Put t=tan^ 1x ⇒ dt=1/1+x^2dx, then ∫_0^1 tan^ 1x/1+x^2dx=∫_0^π/4t dt= [ t^2/2 ]_0^π/4=π^2/32 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Property 1

∫01 tan-1x/1+...

Question

$\int_{0}^{1} \frac{t a n^{- 1} x}{1 + x^{2}} d x =$

\frac{π^{2}}{8}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{π^{2}}{16}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{π^{2}}{4}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{π^{2}}{32}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $\frac{π^{2}}{32}$
$P u t t = t a n^{- 1} x \Rightarrow d t = \frac{1}{1 + x^{2}} d x, t h e n$
$\int_{0}^{1} \frac{t a n^{- 1} x}{1 + x^{2}} d x = \int_{0}^{\frac{π}{4}} t d t = {[\frac{t^{2}}{2}]}_{0}^{\frac{π}{4}} = \frac{π^{2}}{32}$

Suggest Corrections

Similar questions

\int_{0}^{\frac{1}{2}} \frac{x s i n^{- 1} x}{\sqrt{1 - x^{2}}} d x =

\int_{0}^{\frac{1}{2}} \frac{x s i n^{- 1} x}{\sqrt{1 - x^{2}}} d x =

\int_{0}^{1} \frac{x S i n^{- 1} x}{\sqrt{1 - x^{2}}} d x =

\int_{0}^{1} {tan}^{- 1} (\frac{x}{\sqrt{1 - x^{2}}}) d x

\int_{0}^{1} {sin}^{- 1} (x \sqrt{1 - x} - \sqrt{x} \sqrt{1 - x^{2}}) d x, 0 \leq x \leq 1

Join BYJU'S Learning Program