If I-01 x -1-x-1-xdx- the I equals

We can write I=∫_0^1 x(1 x)/√(1 x^2)dx =∫_0^1 (x/√(1 x^2) 1/√(1 x^2)+1 x^2/√(1 x^2))dx =.( √(1 x^2) sin^ 1x)]_0^1 +.(x/2√(1 x^2)+1/2sin^ 1x)]_0^1 = π/2+1+1/2(π/ ...

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Byju's Answer

Standard XII

Mathematics

Integration Using Substitution

If I=∫01 x √1...

Question

If $I = \int_{0}^{1} x \sqrt{\frac{1 - x}{1 + x}} d x$ , the I equals

1 + \frac{π}{4}

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1 - \frac{π}{4}

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π

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π - \sqrt{2}

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Solution

The correct option is B $1 - \frac{π}{4}$
We can write
$I = \int_{0}^{1} \frac{x (1 - x)}{\sqrt{1 - x^{2}}} d x$
$= \int_{0}^{1} (\frac{x}{\sqrt{1 - x^{2}}} - \frac{1}{\sqrt{1 - x^{2}}} + \frac{1 - x^{2}}{\sqrt{1 - x^{2}}}) d x$
$= {(- \sqrt{1 - x^{2}} - s i n^{- 1} x)]}_{0}^{1} + {(\frac{x}{2} \sqrt{1 - x^{2}} + \frac{1}{2} s i n^{- 1} x)]}_{0}^{1}$
$= - \frac{π}{2} + 1 + \frac{1}{2} (\frac{π}{2}) = 1 - \frac{π}{4}$ .

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If I=∫10x√1−x1+xdx, the I equals

The correct option is B 1−π4We can write I=∫10x(1−x)√1−x2dx =∫10(x√1−x2−1√1−x2+1−x2√1−x2)dx =(−√1−x2−sin−1x)]10+(x2√1−x2+12sin−1x)]10 =−π2+1+12(π2)=1−π4.

If $I = \int_{0}^{1} x \sqrt{\frac{1 - x}{1 + x}} d x$ , the I equals