Question

$\\ \displaystyle\Huge\int\sf\tiny\dfrac{\Big\{cos^{-1}x\big[\sqrt{(1-x^2)}\:\big]\Big\}^{-1}}{log_{e}\left\{1+\left[\dfrac{sin\left\lgroup 2x\sqrt{(1-x^2)}\right\rgroup}{\boldsymbol\pi}\right]\right\}}dx$

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Answer 1

Answer :

$\sf{\displaystyle\frac{2}{\pi} \log_{e}\left(\log _{e}\left(\frac{\pi+2 \theta}{\pi}\right)\right)}$

Step by Step Explanation :

$\sf{\int \dfrac{\left\{\cos ^{-1} x\left[\sqrt{\left(1-x^{2}\right)}\right]\right\}^{-1}}{\log_e \left\{1+\left[\frac{\sin \left(2 x \sqrt{\left(1-x^{2}\right)}\right)}{\pi}\right]\right)} d x}$

➳ Put x = sinθ

∴ dx = cosθ dθ

$\therefore\sf{cos^{-1}x\cdot \left\{\sqrt{1-x^2}\right\}=cos^{-1}\left(sin\theta \right)\cdot \left\{\sqrt{1-x^2}\right\} }$

$\to\sf{cos^{-1}x\cdot \left\{\sqrt{1-x^2}\right\}=cos^{-1}\left(sin\theta \right)\cdot \left\{\sqrt{1-\sin ^2\theta}\right\} }$

$\to\sf{cos^{-1}x\cdot \left\{\sqrt{1-x^2}\right\}=cos^{-1}\left(sin\theta \right)\cdot \left\{\sqrt{\cos ^2\left(\theta\right)}}\right\} }$

$\to\sf{cos^{-1}x\cdot \left\{\sqrt{1-x^2}\right\}=cos^{-1}\left(sin\theta \right)\cdot \cos \left(\theta\right) }$

$\to\sf{cos^{-1}x\cdot \left\{\sqrt{1-x^2}\right\}=\cos \:^{-1}\left(\cos \:\left(\dfrac{\pi }{2}+\theta \:\right)\right)\:\cdot \:\cos \theta }$

$\to\sf{cos^{-1}x\cdot \left\{\sqrt{1-x^2}\right\}=\:\left(\dfrac{\pi \:}{2}+\theta \:\right)\:\cdot \:\cos \theta }$

Also

$\sf{\sin ^{-1}\left(2 x \sqrt{1-2 x^{2}}\right)=\sin ^{-1}(2 \sin \theta \cos \theta)}$

$\to\sf{\sin ^{-1}\left(2 x \sqrt{1-2 x^{2}}\right)=sin^{-1}\left(sin2\theta \right)}$

$\to\sf{\sin ^{-1}\left(2 x \sqrt{1-2 x^{2}}\right)=2\theta}$

Now

$\sf{\int \dfrac{\left\{\cos ^{-1} x\left[\sqrt{\left(1-x^{2}\right)}\right]\right\}^{-1}}{\log_e \left\{1+\left[\frac{\sin \left(2 x \sqrt{\left(1-x^{2}\right)}\right)}{\pi}\right]\right)} d x}$

$\sf{=\displaystyle\int \frac{\left(\cos ^{-1} \sin \theta \sqrt{1-\sin ^{2} \theta}\right)^{-1}}{log_e \left(1+\frac{2 \theta}{\pi}\right)} \cos \theta d \theta}$

$\sf{\displaystyle=\int\frac{\left(\frac{\pi }{2}+\theta\right)^{-1}\:\cdot \:\cos \:^{-1}\:\theta \:\cdot \:\cos \:\theta \:d\:\theta }{\log _e\left(\frac{\pi +2\:\theta }{\pi }\right)}}$

$\sf{\displaystyle=\int \:\frac{2d\theta }{\left(\pi +2\theta \right)\:\log_e\left(\frac{\pi +2\theta }{\pi }\right)}}$

➳ Put  $\sf{log\left(\dfrac{\pi +2\theta }{\pi }\right)=t}$

$\to\sf{\dfrac{\pi}{\pi+2 \theta} d \theta=d t}$

$\to\sf{\dfrac{2}{\pi} \int \dfrac{d t}{t}}$

$\to\sf{\dfrac{2}{\pi} \log (t)}$

➳Substitute back the Value of t

$\to\sf{\displaystyle\frac{2}{\pi} \log_{e}\left(\log _{e}\left(\frac{\pi+2 \theta}{\pi}\right)\right)}$

$\boxed{\boxed{\therefore\:\:\sf{\int \dfrac{\left\{\cos ^{-1} x\left[\sqrt{\left(1-x^{2}\right)}\right]\right\}^{-1}}{\log_e \left\{1+\left[\frac{\sin \left(2 x \sqrt{\left(1-x^{2}\right)}\right)}{\pi}\right]\right)} d x}=\sf{\displaystyle\frac{2}{\pi} \log_{e}\left(\log _{e}\left(\frac{\pi+2 \theta}{\pi}\right)\right)}}}$

Answer 2

SOLUTION

TO EVALUATE

$\displaystyle \sf{ \int\limits_{}^{} \frac{{ \bigg[ { \cos}^{ - 1}x( \sqrt{1 - {x}^{2} } ) \bigg]}^{ - 1} }{ \log \: \{ \: 1 + \frac{ { \sin}^{ - 1} (2x \sqrt{1 - {x}^{2} } )}{\pi} \}} \, dx }$

EVALUATION

Let I be the given Integral

Then

$\displaystyle \sf{I = \int\limits_{}^{} \frac{{ \bigg[ { \cos}^{ - 1}x( \sqrt{1 - {x}^{2} } ) \bigg]}^{ - 1} }{ \log \: \{ \: 1 + \frac{ { \sin}^{ - 1} (2x \sqrt{1 - {x}^{2} } )}{\pi} \}} \, dx }$

Let x = sin θ

then dx = cos θ dθ

Then the given Integral becomes

$\displaystyle \sf{I = \int\limits_{}^{} \frac{{ \bigg[ { \cos}^{ - 1}x( \sqrt{1 - {x}^{2} } ) \bigg]}^{ - 1} }{ \log \: \{ \: 1 + \frac{ { \sin}^{ - 1} (2x \sqrt{1 - {x}^{2} } )}{\pi} \}} \, dx }$

$\displaystyle \sf{ \implies \: I = \int\limits_{}^{} \frac{{ \bigg[ { \cos}^{ - 1}( \sin \theta)( \sqrt{1 - {\sin }^{2} \theta } ) \bigg]}^{ - 1} }{ \log \: \{ \: 1 + \frac{ { \sin}^{ - 1} (2\sin \theta \sqrt{1 - { \sin}^{2} \theta } )}{\pi} \}} \, \cos \theta d \theta }$

$\displaystyle \sf{ \implies \: I = \int\limits_{}^{} \frac{{ \bigg[ { \cos}^{ - 1}( \cos \bigg( \frac{\pi}{2} + \theta \bigg)( {\cos }^{} \theta ) \bigg]}^{ - 1} }{ \log \: \{ \: 1 + \frac{ { \sin}^{ - 1} (2\sin \theta \cos\theta )}{\pi} \}} \, \cos \theta d \theta }$

$\displaystyle \sf{ \implies \: I = \int\limits_{}^{} \frac{{ \bigg[ \bigg( \frac{\pi}{2} + \theta \bigg)\bigg]}^{ - 1} }{ {\cos }^{} \theta \: \log \: \{ \: 1 + \frac{ { \sin}^{ - 1} (\sin 2\theta )}{\pi} \}} \, \cos \theta d \theta }$

$\displaystyle \sf{ \implies \: I = \int\limits_{}^{} \frac{{ \bigg( \frac{\pi}{2} + \theta \bigg)}^{ - 1} }{ \: \log \: \bigg(\: 1 + \frac{ 2\theta }{\pi} \bigg)} \, d \theta }$

$\displaystyle \sf{ \implies \: I = \frac{2}{\pi} \int\limits_{}^{} \frac{{ \bigg( 1 + \frac{ 2\theta }{\pi} \bigg)}^{ - 1} }{ \: \log \: \bigg(\: 1 + \frac{ 2\theta }{\pi} \bigg)} \, d \theta }$

$\displaystyle \sf{ \implies \: I = \frac{2}{\pi} \int\limits_{}^{} \frac{{ 1 }^{ } }{ \bigg( 1 + \frac{ 2\theta }{\pi} \bigg) \: \log \: \bigg(\: 1 + \frac{ 2\theta }{\pi} \bigg)} \, d \theta }$

$\displaystyle \sf{ Let \: \: y = \log \bigg(\: 1 + \frac{ 2\theta }{\pi} \bigg) \: \: then \: \: dy = \frac{2}{\pi} \frac{1}{\bigg(\: 1 + \frac{ 2\theta }{\pi} \bigg)} d \theta }$

Now from above we get

$\displaystyle \sf{ \implies \: I = \int\limits_{}^{} \frac{1 }{ \:y} \, dy }$

$\displaystyle \sf{ \implies \: I = \log |y| + c \: }$

$\displaystyle \sf{ \implies \: I = \log \bigg|\log \bigg(\: 1 + \frac{ 2\theta }{\pi} \bigg) \bigg| + c \: }$

$\displaystyle \sf{ \implies \: I = \log \bigg|\log \bigg(\: 1 + \frac{ 2 { \sin}^{ - 1}x }{\pi} \bigg) \bigg| + c \: }$

Where C is integration constant

FINAL ANSWER

$\displaystyle \sf{ \int\limits_{}^{} \frac{{ \bigg[ { \cos}^{ - 1}x( \sqrt{1 - {x}^{2} } ) \bigg]}^{ - 1} }{ \log \: \{ \: 1 + \frac{ { \sin}^{ - 1} (2x \sqrt{1 - {x}^{2} } )}{\pi} \}} \, dx }\displaystyle \sf{ = \log \bigg|\log \bigg(\: 1 + \frac{ 2 { \sin}^{ - 1}x }{\pi} \bigg) \bigg| + c \: }$

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