Question

Integrate. [Cos^-1x (√1-x^2)]^-1 / Log{1+(sin(2x√1-x^2)/π}​

Answer 1

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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Answer 2

Answer:

Step-by-step explanation: