If I1- - 1xdt-1-t2 and I2- - 1-x1dt-1-t2 for x- 0- then

The correct options are A I_1= I_2 C I_2= π /4 tan ^ 1x #160;I_ 1 =∫ _ x ^ 1 dt 1+t^ 2 Substitute #160;t= 1 u #160; I _ 1 =∫ _ 1 x ...

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Byju's Answer

Standard XII

Mathematics

Integration Using Substitution

If I1= ∫ 1x...

Question

If $I_{1} = \int_{x}^{1} \frac{d t}{1 + t^{2}}$ and $I_{2} = \int_{1}^{1 / x} \frac{d t}{1 + t^{2}}$ for $x > 0$ , then

I_{1} = I_{2}

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I_{1} > I_{2}

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I_{2} > I_{1}

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I_{2} = π / 4 - {tan}^{- 1} x

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Solution

The correct options are
A $I_{1} = I_{2}$
C $I_{2} = π / 4 - {tan}^{- 1} x$
$I_{1} = \int_{x}^{1} \frac{d t}{1 + t^{2}}$
Substitute $t = \frac{1}{u}$
$I_{1} = \int_{\frac{1}{x}}^{1} \frac{(\frac{- 1}{u^{2}})}{1 + (\frac{1}{u^{2}})} d u = - \int_{\frac{1}{x}}^{1} \frac{d u}{1 + u^{2}} = \int_{1}^{\frac{1}{x}} \frac{d u}{1 + u^{2}} = I_{2}$
$I_{2} = - \int_{1}^{\frac{1}{x}} \frac{d t}{1 + t^{2}} = {tan}^{- 1} {t |}_{1}^{\frac{1}{x}} = {tan}^{- 1} \frac{1}{x} - \frac{π}{4}$

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If I1=∫1xdt1+t2 and I2=∫1/x1dt1+t2 for x>0, then

The correct options are A I1=I2 C I2=π/4−tan−1xI1=∫1xdt1+t2Substitute t=1uI1=∫11x(−1u2)1+(1u2)du=−∫11xdu1+u2=∫1x1du1+u2=I2I2=−∫1x1dt1+t2=tan−1t|1x1=tan−11x−π4

If $I_{1} = \int_{x}^{1} \frac{d t}{1 + t^{2}}$ and $I_{2} = \int_{1}^{1 / x} \frac{d t}{1 + t^{2}}$ for $x > 0$ , then