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Question: I<sub>2</sub> = \(\int_{\sin^{2}t}^{1 + \cos^{2}t}{f\{ x(2 - x)\} dx.}\)Then \(\frac{I_{1}}{I_{2}}\)...

I2 = ∫sin⁡2t1+cos⁡2tf{x(2−x)}dx.\int_{\sin^{2}t}^{1 + \cos^{2}t}{f\{ x(2 - x)\} dx.}Then I1I2\frac{I_{1}}{I_{2}}is -

A

0

B

1

C

2

D

3

Answer

1

Explanation

Solution

I1 = ∫sin⁡2t1+cos⁡2txf(x(2−x))\int_{\sin^{2}t}^{1 + \cos^{2}t}{xf(x(2 - x))}dx

= ∫sin⁡2t1+cos⁡2t(2−x)\int_{\sin^{2}t}^{1 + \cos^{2}t}{(2 - x)}f(x(2 – x)) dx = 2 . I2 – I1

2I1 = 2I2 Ž I1I2\frac{I_{1}}{I_{2}}= 1.