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Doubt in expanding the integral of given question.Problem in applying tricks.

Doubt in expanding the integral of given question.Problem in applying tricks.

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Grade:11

1 Answers

Piyush Maheshwari
44 Points
5 years ago
Apply this property for the integral 
\int_{b}^{a}f(x) dx= \int_{b}^{a} f(a+b-x)dx
Add both the expressions.
Simplify the denominator to realise that the denominator does not change on replacing x with 1-x.
Then simplify the
 sin^{-1}\sqrt{x} + sin^{-1}\sqrt{1- x} 
in the numerator of 2I to get sin^{-1}1 which is equal to pi/2 with the denominator being unaffected.
Divide the equation by 2 to get
I =\frac{\pi }{4} \int_{0}^{1} \frac{1}{x^2-x+1}dx
Now this is a simple integral and can be solved by completing the square in the denominator.
I =\frac{\pi }{4} \int_{0}^{1} \frac{1}{(x-\frac{1}{2})^2+\frac{3}{4}}dx
Use the standard result now to get
I =\frac{\pi }{4}.\frac{2}{\sqrt{3}}\left ( tan^{-1}\frac{x-\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right )_{0}^{1}
Solving this you will get C as the correct option.
 
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