For t∈R & f be a continuous function. Let I1=∫sin2t1+cos2txf(x(2−x))dx & I2= ∫sin2t1+cos2tf(x(2−x))dx . Then I1I2

∫abf(x)dx=∫abf(a+b−x)dxI1=∫sin2t1+cos2txf(x(2−x))dx=2∫sin2t1+cos2tf(x(2−x))−I,⇒2I1=2∫sin2t1+cos2tf(x(2−x))dx⇒2I1=2I2⇒I1I2=1

For t∈R & f be a continuous function. Let I1=∫sin2t1+cos2txf(x(2−x))dx & I2= ∫sin2t1+cos2tf(x(2−x))dx . Then I1I2

∫abf(x)dx=∫abf(a+b−x)dxI1=∫sin2t1+cos2txf(x(2−x))dx=2∫sin2t1+cos2tf(x(2−x))−I,⇒2I1=2∫sin2t1+cos2tf(x(2−x))dx⇒2I1=2I2⇒I1I2=1

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For $t \in R$ & f be a continuous function. Let
$I_{1} = \int_{\sin^{2} t}^{1 + \cos^{2} t} x f (x (2 - x)) d x & I_{2} = \int_{\sin^{2} t}^{1 + \cos^{2} t} f (x (2 - x)) d x . T h e n \frac{I_{1}}{I_{2}}$

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Detailed Solution

$\begin{array}{l} \int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x \\ I_{1} = \int_{\sin^{2} t}^{1 + \cos^{2} t} x f (x (2 - x)) d x = 2 \int_{\sin^{2} t}^{1 + \cos^{2} t} f (x (2 - x)) - I, \\ \Rightarrow 2 I_{1} = 2 \int_{\sin^{2} t}^{1 + \cos^{2} t} f (x (2 - x)) d x \Rightarrow 2 I_{1} = 2 I_{2} \Rightarrow \frac{I_{1}}{I_{2}} = 1 \end{array}$