Fundamental theorem of calculus - class-XII

Let $I=\int _0^1\tan^{-1}\left (\dfrac {2x-1}{1+x-x^2}\right )dx$
$\Rightarrow I=\int _0^1 \tan^{-1}\left (\dfrac {x-(1-x)}{1+x(1-x)}\right )dx$
$\Rightarrow I=\int _0^1[\tan^{-1}x-\tan^{-1}(1-x)]dx$ ................ (1)
$\Rightarrow I=\int _0^1[\tan^{-1}(1-x)-\tan^{-1}(1-1+x)]dx$
$\Rightarrow I=\int _0^1[\tan^{-1}(1-x)-\tan^{-1}(x)]dx$
$\Rightarrow I=\int _0^1[\tan^{-1}(1-x)-\tan^{-1}(x)]dx$ ........... (2)
Adding (1) and (2), we obtain
$2I=\int _0^1(\tan^{-1}x+\tan^{-1}(1-x)-\tan^{-1}(1-x)-\tan^{-1}x)dx=0$
$\Rightarrow I=0$
Hence, the correct Answer is B.

We have,

$I=\int _{0}^{\dfrac{\pi }{2}}{\sin 2x{{\tan }^{-1}}\left( \sin x \right)dx}$

$=\int _{0}^{\dfrac{\pi }{2}}{2\sin x\cos x{{\tan }^{-1}}\left( \sin x \right)dx}$

Let

$ \sin x=t $

$ \cos xdx=dt $

Change limit

$ \sin 0=t $

$ t=0 $

And,

$ \sin \dfrac{\pi }{2}=t $

$ t=1 $

Then,

$ \int _{0}^{1}{2t{{\tan }^{-1}}t\cos xdx} $

$ =\int _{0}^{1}{2t{{\tan }^{-1}}tdt} $

$ =2\int _{0}^{1}{t{{\tan }^{-1}}tdt} $

On integrating and we get,

$ 2\left[ {{\tan }^{-1}}t\int _{0}^{1}{t}dt-\int _{0}^{1}{\left( \dfrac{d\left( {{\tan }^{-1}}t \right)}{dt}\int _{0}^{1}{tdt} \right)}dt \right] $

$ =2\left[ {{\tan }^{-1}}t\left[ {{\left( \dfrac{{{t}^{2}}}{2} \right)} _{0}}^{1} \right]-\int _{0}^{1}{\dfrac{1}{1+{{t}^{2}}}}\dfrac{{{t}^{2}}}{2}dt \right] $

$ =2{{\tan }^{-1}}t{{\left( \dfrac{{{t}^{2}}}{2} \right)} _{0}}^{1}-\int _{0}^{1}{\dfrac{{{t}^{2}}}{1+{{t}^{2}}}}dt $

$ =2{{\tan }^{-1}}t{{\left( \dfrac{{{t}^{2}}}{2} \right)} _{0}}^{1}-\int _{0}^{1}{\dfrac{{{t}^{2}}+1-1}{1+{{t}^{2}}}}dt $

$ =2{{\tan }^{-1}}t{{\left( \dfrac{{{t}^{2}}}{2} \right)} _{0}}^{1}-\int _{0}^{1}{\dfrac{{{t}^{2}}+1}{1+{{t}^{2}}}}dt+\int _{0}^{1}{\dfrac{1}{1+{{t}^{2}}}}dt $

$ =2{{\tan }^{-1}}t{{\left( \dfrac{{{t}^{2}}}{2} \right)} _{0}}^{1}-\int _{0}^{1}{1}dt+\int _{0}^{1}{\dfrac{1}{1+{{t}^{2}}}}dt $

$ =2{{\tan }^{-1}}t{{\left( \dfrac{{{t}^{2}}}{2} \right)} _{0}}^{1}-{{\left[ t \right]} _{0}}^{1}+{{\left[ {{\tan }^{-1}}t \right]} _{0}}^{1}+C $

$ =2\left[ {{\tan }^{-1}}1-{{\tan }^{-1}}0 \right]\left[ \dfrac{{{1}^{2}}}{2}-\dfrac{{{0}^{2}}}{2} \right]-\left[ 1-0 \right]+\left[ {{\tan }^{-1}}1-{{\tan }^{-1}}0 \right]+C $

$ =2\left[ \dfrac{\pi }{4}-0 \right]\left[ \dfrac{1}{2} \right]-1+\left[ \dfrac{\pi }{4}-0 \right] $

$ =\dfrac{\pi }{4}+\dfrac{\pi }{4}-1 $

$ =\dfrac{2\pi }{4}-1 $

$ =\dfrac{\pi }{2}-1 $

Hence, this is the
answer.

Consider, $\displaystyle I= \int _{0}^{\sqrt{3}}[x^{3} -1] dx$

The integral is the sum of integrals from n to n+1. For x in [tan(n), tan(n+1)], [tan^-1(x)] = n. The sum is 0integral(0 to tan(1)) + 1integral(tan(1) to tan(2)) + ... + 99*integral(tan(99) to tan(100)). This simplifies to 100 - tan(1).

$\int _0^{\pi/2}\sin x\sin 2x\sin 3x dx$

$\int _{1}^{4}\dfrac{x   dx}{\sqrt{2 + 4x}}=\dfrac{1}{2}\int _{1}^{4}\dfrac{x   dx}{\sqrt{x+\dfrac{1}{2}}}$
$=\dfrac{1}{2}\left [ \int _{1}^{4}\dfrac{(x+\dfrac{1}{2})dx}{\sqrt{x+\dfrac{1}{2}}}-\int _{1}^{4}\dfrac{\dfrac{1}{2}dx}{\sqrt{x+\dfrac{1}{2}}} \right ]$
$=\dfrac{1}{2}\left [ \int _{1}^{4} \sqrt{x+\dfrac{1}{2}} dx-\int _{1}^{4}(x+\dfrac{1}{2})^{\dfrac{1}{2}} \int _{1}^{4} \right ]$
$=\dfrac{1}{2} \left [ \dfrac{2}{3} (x+\dfrac{1}{2})^{\dfrac{3}{2}} \int _{1}^{4}-(x+\dfrac{1}{2})^{\dfrac{4}{2}} \int _{1}^{4} \right ]$
$=\dfrac{1}{3} \left [ \left ( \dfrac{9}{2} \right )^{\dfrac{3}{2}}-\left ( \dfrac{3}{2} \right )^{\dfrac{3}{2}} \right ] -\dfrac{1}{2} \left [ \left ( \dfrac{9}{2} \right )^{\dfrac{1}{2}}-\left ( \dfrac{3}{2} \right )^{\dfrac{1}{2}} \right ]$
$=\dfrac{1}{3} \left [ \left ( \dfrac{9}{2} \right )\left ( \dfrac{9}{2} \right )^{\dfrac{1}{2}}-\left ( \dfrac{3}{2} \right )\left ( \dfrac{3}{2} \right )^{\dfrac{1}{2}} \right ] - \dfrac{1}{2} \left [ \left ( \dfrac{3}{\sqrt{2}} \right )-\dfrac{\sqrt{3}}{\sqrt{2}} \right ]$
$=\dfrac{3}{\sqrt{2}}$

$\int _{ 0 }^{ 2 }{ (x-\log _{ 2 }a } )dx=2\log _{ 2 }(\cfrac { 2 }{ a } )$

$I = \displaystyle \int _{0}^{\pi} {ln(\sin x)dx}$

$I = \displaystyle \int _{0}^{\pi} {ln(\sin x)dx}$

$ \int _{\sin x}^1 t^2 f(t) dt = 1 - \sin x \forall x \epsilon (0, \pi/2 ) $
Differentiating both sides we get 
$ \dfrac {d}{dx} (1) [ 1 \cdot f (1)] - \cos x ( \sin^2 x) f ( \sin x) = -\cos x  $
$ \Rightarrow f ( \sin x) = \dfrac {1}{ \sin^2 x} $
$ \therefore f \left( \dfrac {1}{\sqrt3} \right) = f \left( \sin \left( \sin^{-1} \dfrac {1}{\sqrt3} \right) \right) $
$ = \left[ \dfrac {1}{ \sin \left( \sin^{-1} \dfrac {1}{\sqrt3} \right)} \right]^2 = 3 $

$I _1=\int _{0}^{4}e^{-x} cos^2x dx$
$I _2=\int _{0}^{1}e^{-x^2}cos^2x dx$
Both have $cos^2x$ so value get restricted more in (0, 1)
Now in (0, 1) $e^{-\frac {x^2}{2}}>e^{-x}$
$\therefore \int _{0}^{1}e^{-\frac {x^2}{2}}>\int _{0}^{1}e^{-x}$
$\therefore I _4>I _3>I _1>I _2$

First of all integrate the function.

Let $I=\int _{ 0 }^{ 1 }{ \cfrac { \tan ^{ -1 }{ x }  }{ x }  } dx$

$|x\sin(\pi x)|=\begin{cases}x\sin \pi x ,,,,,x\in(-1,1)\  -x\sin\pi x,x\in (1,\dfrac{3}{2})  \end{cases}$

Using Integration by part$:-$