A) \[x-{{\sin }^{-1}}x\]
B) \[{{\sin }^{-1}}x-x\]
C) \[x+{{\sin }^{-1}}x\]
D) \[{{\sin }^{-1}}x-{{x}^{2}}\]
Correct Answer: B
Solution :
| [b] \[I=\int{{{\log }_{e}}\left( \sqrt{1-x}+\sqrt{1+x} \right).1dx}\] |
| Integrating by parts taking 1 as the second function. |
| \[I=\log \left( \sqrt{1-x}+\sqrt{1+x} \right)x-\int{\frac{1}{\sqrt{1-x}+\sqrt{1+x}}}\] |
| \[\left[ -\frac{1}{2\sqrt{1-x}}+\frac{1}{2\sqrt{1+x}} \right](x)\,\,dx\] |
| \[=x\log \left( \sqrt{1-x}+\sqrt{1+x} \right)\] |
| \[=\frac{1}{2}\int{\frac{\sqrt{1-x}-\sqrt{1+x}}{\sqrt{1-x}+\sqrt{1+x}}.\frac{1}{\sqrt{1+{{x}^{2}}}}.xdx}\] |
| \[=x\log (\sqrt{1-x}+\sqrt{1+x})\] |
| \[-\frac{1}{2}\int{\frac{(1-x)+(1+x)-2\sqrt{1-{{x}^{2}}}}{(1-x)-(1+x)}.\frac{1}{\sqrt{1-{{x}^{2}}}}.xdx}\] |
| \[=x\log \left( \sqrt{1-x}+\sqrt{1+x} \right)-\frac{1}{2}\int{\frac{\sqrt{1-{{x}^{2}}}-1}{\sqrt{1-{{x}^{2}}}}dx}\] |
| \[=x\log \left( \sqrt{1-x}+\sqrt{1+x} \right)-\frac{1}{2}\left[ \int{1dx-}\int{\frac{1}{\sqrt{1-{{x}^{2}}}}dx} \right]\]\[=x\log \left( \sqrt{1-x}+\sqrt{1+x} \right)+\frac{1}{2}\left[ {{\sin }^{-1}}x-x \right]+C\] |
| \[\therefore f(x)=x,g(x)={{\sin }^{-1}}x-x\] |
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