Mathematics

If $\displaystyle I=\int \log\left ( \sqrt{1-x}+\sqrt{1+x} \right )dx,$ then I is equal to

$\displaystyle x\log \left ( \sqrt{1-x}+\sqrt{1+x} \right )+\frac{1}{2}\sin^{-1}x-\frac{1}{2}x+C$

SOLUTION
Let $\displaystyle I=\int { \log { \left( \sqrt { 1-x } +\sqrt { 1+x } \right) } } dx$

$\displaystyle =x\log { \left( \sqrt { 1-x } +\sqrt { 1+x } \right) } -\int { \frac { x }{ \left( \sqrt { 1-x } +\sqrt { 1+x } \right) } } \left( \frac { -1 }{ 2\sqrt { 1-x } } +\frac { 1 }{ 2\sqrt { 1+x } } \right) dx$

$\displaystyle =x\log { \left( \sqrt { 1-x } +\sqrt { 1+x } \right) } +\frac { 1 }{ 2 } { I }_{ 1 }$

Where $\displaystyle { I }_{ 1 }=\int { \frac { \sqrt { 1+x } -\sqrt { 1-x } }{ \sqrt { 1+x } +\sqrt { 1-x } } } \frac { x }{ \sqrt { 1-{ x }^{ 2 } } } dx$

$\displaystyle =\int { \frac { { \left( \sqrt { 1+x } -\sqrt { 1-x } \right) }^{ 2 } }{ 1+x-1+x } } \frac { x }{ \sqrt { 1-{ x }^{ 2 } } } dx$

$\displaystyle =\frac { 1 }{ 2 } \int { \frac { \\ 1+x+1-x-2\sqrt { 1-{ x }^{ 2 } } }{ \sqrt { 1-{ x }^{ 2 } } } } dx=\int { \left( \frac { 1 }{ \sqrt { 1-{ x }^{ 2 } } } -1 \right) } dx=\sin ^{ -1 }{ x } -x$

Therefore $\displaystyle I=x\log { \left( \sqrt { 1-x } +\sqrt { 1+x } \right) } +\frac { 1 }{ 2 } \left( \sin ^{ -1 }{ x } -x \right) +c$

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Single Correct Hard Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 111

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