Mathematics

# Prove that : $\displaystyle \int_{0}^{1} \tan^{-1} x dx = \dfrac {\pi}{4} - \dfrac {1}{2}\log 2$.

##### SOLUTION
Suppose that
$I = \displaystyle \int _{0}^{1}\tan^{-1} x dx .... (1)$
By putting $x = \tan \theta$
$\therefore dx = \sec^{2} \theta . d\theta$
When $x = 0$ then $\theta = 0$
When $x = 1$ then $\theta = \dfrac {\pi}{4}$
$\therefore$ By equation $(1)$
$I = \displaystyle \int _{0}^{\tfrac {\pi}{4}} \tan^{-1} (\tan \theta)\sec^{2}\theta . d\theta$
$= \displaystyle \int _{0}^{\tfrac {\pi}{4}}\theta . \sec^{2} \theta. d\theta$
$= [\theta . \displaystyle \int \sec^{2} \theta d\theta]_{0}^{\tfrac {\pi}{4}} - \displaystyle \int _{0}^{\tfrac {\pi}{4}} \left [\dfrac {d}{d\theta} . \theta \displaystyle \int \sec^{2} \theta d\theta \right ]d\theta$
$= [\theta . \tan \theta]_{0}^{\tfrac {\pi}{4}} - \displaystyle \int _{0}^{\tfrac {\pi}{4}} \tan \theta d\theta$
$= [\theta . \tan \theta ]_{0}^{\tfrac {\pi}{4}} + [\log \cos \theta ]_{0}^{\tfrac {\pi}{4}}$
$= \dfrac {\pi}{4},\tan \dfrac {\pi}{4} - 0 + \log \cos \dfrac {\pi}{4} - \log \cos 0$
$= \dfrac {\pi}{4}\times 1 + \log \dfrac {1}{\sqrt {2}} - \log (1)$
$= \dfrac {\pi}{4} + \log (1) - \log \sqrt {2} - \log (1)$
$= \dfrac {\pi}{4} - \log (2)^{\dfrac {1}{2}}$
$= \dfrac {\pi}{4} - \dfrac {1}{2}\log 2$.

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Subjective Hard Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 111

#### Realted Questions

Q1 Subjective Medium
Evaluate: $\displaystyle\int \dfrac { \cos x + x \sin x } { x ( x + \cos x ) } d x$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Hard
$\int {\frac{{dx}}{{\sqrt {{x^2} - 9} + \sqrt {{x^2} - 4} }} = } ?$
• A. $- \frac{1}{5}\left[ {\frac{x}{2}\left( {\sqrt {{x^2} - 9} - \sqrt {{x^2} - 4} } \right) + \log \left[ {\frac{{{{\left( {x + \sqrt {{x^2} - 4} } \right)}^2}}}{{\left( {x + \sqrt {{x^2} - 4} } \right)}}} \right]} \right]$
• B. $\frac{1}{5}\left[ {\frac{x}{2}\left( {\sqrt {{x^2} - 9} + \sqrt {{x^2} - 4} } \right) + \log \left[ {\frac{{{{\left( {x + \sqrt {{x^2} - 4} } \right)}^2}}}{{{{\left( {x + \sqrt {{x^2} - 4} } \right)}^{9/2}}}}} \right]} \right]$
• C. None of these
• D. $\frac{1}{5}\left[ {\frac{x}{2}\left( {\sqrt {{x^2} - 9} - \sqrt {{x^2} - 4} } \right) + \log \left[ {\frac{{{{\left( {x + \sqrt {{x^2} - 4} } \right)}^2}}}{{{{\left( {x + \sqrt {{x^2} - 4} } \right)}^{9/2}}}}} \right]} \right]$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
Let  $\theta$  be the angle between the lines  ${ L }_{ { 1 } }:\left[ \begin{array}{l} { { x }=2{ t }+{ 1 } } \\ { { y }={ t }+{ 1 } } \\ { { z }=3{ t }+{ 1 } } \end{array} \right.$  and  ${ L }_{ { 2 } }:\left[ \begin{array}{l} { { x }=3{ s }+2 } \\ { { y }=6{ s }-1 } \\ { { z }=4 } \end{array} \right.$  where  $s , t \in { R }.$  Then the value of  $\int _ { 0 } ^ { \theta } \dfrac { 1 } { 1 + \tan x } d x =$
• A. $\pi / 4$
• B. $\pi / 2$
• C. $\pi / 3$
• D. $\pi / 6$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
Evaluate $\displaystyle\int^{1}_0\dfrac{(1-x)}{(1+x)}dx$
• A. $\dfrac{1}{2}log 2$
• B. $(2log 2+1)$
• C. $\left(\dfrac{1}{2}log 2-1\right)$
• D. $(2log 2-1)$

1 Verified Answer | Published on 17th 09, 2020

Q5 Passage Hard

In calculating a number of integrals we had to use the method of integration by parts several times in succession. The result could be obtained more rapidly and in a more concise form by using the so-called generalized formula for integration by parts.

$\int u(x)\, v(x)dx\, =\, u(x)\, v_{1}(x)\, -\, u^{}(x)v_{2}(x)\, +\, u^{}(x)\, v_{3}(x)\, -\, .\, +\, (-1)^{n\, -\, 1}u^{n\, -\, 1}(x)v_{n}(x)\, -\, (-1)^{n\, -\, 1}$ $\int\, u^{n}(x)v_{n}(x)\, dx$ where $v_{1}(x)\, =\, \int v(x)dx,\, v_{2}(x)\, =\, \int v_{1}(x)\, dx\, ..\, v_{n}(x)\, =\, \int v_{n\, -\, 1}(x) dx$

Of course, we assume that all derivatives and integrals appearing in this formula exist. The use of the generalized formula for integration  by parts is especially useful when calculating $\int P_{n}(x)\, Q(x)\, dx$, where $P_{n}(x)$, is polynomial of degree n and the factor Q(x) is such that it can be integrated successively n + 1 times.