Put x=tan θ⇒ dx=sec^2 θ d θ Also as x=0, θ = 0 and x=1, θ=π/4 Therefore, ∫_0^1 tan^ 1x dx=∫_0^π/4θ sec^2 θ d θ =π/4 log √(2)=π/4 1/2log 2 ...

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Property 1

∫01 tan-1x dx...

Question

$\int_{0}^{1} t a n^{- 1} x d x =$

\frac{π}{4} - \frac{1}{2} l o g 2

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π - \frac{1}{2} l o g 2

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\frac{π}{4} - l o g 2

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π - l o g 2

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Solution

The correct option is A $\frac{π}{4} - \frac{1}{2} l o g 2$
$P u t x = t a n θ \Rightarrow d x = s e c^{2} θ d θ$
$A l s o a s x = 0, θ = 0 a n d x = 1, θ = \frac{π}{4}$
$T h e r e f o r e, \int_{0}^{1} t a n^{- 1} x d x = \int_{0}^{\frac{π}{4}} θ s e c^{2} θ d θ$
$= \frac{π}{4} - l o g \sqrt{2} = \frac{π}{4} - \frac{1}{2} l o g 2$

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