Definite Integration

$F(x) = \int\limits_1^x {{t^2}g(t)dt} $

$ \Rightarrow $ F'(x) = x²g(x) = x²$\int\limits_1^t {f(u)du} $

$ \therefore $ F'(1) = (1)(0) = 0

Now, F''(x) = 2xg(x) + x²g'(x)

F''(1) = 2g(1) + g'(1) = 0 + g'(1) = 3

[ As g'(t) = f(t); g'(1) = f'(1) = 3 ]

So, at x = 1, F'(1) = 0 and F"(1) = 3 > 0

$ \therefore $ For the function f(x), x = 1 is a point of local minima.

I = $\int\limits_0^{2\pi } {{{x{{\sin }^8}x} \over {{{\sin }^8}x + {{\cos }^8}x}}} dx$ .....(1)

I = $\int\limits_0^{2\pi } {{{\left( {2\pi - x} \right){{\sin }^8}x} \over {{{\sin }^8}x + {{\cos }^8}x}}} dx$ ......(2)

Adding (1) and (2)

2I = $\int\limits_0^{2\pi } {{{2\pi {{\sin }^8}x} \over {{{\sin }^8}x + {{\cos }^8}x}}} dx$

$ \Rightarrow $ I = $2\pi \int\limits_0^\pi {{{{{\sin }^8}x} \over {{{\sin }^8}x + {{\cos }^8}x}}} dx$

$ \Rightarrow $ I = $2\pi \left[ {\int\limits_0^{\pi /2} {{{{{\sin }^8}x} \over {{{\sin }^8}x + {{\cos }^8}x}}} dx + \int\limits_0^{\pi /2} {{{{{\cos }^8}x} \over {{{\sin }^8}x + {{\cos }^8}x}}} dx} \right]$

= ${2\pi \int\limits_0^{\pi /2} 1 dx}$

= $2\pi .{\pi \over 2}$ = ${\pi ^2}$

ƒ(x) = a + bx + cx²

$\int\limits_0^1 {f\left( x \right)dx} $ = $\left[ {ax + {{b{x^2}} \over 2} + {{c{x^3}} \over 3}} \right]_0^1$

= ${a + {b \over 2} + {c \over 3}}$

= ${1 \over 6}\left[ {6a + 3b + c} \right]$

f(1) = a + b + c

f(0) = a

$f\left( {{1 \over 2}} \right) = a + {b \over 2} + {c \over 4}$

By checking each option you have to find the solution.

${1 \over 6}\left\{ {f(0) + f(1) + 4f\left( {{1 \over 2}} \right)} \right\}$

= ${1 \over 6}\left[ {a + a + b + c + 4\left( {a + {b \over 2} + {c \over 4}} \right)} \right]$

= ${1 \over 6}\left[ {6a + 3b + c} \right]$

$ \therefore $ Option (A) is correct option.

Let f(x) = ${1 \over {\sqrt {2{x^3} - 9{x^2} + 12x + 4} }}$

f'(x) = $ - {1 \over 2}{{ - 6{x^2} - 18x + 12} \over {2{{\left( {2{x^3} - 9{x^2} + 12x + 4} \right)}^{3/2}}}}$

= ${{ - 6\left( {x - 1} \right)\left( {x - 2} \right)} \over {2{{\left( {2{x^3} - 9{x^2} + 12x + 4} \right)}^{3/2}}}}$



f_max = f(1)

f_min = f(2)

f(1) = ${1 \over {\sqrt {2 - 9 + 12 + 4} }}$ = ${1 \over {\sqrt 9 }}$ = ${1 \over 3}$

f(2) = ${1 \over {\sqrt {16 - 36 + 24 + 4} }}$ = ${1 \over {\sqrt 8 }}$

$ \therefore $ ${1 \over 3} < {I} <$ ${1 \over {\sqrt 8 }}$

So ${1 \over 9} < {I^2} < {1 \over 8}$

$\mathop {\lim }\limits_{x \to 0} {{\int_0^x {t\sin \left( {10t} \right)dt} } \over x}$

This is in ${0 \over 0}$ form.

So apply newton leibniz rule

$\mathop {\lim }\limits_{x \to 0} {{x.\sin \left( {10x} \right) - 0} \over 1}$ = 0

$4\alpha \int\limits_{ - 1}^2 {{e^{ - \alpha \left| x \right|}}dx} = 5$

$ \Rightarrow $ $4\alpha \int\limits_{ - 1}^0 {{e^{\alpha x}}dx} + 4\alpha \int\limits_0^2 {{e^{ - \alpha x}}dx} $ = 5

$ \Rightarrow $ $4\alpha \left[ {{{{e^{\alpha x}}} \over \alpha }} \right]_{ - 1}^0 + 4\alpha \left[ {{{{e^{ - \alpha x}}} \over { - \alpha }}} \right]_0^2$ = 5

$ \Rightarrow $ $4{e^{ - 2\alpha }} + 4{e^{ - \alpha }} - 3$ = 0

Let e^{-$\alpha $} = t

$ \therefore $ 4t² + 4t - 3 = 0

$ \Rightarrow $ t = ${1 \over 2}$

$ \therefore $ e^{-$\alpha $} = ${1 \over 2}$

$ \Rightarrow $ $\alpha $ = ${\log _e}2$

2cot²$\theta $ - ${5 \over {\sin \theta }}$ + 4 = 0

$ \Rightarrow $ $2{{{{\cos }^2}\theta } \over {{{\sin }^2}\theta }} - {5 \over {\sin \theta }} + 4$ = 0

$ \Rightarrow $ 2sin² $\theta $ – 5sin$\theta $ + 2 = 0

$ \Rightarrow $ (2sin$\theta $ – 1)(sin$\theta $ – 2) = 0

$ \therefore $ sin$\theta $ = ${1 \over 2}$ [ sin$\theta $ = 2 not possible]

$ \therefore $ $\theta $₁ = ${\pi \over 6}$ and $\theta $₂ = ${{5\pi } \over 6}$ as $\theta $₁ $<$ $\theta $₂

$ \therefore $ I = $\int\limits_{{\pi \over 6}}^{{{5\pi } \over 6}} {{{\cos }^2}3\theta d\theta } $

= $\int\limits_{{\pi \over 6}}^{{{5\pi } \over 6}} {{{1 + \cos 6\theta } \over 2}d\theta } $

= ${1 \over 2}\left[ {\theta + {{\sin 6\theta } \over 2}} \right]_{{\pi \over 6}}^{{{5\pi } \over 6}}$

= ${{\pi \over 3}}$

I = ${1 \over {a + b}}\int_a^b {x\left( {f(x) + f(x + 1)} \right)} dx$ ...(1)

x $ \to $ a + b - x

I = ${1 \over {a + b}}\int\limits_a^b {\left( {a + b - x} \right)\left( {f\left( {a + b - x} \right) + f\left( {a + b - x + 1} \right)} \right)dx} $

I = ${1 \over {a + b}}\int\limits_a^b {\left( {a + b - x} \right)\left( {f\left( {x + 1} \right) + f\left( x \right)} \right)dx} $ .....(2)

[As ƒ(x) = ƒ(a + b + 1 - x)

$ \Rightarrow $ ƒ(x + 1) = ƒ(a + b - x)]

Adding (1) and (2) we get

2I = ${{a + b} \over {a + b}}\int\limits_a^b {\left( {f\left( {x + 1} \right) + f\left( x \right)} \right)dx} $

$ \Rightarrow $ I = ${1 \over 2}\int\limits_a^b {f\left( x \right)dx} + {1 \over 2}\int\limits_a^b {f\left( {x + 1} \right)dx} $

$ \Rightarrow $ I = ${1 \over 2}\int\limits_a^b {f\left( x \right)dx} + {1 \over 2}\int\limits_a^b {f\left( {a + b - x + 1} \right)dx} $

$ \Rightarrow $ I = ${1 \over 2}\int\limits_a^b {f\left( x \right)dx} + {1 \over 2}\int\limits_a^b {f\left( x \right)dx} $

$ \Rightarrow $ I = $\int\limits_a^b {f\left( x \right)dx} $

Let x = t + 1

$ \therefore $ I = $\int\limits_{a - 1}^{b - 1} {f\left( {t + 1} \right)dt} $

= $\int\limits_{a - 1}^{b - 1} {f\left( {x + 1} \right)dx} $

$I = \int\limits_\alpha ^{\alpha + 1} {{{dx} \over {(x + \alpha )(x + \alpha + 1)}}} $

Let x + $\alpha $ = t and dx = dt

$I = \int\limits_{2\alpha }^{2\alpha + 1} {{{dt} \over {t(t + 1)}}} = \int\limits_{2\alpha }^{2\alpha + 1} {\left( {{1 \over t} - {1 \over {t + 1}}} \right)dx} $

$ \Rightarrow \left( {\ln t - \ln (t + 1)} \right)|_{2\alpha }^{2\alpha + 1} = \ln \left( {{t \over {t + 1}}} \right)|_{2\alpha }^{2\alpha + 1}$

$ \Rightarrow \ln \left( {{{2\alpha + 1} \over {2\alpha + 2}}} \right) - \ln \left( {{{2\alpha } \over {2\alpha + 1}}} \right)$

$ \Rightarrow \ln \left( {{{2\alpha + 1} \over {2\alpha + 2}} \times {{2\alpha + 1} \over {2\alpha }}} \right) = \ln \left( {{9 \over 8}} \right)$

$ \Rightarrow {{{{\left( {2\alpha + 1} \right)}^2}} \over {\left( {2\alpha + 2} \right)}} = {{9\alpha } \over 4}$

$4(4{\alpha ^2} + 1 + 4\alpha ) = 18{\alpha ^2} + 18\alpha $

$ \Rightarrow 2{\alpha ^2} - 2\alpha - 4 = 0$

$ \Rightarrow {\alpha ^2} + \alpha - 2 = 0$

$ \Rightarrow (\alpha + 2)(\alpha - 1) = 0$

$\alpha $ = 1 or $\alpha $ = -2

$\int\limits_0^{\pi /2} {{{\cot x} \over {\cot x + {\mathop{\rm cosec}\nolimits} \,x}}dx = } \int\limits_0^{\pi /2} {{{\cos x} \over {\cos x + 1}}dx} $

$ \Rightarrow \int\limits_0^{\pi /2} {{{2{{\cos }^2}{x \over 2} - 1} \over {2{{\cos }^2}{x \over 2}}}dx = } \int\limits_0^{\pi /2} {\left( {1 - {1 \over 2}{{\sec }^2}{x \over 2}} \right)dx} $

$ \Rightarrow \left( {x - \tan {x \over 2}} \right)_0^{\pi /2} = {\pi \over 2} - 1 = {1 \over 2}\left( {\pi - 2} \right)$

mn = ${1 \over 2}x - 2 = - 1$

Given $\int\limits_6^{f(x)} {4{x^3}dx} = g(x).(x - 2)$

$ \Rightarrow g(x) = $ ${{\int\limits_0^{f\left( x \right)} {4{x^3}dx} } \over {x - 2}}$

$ \therefore $ $\mathop {\lim }\limits_{x \to 2} g(x) = $$\mathop {\lim }\limits_{x \to 2} {{\int\limits_0^{f\left( x \right)} {4{x^3}dx} } \over {x - 2}}$

At x = 2 this limit is in ${0 \over 0}$ form.

So we can use L'Hopital's rule. Use leibniz intgral rule to differentiate the integration.

$ \Rightarrow \mathop {\lim }\limits_{x \to 2} {{4{f^3}(x).f'(x)} \over 1} = 4 \times {6^3} \times {1 \over {48}} = 18$

$\int\limits_{\pi /6}^{\pi /3} {{{\sec }^{2/3}}x\,{{{\mathop{\rm cosec}\nolimits} }^{4/3}}x\,dx} $

$ = \int {{{{{\sec }^2}x} \over {{{\tan }^{4/3}}x}}dx} $

Let tan x = t, sec² x dx = dt

$ = \int {{{dt} \over {{t^{4/3}}}}} $

$I = - 3\left( {{t^{ - 1/3}}} \right)$

$ \Rightarrow \left( { - 3{{\left( {\tan x} \right)}^{{{ - 1} \over 3}}}} \right)_{\pi /6}^{\pi /3}$

$ \Rightarrow 3\left( {{3^{{1 \over 3}}} - {1 \over {{3^{{1 \over 6}}}}}} \right)$ = 3^7/6 - 3^5/6

$I = \int\limits_0^{2\pi } {\left[ {\sin 2x(1 + \cos 3x)} \right]} dx$ .... (i)

$ \therefore\int\limits_0^a {f(x)} = \int\limits_0^a {f(a - x)} dx$

$ \because I = \int\limits_0^{2\pi } {\left[ { - \sin 2x(1 + \cos 3x)} \right]} dx$

By (i) + (ii)

$ \Rightarrow 2I = \int\limits_0^{2\pi } {( - 1)dx} $

$ \Rightarrow 2I = - (x)_0^{2\pi }$

$ \Rightarrow $ I = –$\pi $

$\mathop {\lim }\limits_{n \to \infty } {{{{(n + 1)}^{{1 \over 3}}} + {{(n + 2)}^{{1 \over 3}}} + ..... + {{\left( {n + n} \right)}^{{1 \over 3}}}} \over {n{{(n)}^{{1 \over 3}}}}}$

$ \Rightarrow \mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^n {{{{{(n + r)}^{{1 \over 3}}}} \over {n.{n^{{1 \over 3}}}}}\,\,\,{r \over n} \to x\,and\,{1 \over n} \to dx\,} $

$ \Rightarrow \int\limits_0^1 {{{(1 + x)}^{{1 \over 3}}}dx} $

$ \Rightarrow \left[ {{3 \over 4}{{(1 + x)}^{{4 \over 3}}}} \right]_0^1 = {3 \over 4}{(2)^{{4 \over 3}}} - {3 \over 4}$

$\mathop {\lim }\limits_{x \to 2} {{\int\limits_6^{f\left( x \right)} {2tdt} } \over {\left( {x - 2} \right)}}$

This is ${0 \over 0}$ form so we use L – Hospital Rule.

= $\mathop {\lim }\limits_{x \to 2} {{2f\left( x \right).f'\left( x \right) - 0} \over 1}$

= ${2f\left( 2 \right).f'\left( 2 \right)}$

= 12f'(2)

Note : Newton - Leibnitz rule of differentiation under integration

${d \over {dx}}\int\limits_{\alpha \left( x \right)}^{\beta \left( x \right)} {f\left( u \right)} du$

= $f\left( {\beta \left( x \right)} \right)\beta '\left( x \right) - f\left( {\alpha \left( x \right)} \right)\alpha '\left( x \right)$

I = $\int\limits_0^1 {x{{\cot }^{ - 1}}(1 - {x^2} + {x^4})dx} $

Let x² = t

$ \Rightarrow $ 2xdx = dt

At x = 0, t = 0

At x = 1, t = 1

I = ${1 \over 2}\int\limits_0^1 {{{\cot }^{ - 1}}\left( {1 - t + {t^2}} \right)dt} $

= ${1 \over 2}\int\limits_0^1 {{{\tan }^{ - 1}}\left( {{1 \over {1 - t + {t^2}}}} \right)dt} $

= ${1 \over 2}\int\limits_0^1 {{{\tan }^{ - 1}}\left( {{{t + \left( {1 - t} \right)} \over {1 - t + {t^2}}}} \right)dt} $

= ${1 \over 2}\int\limits_0^1 {{{\tan }^{ - 1}}\left( {{{t + \left( {1 - t} \right)} \over {1 - t\left( {1 - t} \right)}}} \right)dt} $

= ${1 \over 2}\int\limits_0^1 {\left[ {{{\tan }^{ - 1}}\left( {1 - t} \right) + {{\tan }^{ - 1}}\left( t \right)} \right]dt} $

= ${1 \over 2}\int\limits_0^1 {{{\tan }^{ - 1}}\left( t \right)dt} + {1 \over 2}\int\limits_0^1 {{{\tan }^{ - 1}}\left( {1 - t} \right)dt} $

= ${1 \over 2}\int\limits_0^1 {{{\tan }^{ - 1}}\left( t \right)dt} + {1 \over 2}\int\limits_0^1 {{{\tan }^{ - 1}}\left( t \right)dt} $

[ As ${1 \over 2}\int\limits_0^1 {{{\tan }^{ - 1}}\left( {1 - t} \right)dt} $

= ${1 \over 2}\int\limits_0^1 {{{\tan }^{ - 1}}\left[ {0 + 1 - \left( {1 - t} \right)} \right]dt} $

= ${1 \over 2}\int\limits_0^1 {{{\tan }^{ - 1}}\left( t \right)dt} $ ]

= $2 \times {1 \over 2}\int\limits_0^1 {{{\tan }^{ - 1}}\left( t \right)dt} $

= $\int\limits_0^1 {{{\tan }^{ - 1}}\left( t \right)dt} $

Using integration by parts rule

= $\left[ {t.{{\tan }^{ - 1}}\left( t \right)} \right]_0^1$$ - \int\limits_0^1 {t.{1 \over {1 + {t^2}}}} dt$

= ${\pi \over 4} - {1 \over 2}\left[ {\log \left( {1 + {t^2}} \right)} \right]_0^1$

= ${\pi \over 4} - {1 \over 2}{\log _e}2$

I = $\int\limits_0^{\pi /2} {{{{{\sin }^3}x} \over {\sin x + \cos x}}dx} $ .....(1)

$ \Rightarrow $ I = $\int\limits_0^{{\pi \over 2}} {{{{{\cos }^3}x} \over {\sin x + \cos x}}} dx$ ......(2)

Adding those two

2I = $\int\limits_0^{{\pi \over 2}} {{{si{n^3}x + {{\cos }^3}x} \over {\sin x + \cos x}}} dx$

= $\int\limits_0^{{\pi \over 2}} {{{\left( {\sin x + \cos x} \right)\left( {si{n^2}x + {{\cos }^2}x - \sin x\cos x} \right)} \over {\sin x + \cos x}}} dx$

= $\int\limits_0^{{\pi \over 2}} {\left( {si{n^2}x + {{\cos }^2}x - \sin x\cos x} \right)} dx$

= $\int\limits_0^{{\pi \over 2}} {\left( {1 - {{\sin 2x} \over 2}} \right)} dx$

= $\left[ {x + {{\cos 2x} \over 4}} \right]_0^{{\pi \over 2}}$

= $\left( {{\pi \over 2} - {1 \over 4}} \right) - \left( {{1 \over 4}} \right)$

$ \therefore $ 2I = $\left( {{\pi \over 2} - {1 \over 2}} \right)$

$ \therefore $ I = ${{\pi - 1} \over 4}$

$f(x) = \int\limits_0^x {g(t)dt} $

$f\left( { - x} \right) = \int\limits_0^{ - x} {g\left( t \right)} dt$

Put t = -v

= $ - \int\limits_0^x {g\left( { - v} \right)} dv$

= $ - \int\limits_0^x {g\left( v \right)} d(v)$ [ as g(v) is an even function.]

= - f(x)

$ \Rightarrow $ f(-x) = -f(x)

$ \therefore $ f(x) is an odd function.

Given ƒ(x + 5) = g(x)

$ \therefore $ g(- x) = ƒ(- x + 5)

$ \Rightarrow $ g(x) = - f(x - 5) [as g(x) is even and f(x) is an odd function]

Replacing x by x + 5, we get

f(x) = - g(x + 5) ......(1)

Now

$ \int\limits_0^x {f(t)dt} $

= $ - \int\limits_0^x {g\left( {t + 5} \right)} dt$

= $ - \int\limits_5^{x + 5} {g\left( t \right)} dt$

= $\int\limits_{x + 5}^5 {g(t)dt} $

$g\left( {f\left( x \right)} \right)$ = $\ln \left( {f\left( x \right)} \right)$ = $\ln \left( {{{2 - x\cos x} \over {2 + x\cos x}}} \right)$

$ \therefore $ I = $\int\limits_{ - {\pi \over 4}}^{{\pi \over 4}} {\ln \left( {{{2 - x\cos x} \over {2 + x\cos x}}} \right)dx} $

( Using property $\int\limits_{ - a}^a {f\left( x \right)} dx = \int\limits_0^a {\left( {f\left( x \right) + f\left( { - x} \right)} \right)} dx$ )

I = $\int\limits_0^{{\pi \over 4}} {\left( {\ln \left( {{{2 - x\cos x} \over {2 + x\cos x}}} \right) + \ln \left( {{{2 + x\cos x} \over {2 - x\cos x}}} \right)} \right)dx} $

= $\int\limits_0^{{\pi \over 4}} {\left( {\ln \left( 1 \right)} \right)dx} $

= 0 = log_e1

$\mathop {\lim }\limits_{x \to \infty } \sum\limits_{r = 1}^{2n} {{n \over {{n^2} + {r^2}}}} $

$\mathop {\lim }\limits_{x \to \infty } \sum\limits_{r = 1}^{2n} {{1 \over {n\left( {1 + {{{r^2}} \over {{n^2}}}} \right)}} = \int\limits_0^2 {{{dx} \over {1 + {x^2}}}} } = {\tan ^{ - 1}}2$

$\int\limits_1^e {\left\{ {{{\left( {{x \over e}} \right)}^{2x}} - {{\left( {{e \over x}} \right)}^x}} \right\}} \,$

Let  ${\left( {{x \over e}} \right)^{2x}} = t,{\left( {{e \over x}} \right)^x} = v$

$ = {1 \over 2}\int\limits_{{{\left( {{1 \over e}} \right)}^2}}^1 {dt + \int\limits_e^1 {dv} } $

$ = {1 \over 2}\left( {1 - {1 \over {{e^2}}}} \right) + \left( {1 - e} \right) = {3 \over 2} - {1 \over {2{e^2}}} - e$

${\rm I} = \int_0^a {f\left( x \right)g\left( x \right)dx} $

${\rm I} = \int_0^a {f\left( {a - x} \right)g\left( {a - x} \right)dx} $

${\rm I} = \int_0^a {f\left( x \right)\left( {4 - g\left( x \right.} \right)dx} $

${\rm I} = 4\int_0^a {f\left( x \right)dx - {\rm I}} $

$ \Rightarrow {\rm I} = 2\int_0^a {f\left( x \right)dx} $

I $=$ $\int\limits_{\pi /6}^{\pi /4} {{{dx} \over {\sin 2x\left( {{{\tan }^5}x + {{\cot }^5}x} \right)}}} $

${\rm I} = {1 \over 2}\int\limits_{\pi /6}^{\pi /4} {{{{{\tan }^4}x{{\sec }^2}xdx} \over {\left( {1 + {{\tan }^{10}}x} \right)}}} $   Put tan⁵x $=$ t

${\rm I} = {1 \over {10}}\int\limits_{{{\left( {{1 \over {\sqrt 3 }}} \right)}^5}}^1 {{{dt} \over {1 + {t^2}}}} = {1 \over {10}}\left( {{\pi \over 4} - {{\tan }^{ - 1}}{1 \over {9\sqrt 3 }}} \right)$

I $=$ $\int\limits_{ - 2}^2 {{{{{\sin }^2}x} \over { \left[ {{x \over \pi }} \right] + {1 \over 2}}}} \,dx$

${\rm I} = \int\limits_0^2 {\left( {{{{{\sin }^2}x} \over {\left[ {{x \over \pi }} \right] + {1 \over 2}}} + {{{{\sin }^2}\left( { - x} \right)} \over {\left[ { - {x \over \pi }} \right] + {1 \over 2}}}} \right)dx} $

$\left( {\left[ {{x \over \pi }} \right] + \left[ { - {x \over \pi }} \right] = - 1\,\,} \right.$   as   $\left. {\matrix{ \, \cr \, \cr } x \ne n\pi } \right)$

${\rm I} = \int\limits_0^2 {\left( {{{{{\sin }^2}x} \over {\left[ {{x \over \pi }} \right] + {1 \over 2}}} + {{{{\sin }^2}x} \over { - 1 - \left[ {{x \over \pi }} \right] + {1 \over 2}}}} \right)dx = 0} $

${\rm I} = \int\limits_{{{ - \pi } \over 2}}^{{\pi \over 2}} {{{dx} \over {\left[ x \right] + \left[ {\sin x} \right] + 4}}} $

$ = \int\limits_{{{ - \pi } \over 2}}^{ - 1} {{{dx} \over { - 2 - 1 + 4}}} + \int\limits_{ - 1}^0 {{{dx} \over { - 1 - 1 + 4}}} $

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \int\limits_0^1 {{{dx} \over {0 + 0 + 4}} + \int\limits_1^{{\pi \over 2}} {{{dx} \over {1 + 0 + 4}}} } $

$\int\limits_{{{ - \pi } \over 2}}^{ - 1} {{{dx} \over 1} + \int\limits_{ - 1}^0 {{{dx} \over 2}} } + \int\limits_0^1 {{{dx} \over 4}} + \int\limits_1^{{\pi \over 2}} {{{dx} \over 5}} $

$\left( { - 1 + {\pi \over 2}} \right) + {1 \over 2}\left( {0 + 1} \right) + {1 \over 4} + {1 \over 5}\left( {{\pi \over 2} - 1} \right)$

$ - 1 + {1 \over 2} + {1 \over 4} - {1 \over 5} + {\pi \over 2} + {\pi \over {10}}$

${{ - 20 + 10 + 5 - 4} \over {20}} + {{6\pi } \over {10}}$

${{ - 9} \over {20}} + {{3\pi } \over 5}$

$\int\limits_0^x \, $f(t) dt = x² + $\int\limits_x^1 \, $ t²f(t) dt           f '$\left( {{1 \over 2}} \right)$ = ?

Differentiate w.r.t. 'x'

f(x) = 2x + 0 $-$ x² f(x)

f(x) = ${{2x} \over {1 + {x^2}}}$ $ \Rightarrow $ f '(x) = ${{\left( {1 + {x^2}} \right)2 - 2x\left( {2x} \right)} \over {{{\left( {1 + {x^2}} \right)}^2}}}$

f '(x) = ${{2{x^2} - 4{x^2} + 2} \over {{{\left( {1 + {x^2}} \right)}^2}}}$

f '$\left( {{1 \over 2}} \right) = {{2 - 2\left( {{1 \over 4}} \right)} \over {{{\left( {1 + {1 \over 4}} \right)}^2}}} = {{\left( {{3 \over 2}} \right)} \over {{{25} \over {16}}}} = {{48} \over {50}} = {{24} \over {25}}$

Let f(x) = x²(x² $-$ 2)


As long as f(x) lie below the x-axis, define integral will remain negative,
so correct value of (a, b) is ($-$ $\sqrt 2 $, $\sqrt 2 $) for minimum of I

$\int\limits_0^{{\pi \over 3}} {{{\tan \theta } \over {\sqrt {2k\,\sec \theta } }}} \,d\theta = 1 - {1 \over {\sqrt 2 }},\left( {k > 0} \right),$

$ \Rightarrow \,\,\int\limits_0^{\pi /3} {{{\tan \theta } \over {\sqrt {2k\sec \theta } }}} \,d\theta = 1 - {1 \over {\sqrt 2 }}\left( {k > 0} \right)$

$ \Rightarrow \,\,{1 \over {\sqrt {2k} }}\left( { - 2\sqrt {\cos \theta } } \right)_0^{\pi /3} = 1 - {1 \over {\sqrt 2 }}$

$ \Rightarrow \,\,{1 \over {\sqrt {2k} }}\left( { - \sqrt 2 + 2} \right) = 1 - {1 \over {\sqrt 2 }}$

$ \Rightarrow \,\,{{\sqrt 2 \left( {\sqrt 2 - 1} \right)} \over {\sqrt {2k} }} = {{\sqrt 2 - 1} \over {\sqrt 2 }}$

$ \Rightarrow \,\,k = 2$

$\left| {f(x) - f(y)} \right| \le 2{\left[ {x - y} \right]^{3/2}}$

$\left| {{{f(x) - f(y)} \over {x - y}}} \right| \le 2{\left| {x - y} \right|^{1/2}}$

$\mathop {\lim }\limits_{y \to x} \left| {{{f(x) - f(y)} \over {x - y}}} \right| \le \mathop {\lim }\limits_{y \to x} 2{\left| {x - y} \right|^{1/2}}$

$ \Rightarrow \left| {f'\left( x \right)} \right| \le 0$  $ \Rightarrow f'\left( x \right) = 0$

$ \Rightarrow f\left( x \right) = $ constant

as  $f\left( 0 \right) = 1 \Rightarrow f\left( x \right) = 1$

$\int\limits_0^1 {{f^2}} \left( x \right)dx = 1$

$\int\limits_0^\pi {{{\left| {\cos x} \right|}^3}} \,dx$

The period of $\left| {\cos x} \right|$ = ${\pi \over 2}$

$ \therefore $  I = 2 $\int\limits_0^{{\pi \over 2}} {{{\left| {\cos x} \right|}^3}} \,dx$

as in the range 0 to ${\pi \over 2}$ $\left| {\cos x} \right|$ is positive.

So,     $\left| {\cos x} \right|$ = $cosx$

$ \therefore $  I = 2 $\int\limits_0^{{\pi \over 2}} {{{\cos }^3}x\,dx} $

= 2$\int\limits_0^{{\pi \over 2}} {\left( {{{\cos 3x + 3\cos x} \over 4}} \right)} dx$

I = ${1 \over 2}\left[ {{{\sin 3x} \over 3} + 3\sin x} \right]_0^{{\pi \over 2}}$

I = ${1 \over 2}\left[ {{1 \over 3}\left( {{{3\pi } \over 2}} \right) + 3.\sin {\pi \over 2}} \right]$

I = ${1 \over 2}\left[ { - {1 \over 3} + 3} \right]$

= ${1 \over 2}\left( {{8 \over 3}} \right)$

= ${4 \over 3}$

f(x) = $\int_0^x {t(\sin x - \sin t).dt} $

= sin x$\int_0^x {t.dt - \int_0^x {t\sin t.dt} } $

= ${{{x^2}} \over 2}$ sin x +$\left[ {t\cos t} \right]_0^x$ + sin x

$ \Rightarrow $f(x) = ${{{x^2}} \over 2}$ sinx + xcosx + sinx

f'(x) = ${{{x^2}} \over 2}$ cosx + 2cos x

f''(x) = x cos x $-$ ${{{x^2}} \over 2}$ sin x $-$ 2sin x

f'''(x) = cos x $-$ 2x sin x $-$ ${{{x^2}} \over 2}$ cos x $-$ 2cos x

$\therefore\,\,\,$ f'''(x) + f'(x) = cos x $-$ 2x sin x

As we know,

$\int\limits_a^b { + \left( x \right)dx} = \int\limits_a^b {f\left( {a + b - x} \right)} dx$

Let $I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {\,\,{{{{\sin }^2}x} \over {1 + {2^x}}}} \,dx$

$ \Rightarrow \,\,\,\,I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {\,\,{{{{\sin }^2}\left( { - {\pi \over 2} + {\pi \over 2} - x} \right)} \over {1 + {2^{ - {\pi \over 2} + {\pi \over 2} - x}}}}} $

$ \Rightarrow\,\,\,\, I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{{{\sin }^2}x} \over {1 + {2^{ - x}}}}} dx$

$ \Rightarrow \,\,\,\,I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{{2^x}{{\sin }^2}x} \over {1 + {2^x}}}} \,\,dx$

$\therefore\,\,\,$ $2I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {\left( {{{{{\sin }^2}x} \over {1 + {2^x}}} + {{{2^x}{{\sin }^2}x} \over {1 + {2^x}}}} \right)} \,\,dx$

$ \Rightarrow \,\,\,\,2I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{\sin }^2}x\left( {{{1 + {2^x}} \over {1 + {2^x}}}} \right)} \,\,dx$

$ \Rightarrow \,\,\,\,2I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{\sin }^2}x} \,\,dx$

$ \Rightarrow \,\,\,\,2I = 2\int\limits_0^{{\pi \over 2}} {{{\sin }^2}x} \,\,dx\,\,\,$ [ as sin² x is an even function ]

$ \Rightarrow \,\,\,\,I = \int\limits_0^{{\pi \over 2}} {{{\sin }^2}x} \,\,dx\,\,\,\,$

[ as $\,\,\,\,$ $\int\limits_0^a {f\left( a \right)\,\,dx = \int\limits_0^a {f\left( {a - x} \right)\,dx]} } $

So, $\,\,\,$ $I = \int\limits_0^{{\pi \over 2}} {{{\sin }^2}\left( {{b \over 2} - x} \right)} \,\,dx\,\,\,\,$

$\,\,\,I = \int\limits_0^{{\pi \over 2}} {{{\cos }^2}x} \,\,dx\,\,\,\,$

$\therefore\,\,\,\,$ $\,\,\,\,2I = \int\limits_0^{{\pi \over 2}} {\left( {{{\sin }^2}x + {{\cos }^2}x} \right)} \,\,dx\,\,\,\,$

$ \Rightarrow \,\,\,\,2I = \int\limits_0^{{\pi \over 2}} \, \,\,dx\,\,\,\,$

$ \Rightarrow \,\,\,\,2I = {\left[ x \right]_0}^{{\pi \over 2}}$

$ \Rightarrow \,\,\,\,2I = {\pi \over 2}$

$ \Rightarrow \,\,\,\,I = {\pi \over 4}$

Given,

${I_1} = \int_0^1 {{e^{ - x}}} {\cos ^2}x{\mkern 1mu} dx;$

${I_2} = \int_0^1 {{e^{ - {x^2}}}} {\cos ^2}x{\mkern 1mu} dx$ and

${I_3} = \int_0^1 {{e^{ - {x^3}}}} dx$

For x $ \in $ (0, 1)

$ \Rightarrow $  x > x² or $-$ x < $-$ x²

and x² > x³ or $-$ x² < $-$ x³

$ \therefore $ ${e^{ - {x^2}}}$ < ${e^{ - {x^3}}}$ and ${e^{ - x}}$ < ${e^{ - {x^2}}}$

$ \Rightarrow $ ${e^{ - x}} < {e^{ - {x^2}}} < {e^{ - {x^3}}}$

$ \Rightarrow $ ${e^{ - {x^3}}} > {e^{ - {x^2}}} > {e^{ - x}}$

$ \Rightarrow $ ${I_3} > {I_2} > {I_1}$

Let  $I = \int_{{\pi \over 4}}^{{{3\pi } \over 4}} {{x \over {1 + \sin x}}dx} $

also let K = ${x \over {1 + \sin x}}$

Multiplying numerator and denominator by
(1 $-$ sin x) we get;

$K = {{x\left( {1 - \sin x} \right)} \over {1 - {{\left( {\sin x} \right)}^2}}} = {{x(1 - \sin x)} \over {{{\left( {\cos x} \right)}^2}}}$

= x(1 $-$ sin x) sec²x

= x sec²x $-$ x sin x sec²x = x sec²x $-$ x tan x sec x

Now, $I = \int_{{\pi \over 4}}^{{{3\pi } \over 4}} {x{{\sec }^2}xdx - \int_{{\pi \over 4}}^{{{3\pi } \over 4}} {x\sec x\,\tan \,xdx} } $

$ \Rightarrow $ I = $\left| {x\tan x} \right|_{{\pi \over 4}}^{{{3\pi } \over 4}} - \left[ {\log \left( {\sec x} \right)} \right]_{{\pi \over 4}}^{{{3\pi } \over 4}}$

- $\left| {{x \over {\cos x}}} \right|_{{\pi \over 4}}^{{{3\pi } \over 4}} - \int\limits_{{\pi \over 4}}^{{{3\pi } \over 4}} {{{dx} \over {\cos x}}} $

$ \Rightarrow $ I = $\left| {x\tan x} \right|_{{\pi \over 4}}^{{{3\pi } \over 4}}$ - $\left| {{x \over {\cos x}}} \right|_{{\pi \over 4}}^{{{3\pi } \over 4}}$

$ - \left[ {\log \left( {\sec x + \tan x} \right)} \right]_{{\pi \over 4}}^{{{3\pi } \over 4}}$

$ \Rightarrow $ I = $ - {{3\pi } \over 4} - {\pi \over 4}$$ + {{3\pi } \over 4}\left( {\sqrt 2 } \right)$$ + {\pi \over 4}\left( {\sqrt 2 } \right)$

$ - \log \left( {\sqrt 2 + 1} \right)$$ + \log \left( {\sqrt 2 + 1} \right)$

$ \Rightarrow $ I = -$\pi $ + ${\sqrt 2 \pi }$

$ \Rightarrow $ I = $\pi \left( {\sqrt 2 - 1} \right)$

Let

I = $\int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{\sin }^4}} x\left( {1 + \log \left( {{{2 + \sin x} \over {2 - \sin x}}} \right)} \right)dx$ ......(1)

$ \Rightarrow $ I = $\int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{\sin }^4}}(- x)\left( {1 + \log \left( {{{2 + \sin (- x)} \over {2 - \sin(- x)}}} \right)} \right)dx$

[ As $\int\limits_a^b { f \left( x \right)dx} = \int\limits_a^b {f\left( {a + b - x} \right)} dx$ ]

     = $\int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{\sin }^4}} x\left( {1 + \log \left( {{{2 - \sin x} \over {2 + \sin x}}} \right)} \right)dx$

     = $\int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{\sin }^4}} x\left( {1 - \log \left( {{{2 + \sin x} \over {2 - \sin x}}} \right)} \right)dx$ .......(2)

Adding equation (1) and (2) we get,

2I = 2$\int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{\sin }^4}} xdx$

2I = 4$\int\limits_0^{{\pi \over 2}} {{{\sin }^4}} xdx$

I = 2$\int\limits_0^{{\pi \over 2}} {{{\sin }^4}} xdx$

  = ${{2 \times {3 \over 2} \times {1 \over 2} \times \pi } \over {2 \times 2}}$ [ Using Gamma function ]

  = ${{3\pi } \over 8}$

$\mathop {\lim }\limits_{n \to \infty } {{{1 \over {\left( {a + 1} \right)}}\,.\,{n^{a + 1}} + {a_1}{n^a} + {a_2}{n^{a - 1}} + .\,.\,.\,.} \over {{{\left( {n + 1} \right)}^{a - 1}}.\,{n^2}\left( {a + {{1 + {1 \over n}} \over 2}} \right)}} = {1 \over {60}}$

$ \Rightarrow $   $\mathop {\lim }\limits_{n \to \infty } {{{{\left( {{1 \over n}} \right)}^2} + {{\left( {{2 \over n}} \right)}^a} + ....... + {{\left( {{n \over n}} \right)}^a}} \over {{{\left( {n + 1} \right)}^{a - 1}}\left[ {{a^2}a + {{n\left( {n + 1} \right)} \over 2}} \right]}} = {1 \over {60}}$

        $ = {{\mathop {\lim }\limits_{n \to \infty } {1 \over n}\sum\limits_{r = 1}^n {{{\left( {{r \over n}} \right)}^a}} } \over {{{\left( {1 + {1 \over n}} \right)}^{a - 1}}\left[ {a + {1 \over 2}\left( {1 + {1 \over n}} \right)} \right]}} = {1 \over {60}}$

        $ = {{\int_0^1 {{x^a}dx} } \over {\left( {a + {1 \over 2}} \right)}} = {1 \over {60}} = {{{1 \over {a + 1}}} \over {a + {1 \over 2}}} = {1 \over {60}}$

$ \Rightarrow $  ${{{1 \over {a + 1}}} \over {\left( {a + {1 \over 2}} \right)}} = {1 \over {60}} \Rightarrow \left( {a + 1} \right)\left( {2a + 1} \right) = 120$

$ \Rightarrow $  2a² + 3a $-$ 119 $=$ 0

$ \Rightarrow $   2a² + 17a $-$ 14a $-$ 119 $=$ 0

$ \Rightarrow $   (a $-$ 7) (2a + 17) $=$ 0

$ \Rightarrow $   a $=$ 7, $-$ ${{17} \over 2}$

Given, I = $\int\limits_1^2 {{{dx} \over {{{\left( {{x^2} - 2x + 4} \right)}^{{3 \over 2}}}}}} $

= $\int\limits_1^2 {{{dx} \over {{{\left[ {{{\left( {x - 1} \right)}^2} + 3} \right]}^{{3 \over 2}}}}}} $

Let x $-$ 1 = $\sqrt 3 $ tan$\theta $

$ \Rightarrow $$\,\,\,$ dx = $\sqrt 3 $ sec²$\theta $ d$\theta $

When x = 1, then $\theta $ = 0

and when x = 2, $\theta $ = ${\pi \over 6}$

I = $\int\limits_0^{{\pi \over 6}} {{{\sqrt 3 {{\sec }^2}\theta \,d\theta } \over {{{\left( {3{{\tan }^2}\theta + 3} \right)}^{{3 \over 2}}}}}} $

= $\int\limits_0^{{\pi \over 6}} {{{\sqrt 3 {{\sec }^2}\theta \,d\theta } \over {{{\left( {3{{\sec }^2}\theta } \right)}^{{3 \over 2}}}}}} $

= $\int\limits_0^{{\pi \over 6}} {{{\sqrt 3 {{\sec }^2}\theta \,d\theta } \over {3\sqrt 3 {{\sec }^3}\theta }}} $

= ${1 \over 3}$ $\int\limits_0^{{\pi \over 6}} {{{d\theta } \over {\sec \theta }}} $

= ${1 \over 3}\int\limits_0^{{\pi \over 6}} {\cos \theta \,d\theta } $

= ${1 \over 3}{\left[ {\sin \theta } \right]_\theta }^{{\pi \over 6}}$

= ${1 \over 3} \times {1 \over 2}$

= ${1 \over 6}$

$\therefore\,\,\,$ According to the question,

${k \over {k + 5}}$ = ${1 \over 6}$

$ \Rightarrow $$\,\,\,$ 6k = k + 5

$ \Rightarrow $$\,\,\,$ k = 1

tan x  +  cot x

= ${{\sin x} \over {\cos x}}$ + ${{\cos x} \over {\sin x}}$

= ${{{{\sin }^2}x + {{\cos }^2}x} \over {\sin x\,\,\cos x}}$

= ${1 \over {\sin x\,\,\cos x}}$

= ${2 \over {\sin 2x}}$

$\therefore\,\,\,$ $\int\limits_{{\pi \over {12}}}^{{\pi \over 4}} {{{8\cos 2x} \over {{{\left( {\tan x + \cot x} \right)}^3}}}} \,\,dx$

= $\int\limits_{{\pi \over {12}}}^{{\pi \over 4}} {{{8\cos 2x} \over {{{\left( {{2 \over {\sin 2x}}} \right)}^3}}}} \,\,dx$

= $\int\limits_{{\pi \over {12}}}^{{\pi \over 4}} {{{8{{\sin }^3}\,2x.\cos \,\,2x} \over 8}} \,\,dx$

= $\int\limits_{{\pi \over {12}}}^{{\pi \over 4}} {{{\sin }^3}2x.\cos \,2x} \,\,dx$

Let sin 2x = t

$ \Rightarrow $$\,\,\,$ 2 cos2x dx = dt

At    x = ${{\pi \over {12}}}$, t = sin ${{\pi \over {6}}}$ = ${1 \over 2}$

At   x = ${{\pi \over 4}}$, t = sin ${{\pi \over 2}}$ = 1.

= ${1 \over 2}$ $\int\limits_{{1 \over 2}}^1 {{t^3}.dt} $

= ${{1 \over 2}}$ $\left[ {{{{t^4}} \over 4}} \right]_{{1 \over 2}}^1$

= ${{1 \over 2}}$ $ \times $ ${{1 \over 4}}$ $ \times $ [ 1⁴ $-$ (${{1 \over 2}}$)⁴]

= ${{1 \over 8}}$ $ \times $ ${{15 \over 16}}$ = ${{15 \over 128}}$

$\int\limits_{{\pi \over 4}}^{{{3\pi } \over 4}} {{{dx} \over {1 + \cos x}}} $

= $\int\limits_{{\pi \over 4}}^{{{3\pi } \over 4}} {{{dx} \over {2{{\cos }^2}{x \over 2}}}} $

= ${1 \over 2}\int\limits_{{\pi \over 4}}^{{{3\pi } \over 4}} {{{\sec }^2}} {x \over 2}\,dx$

${1 \over 2}\left[ {{{\tan {x \over 2}} \over {{1 \over 2}}}} \right]_{{\pi \over 4}}^{{{3\pi } \over 4}}$

= $\left[ {\tan {x \over 2}} \right]_{{\pi \over 4}}^{{3 \over 4}}$

= tan ${{3\pi } \over 8}$ $-$ tan${\pi \over 8}$

= $\left( {\sqrt 2 + 1} \right) - \left( {\sqrt 2 - 1} \right)$

= 2

$x\int\limits_1^x {y\left( t \right)dt} = x\int\limits_1^x {ty} \left( t \right)dt + \int\limits_1^x {ty\left( t \right)} \,\,dt$

Differentiate w.r. to x.

$\int\limits_1^x {y\left( t \right)dt + x\left[ {y\left( x \right) - y\left( 1 \right)} \right]} $

$ = \int\limits_1^x {ty\left( t \right)dt + x\left[ {xy\left( x \right) - y\left( 1 \right)} \right] + xy\left( x \right) - y\left( 1 \right)} $

$\int\limits_1^x {y\left( t \right)dt = \int\limits_1^x {ty\left( t \right)dt + {x^2}y\left( x \right) - y\left( 1 \right)} } $

Diff. again w.r.t to x

$y\left( x \right) - y\left( a \right) = xy\left( x \right) - y\left( a \right) + 2x\,y\left( x \right) + {x^2}{y^3}\left( x \right)$
$\left( {1 - 3x} \right)y\left( x \right) = {x^2}{y^3}\,\left( x \right)$

${{{y^3}\left( x \right)} \over {y\left( x \right)}} = {{1 - 3x} \over {{x^2}}}$

${{1dy} \over {ydx}} = {{1 - 3x} \over {{x^2}}} \Rightarrow \ln \,y = - {1 \over x} - 3\ln x$

$\ln \left( {y{x^3}} \right) = - {1 \over x}$

$y{x^3} = - {e^{ - 1/x}}$

$y = {{{e^{ - {1 \over x}}}} \over {{x^3}}}$

or   $y = {{c{e^{ - {1 \over x}}}} \over {{x^3}}}$

Let   I   =   $\int\limits_4^{10} {{{\left[ {{x^2}} \right]\,dx} \over {\left[ {{x^2} - 28x + 196} \right] + \left[ {{x^2}} \right]}}} $

  =   $\int\limits_4^{10} {{{\left[ {{x^2}} \right]dx} \over {\left[ {{{\left( {x - 14} \right)}^2}} \right] + \left[ {{x^2}} \right]}}} \,\,\,.....(1)$

Using,

$\int\limits_a^b {f\left( {a + b - x} \right)dx\,} $ = $\,\,\int\limits_a^b {f(x)\,\,dx} $

I   =   $\int\limits_4^{10} {{{{{\left( {x - 14} \right)}^2}} \over {\left[ {{x^2}} \right] + \left[ {{{\left( {x - 14} \right)}^2}} \right]}}} \,dx\,\,....(2)$

Adding (1) and (2)

2I   =   $\int\limits_4^{10} {{{\left[ {{{\left( {x - 14} \right)}^2}} \right] + \left[ {{x^2}} \right]} \over {\left[ {{x^2}} \right] + \left[ {{{\left( {x - 14} \right)}^2}} \right]}}} \,dx$

$ \Rightarrow $$\,\,\,$ 2I   =  $\int\limits_4^{10} {dx} = \left[ x \right]_4^{10}$ = 6

=   I = 3

Given,

$2\int\limits_0^1 {{{\tan }^{ - {1_x}\,{d_x}}}} = \int\limits_0^1 {{{\cot }^{ - 1}}} \left( {1 - x + {x^2}} \right)dx$

=  $\int\limits_0^1 {\left( {{\pi \over 2} - {{\tan }^{ - 1}}\left( {1 - x + {x^2}} \right)} \right)} dx$

$ \Rightarrow $$\,\,\,$$2\int\limits_0^1 {{{\tan }^{ - 1}}} x\,dx = \int\limits_0^1 {{\pi \over 2}} dx - \int\limits_0^1 {{{\tan }^{ - 1}}} \left( {1 - x + {x^2}} \right)dx$

$ \Rightarrow $$\,\,\,$ $\int\limits_0^1 {{{\tan }^{ - 1}}} \left( {1 - x + {x^2}} \right)dx = \int\limits_0^1 {{\pi \over 2}} dx - 2\int\limits_0^1 {{{\tan }^{ - 1}}x\,dx} $

$ \Rightarrow $$\,\,\,$ $\int\limits_0^1 {{{\tan }^{ - 1}}} \left( {1 - x + {x^2}} \right)dx = {\pi \over 2} - 2\int\limits_0^1 {{{\tan }^{ - 1}}x\,\,dx} $

Let  I = $\int\limits_0^1 {{{\tan }^{ - 1}}} x\,dx$

=   $\left[ {\left( {{{\tan }^{ - 1}}x} \right)x} \right]_0^1 - \int\limits_0^1 {{1 \over {1 + {x^2}}}} x\,dx$

=  ${\pi \over 4} - {1 \over {20}}\left[ {\log \left| {1 + {x^2}} \right|} \right]_0^1$

=  ${\pi \over 4} - {1 \over 2}\log 2$

$\therefore\,\,\,$ $\int\limits_0^1 {{{\tan }^{ - 1}}} \left( {1 - x + {x^2}} \right)\,dx$

=  ${\pi \over 2} - 2\left( {{\pi \over 4} - {1 \over 2}\log 2} \right)$

=  ${\pi \over 2} - {\pi \over 2} + \log 2$

=   log2

$y = \mathop {\lim }\limits_{n \to \infty } {\left( {{{\left( {n + 1} \right)\left( {n + 2} \right)...3n} \over {{n^{2n}}}}} \right)^{{1 \over n {}}}}$

$\ln \,y = \mathop {\lim }\limits_{n \to \infty } {1 \over n}\ln \left( {1 + {1 \over n}} \right)\left( {1 + {2 \over n}} \right)$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,....\left( {1 + {2 \over n}} \right)$

$\ln {\mkern 1mu} y = \mathop {\lim }\limits_{n \to \infty } {1 \over n}\left[ {\ln \left( {1 + {1 \over n}} \right) + \ln \left( {1 + {2 \over n}} \right)} \right.$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left. { + .... + \ln \left( {1 + {{2n} \over n}} \right)} \right]$

$ = \mathop {\lim }\limits_{n \to \infty } {1 \over n}\sum\limits_{r = 1}^{2n} {\ln \left( {1 + {r \over n}} \right)} $

$ = \int_0^2 {\ln \left( {1 + x} \right)dx} $

Let $1 + x = t \Rightarrow dx = dt$

when $x = 0,t = 1$

$\,\,\,\,\,\,\,\,\,\,\,\,\,x = 2,\,\,t = 3$

$\ln \,y = \int_1^3 {\ln t\,d\,t = } \left[ {t\,\ln \,t - \left. {t_1^3} \right]} \right.$

$\,\,\,\,\,\,\,\,\,\,\, = \ln \left( {{{{3^3}} \over {{e^2}}}} \right) = \ln \left( {{{27} \over {{e^2}}}} \right)$

$ \Rightarrow y = {{27} \over {{e^2}}}$

$I = \int\limits_2^4 {{{\log {x^2}} \over {\log {x^2} + \log \left( {36 - 12x + {x^2}} \right)}}} $

$I = \int\limits_2^4 {{{\log {x^2}} \over {\log {x^2} + \log {{\left( {6 - x} \right)}^2}}}} \,\,\,\,\,\,\,\,\,\,...\left( i \right)$

$I = \int\limits_2^4 {{{\log {{\left( {6 - x} \right)}^2}} \over {\log {{\left( {6 - x} \right)}^2} + \log {x^2}}}} \,\,\,\,\,\,\,\,...\left( {ii} \right)$

Adding $(1)$ and $(2)$

$2I = \int\limits_2^4 {dx = \left[ x \right]_2^4} = 2 \Rightarrow 1 - 1$

Let $I = \int\limits_0^\pi {\sqrt {1 + 4{{\sin }^2}{x \over 2} - 4\sin {x \over 2}} } dx$

$ = \int\limits_0^\pi {\left| {2\sin {x \over 2} - 1} \right|} dx$

$ = \int\limits_0^{\pi /3} {\left( {1 - 2\sin {x \over 2}} \right)} dx + \int\limits_{\pi /3}^\pi {\left( {2\sin {x \over 2} - 1} \right)} dx$

$\left[ {} \right.$ As $\sin {x \over 2} = {1 \over 2} \Rightarrow {x \over 2} = {\pi \over 6} \Rightarrow x = {\pi \over 3},$ ${x \over 2}$

$\,\,\,\,\,\,\,\,\,\,\, = {{5\pi } \over 6} \Rightarrow x = {{5\pi } \over 3}$ $\left. {} \right]$

$ = \left[ {x + 4\cos {x \over 2}} \right]_0^{\pi /3} + \left[ { - 4\cos {x \over 2} - x} \right]_{\pi /3}^\pi $

$ = {\pi \over 3} + 4{{\sqrt 2 } \over 2} - 4 + \left( {0 - \pi + 4{{\sqrt 3 } \over 2} + {\pi \over 3}} \right)$

$ = 4\sqrt 3 - 4 - {\pi \over 3}$

Let $I = \int\limits_{\pi /6}^{\pi /3} {{{dx} \over {1 + \sqrt {\tan \,x} }}} $

$ = \int\limits_{\pi /6}^{\pi /3} {{{dx} \over {\sqrt {\tan \left( {{\pi \over 2} - x} \right)} }}} $

$ = \int\limits_{\pi /6}^{\pi /3} {{{\sqrt {\tan \,x} \,dx} \over {1 + \sqrt {\tan \,x} }}} \,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$

Also, Given,

$I$ $ = \int\limits_{\pi /6}^{\pi /3} {{{\sqrt {\tan \,x} \,dx} \over {1 + \sqrt {\tan \,x} }}} \,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$

By adding $(1)$ and $(2),$ we get

$2I = \int\limits_{\pi /6}^{\pi /3} {dx} $

$ \Rightarrow I = {1 \over 2}\left[ {{\pi \over 3} - {\pi \over 6}} \right]$

$ = {\pi \over {12}},$ statements -$1$ is false

$\int\limits_a^b {f\left( x \right)dx = \int\limits_a^b {f\left( {a + b - x} \right)dx} } $

It is fundamental property.

$\left( {b,c} \right)g\left( {x + \pi } \right) = \int\limits_0^{x + \pi } {\cos \,4t\,dt} $

$ = \int\limits_0^\pi {\cos 4tdt + \int\limits_\pi ^{\pi + x} {\cos 4t\,dt} } $

$ = g\left( \pi \right) + \int\limits_0^x {\cos \,4t\,dt} $

Putting $t = \pi + y$ in second integral, we get

$\int\limits_x^{\pi + x} {\cos \,4t\,dt = \int\limits_0^\pi {\cos \,4t\,dt} } $

$ = g\left( \pi \right) + g\left( x \right) = g\left( x \right) - g\left( \pi \right)$

$\therefore$ $\left. {g\left( \pi \right) = 0} \right)$