>\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} </p><p>⟹I=(θ) </p><p>2</p><p> ∫cos(θ).dθ−∫[ </p><p>dθ</p><p>d(θ) </p><p>2</p><p> </p><p> </p><p> .∫cos(θ)].dθ</p><p> </p><p> </p><p> </p><p> </p><p></p><p>\begin{gathered}\begin{gathered}\\ \implies { \bold{I = \: ( \theta) ^{2} \sin( \theta) - \int (2 \theta). ( \sin( \theta)) .d \theta}}\\\end{gathered}\end{gathered} </p><p>⟹I=(θ) </p><p>2</p><p> sin(θ)−∫(2θ).(sin(θ)).dθ</p><p> </p><p> </p><p> </p><p> </p><p></p><p>\begin{gathered}\begin{gathered}\\ \implies { \bold{I = \: ( \theta) ^{2} \sin( \theta) - 2 \int (\theta). ( \sin \theta) .d \theta }}\\\end{gathered}\end{gathered} </p><p>⟹I=(θ) </p><p>2</p><p> sin(θ)−2∫(θ).(sinθ).dθ</p><p> </p><p> </p><p> </p><p> </p><p></p><p>• Applying again by parts –</p><p></p><p>\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} </p><p>⟹I=(θ) </p><p>2</p><p> sin(θ)−2[θ.∫sin(θ)−∫{ </p><p>dθ</p><p>dθ</p><p> </p><p> ×∫sin(θ).dθ}.dθ]</p><p> </p><p> </p><p> </p><p> </p><p></p><p>\begin{gathered}\begin{gathered}\\ \implies { \bold{I = \: ( \theta) ^{2} \sin( \theta) - 2 [ \theta. ( - \cos( \theta)) - \int \{ - \cos( \theta) \} .d \theta ] }}\\\end{gathered}\end{gathered} </p><p>⟹I=(θ) </p><p>2</p><p> sin(θ)−2[θ.(−cos(θ))−∫{−cos(θ)}.dθ]</p><p> </p><p> </p><p> </p><p> </p><p></p><p>\begin{gathered}\begin{gathered}\\ \implies { \bold{I = \: ( \theta) ^{2} \sin( \theta) - 2 [ \theta. ( - \cos( \theta)) + \int \{ \cos( \theta) \} .d \theta ] }}\\\end{gathered}\end{gathered} </p><p>⟹I=(θ) </p><p>2</p><p> sin(θ)−2[θ.(−cos(θ))+∫{cos(θ)}.dθ]</p><p> </p><p> </p><p> </p><p> </p><p></p><p>\begin{gathered}\begin{gathered}\\ \implies { \bold{I = \: ( \theta) ^{2} \sin( \theta) - 2 [ \theta. ( - \cos( \theta)) + \sin( \theta) ] + c}}\\\end{gathered}\end{gathered} </p><p>⟹I=(θ) </p><p>2</p><p> sin(θ)−2[θ.(−cos(θ))+sin(θ)]+c</p><p> </p><p> </p><p> </p><p> </p><p></p><p>\begin{gathered}\begin{gathered}\\ \implies { \bold{I = \: ( \theta) ^{2} \sin( \theta) + 2 (\theta). \cos( \theta) - 2 \sin( \theta) + c }}\\\end{gathered}\end{gathered} </p><p>⟹I=(θ) </p><p>2</p><p> sin(θ)+2(θ).cos(θ)−2sin(θ)+c</p><p> </p><p> </p><p> </p><p> </p><p></p><p>\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} </p><p>⟹I=(θ) </p><p>2</p><p> sin(θ)+2(θ). </p><p>1−sin </p><p>2</p><p> (θ)</p><p> </p><p> −2sin(θ)+c</p><p> </p><p> </p><p> </p><p> </p><p></p><p>\begin{gathered}\begin{gathered}\\ \: \: { \huge{.}} \: \: { \bold{Now \: \: replace \: \: \theta \: - }}\\\end{gathered}\end{gathered} </p><p>.Nowreplaceθ−</p><p> </p><p> </p><p> </p><p> </p><p></p><p>\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} </p><p>⟹ </p><p>I={sin </p><p>−1</p><p> (x)} </p><p>2</p><p> (x)+2{sin </p><p>−1</p><p> (x)}. </p><p>1−x </p><p>2</p><p> </p><p> </p><p> −2x+c</p><p> </p><p> </p><p> </p><p> </p><p> </p><p> </p><p></p><p>\huge\boxed{\fcolorbox{black}{green}{HOPE\: IT \: HELPS}}$

HOPEIT HELPS