Definite Integration

$f(x) = \int_0^{{x^2}} {{{{t^2} - 5t + 4} \over {2 + {e^t}}}dt} $

$f'(x) = 2x\left( {{{{x^4} - 5{x^2} + 4} \over {2 + {e^{{x^2}}}}}} \right) = 0$

$x = 0$, or $({x^2} - 4)({x^2} - 1) = 0$

$x = 0,$ $x = \pm 2,\, \pm 1$

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

f'(x) changes sign from positive to negative at x = $-$1, 1 So, number of local maximum points = 2

f'(x) changes sign from negative to positive at x = $-$2, 0, 2 So, number of local minimum points = 3

$\therefore$ m = 2, n = 3

$\int\limits_{\cos x}^1 {{t^2}f(t)dt = {{\sin }^3}x + \cos x} $

$ \Rightarrow \sin x{\cos ^2}x\,f(\cos x) = 3{\sin ^2}x\cos x - \sin x$

$ \Rightarrow f(\cos x) = 3\tan x - {\sec ^2}x$

$ \Rightarrow f'(\cos x).\,( - \sin x) = 3{\sec ^2}x - 2{\sec ^2}x\tan x$

Put $\cos x = {1 \over {\sqrt 3 }}$,

$\therefore$ $f'\left( {{1 \over {\sqrt 3 }}} \right)\left( { - {{\sqrt 2 } \over {\sqrt 3 }}} \right) = 9 - 6\sqrt 2 $

${1 \over {\sqrt 3 }}f'\left( {{1 \over {\sqrt 3 }}} \right) = 6 - {9 \over {\sqrt 2 }}$

$\int_0^1 {{1 \over {{7^{\left[ {{1 \over x}} \right]}}}}dx} $, let ${1 \over x} = t$

${{ - 1} \over {{x^2}}}dx = dt$

$ = \int_\infty ^1 {{1 \over { - {t^2}{7^{[t]}}}}dt = \int_1^\infty {{1 \over {{t^2}{7^{[t]}}}}dt} } $

$ = \int_1^2 {{1 \over {7{t^2}}}dt + \int_2^3 {{1 \over {{7^2}{t^2}}}dt + \,\,....} } $

$ = {1 \over 7}\left[ { - {1 \over t}} \right]_1^2 + {1 \over {{7^2}}}\left[ {{{ - 1} \over t}} \right]_2^3 + {1 \over {{7^3}}}\left[ { - {1 \over t}} \right]_2^3 + \,\,....$

$ = \sum\limits_{n = 1}^\infty {{1 \over {{7^n}}}\left( {{1 \over n} - {1 \over {n + 1}}} \right)} $

$ = \sum\limits_{n = 1}^\infty {{{{{\left( {{1 \over 7}} \right)}^n}} \over n} - 7\sum\limits_{n = 1}^\infty {{{{{\left( {{1 \over 7}} \right)}^{n + 1}}} \over {n + 1}}} } $

$ = - \log \left( {1 - {1 \over 7}} \right) + 7\log \left( {1 - {1 \over 7}} \right) + 1$

$ = 1 + 6\log {6 \over 7}$

$I = \int\limits_{ - 2}^2 {{{|{x^3} + x|} \over {{e^{x|x|}} + 1}}dx} $ ..... (i)

$I = \int\limits_{ - 2}^2 {{{|{x^3} + x|} \over {{e^{ - x|x|}} + 1}}dx} $ ..... (ii)

$2I = \int\limits_{ - 2}^2 {|{x^3} + x|dx} $

$2I =2 \int\limits_0^2 {({x^3} + x)dx} $

$I = \int\limits_0^2 {({x^3} + x)dx} $

$ = \left. {{{{x^4}} \over 4} + {{{x^2}} \over 2}} \right]_0^2$

$ = \left( {{{16} \over 4} + {4 \over 2}} \right) - 0$

$ = 4 + 2 = 6$

${b_n} - {b_{n - 1}} = \int_0^{{\pi \over 2}} {{{{{\cos }^2}nx - {{\cos }^2}(n - 1)x} \over {\sin x}}dx} $

$ = \int_0^{{\pi \over 2}} {{{ - \sin (2n - 1)x\,.\,\sin x} \over {\sin x}}dx} $

$ = \left. {{{\cos (2n - 1)x} \over {2n - 1}}} \right|_0^{\pi /2} = - {1 \over {2n - 1}}$

So, ${b_3} - {b_2}$, ${b_4} - {b_3}$, ${b_5} - {b_4}$ are in H.P.

$ \Rightarrow {1 \over {{b_3} - {b_2}}},\,{1 \over {{b_4} - {b_3}}},\,{1 \over {{b_5} - {b_4}}}$ are in A.P. with common difference $-$2.

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

Let $\cos x = t$

$\sin xdx = dt$

$ = \int\limits_1^{ - 1} {{{ - {e^t}dt} \over {\left( {1 + {t^2}} \right)\left( {{e^t} + {e^{ - t}}} \right)}}} $

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

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

Adding (i) and (ii)

$2I = \int\limits_{ - 1}^1 {{{dt} \over {1 + {t^2}}}} $

$2I = \left. {{{\tan }^{ - t}}} \right|_{ - 1}^1$

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

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

$I = {\pi \over 4}$

$I = \int\limits_{{{ - \pi } \over 2}}^{{\pi \over 2}} {{{dx} \over {(1 + {e^x})({{\sin }^6}x + {{\cos }^6}x)}}} $ ...... (i)

$I = \int\limits_{{{ - \pi } \over 2}}^{{\pi \over 2}} {{{dx} \over {(1 + {e^{ - x}})(si{n^6}x + {{\cos }^6}x)}}} $ ..... (ii)

(i) and (ii)

From equation (i) & (ii)

$2I = \int\limits_{{{ - \pi } \over 2}}^{{\pi \over 2}} {{{dx} \over {{{\sin }^6}x + {{\cos }^6}x}}} $

$ \Rightarrow I = \int\limits_0^{{\pi \over 2}} {{{dx} \over {{{\sin }^6}x + {{\cos }^6}x}} = \int\limits_0^{{\pi \over 2}} {{{dx} \over {1 - {3 \over 4}{{\sin }^2}2x}}} } $

$ \Rightarrow I = \int\limits_0^{{\pi \over 2}} {{{4{{\sec }^2}2xdx} \over {4 + {{\tan }^2}2x}} = 2\int\limits_0^{{\pi \over 4}} {{{4{{\sec }^2}2x} \over {4 + {{\tan }^2}2x}}dx} } $

when x = 0, t = 0

Now, $\tan 2x = t$

when $x = {\pi \over 4},\,\,t \to \infty $

$2{\sec ^2}2x\,dx = dt$

$\therefore$ $I = 2\int\limits_0^\infty {{{2dt} \over {4 + {t^2}}} = 2\left( {{{\tan }^{ - 1}}{t \over 2}} \right)_0^\infty } $

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

$\mathop {\lim }\limits_{n \to \infty } \left( {{{{n^2}} \over {({n^2} + 1)(n + 1)}} + {{{n^2}} \over {({n^2} + 4)(n + 2)}} + \,\,...\,\, + \,\,{{{n^2}} \over {({n^2} + {n^2})(n + n)}}} \right)$

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

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

$ = \int\limits_0^1 {{1 \over {(1 + {x^2})(1 + x)}}dx} $

$ = {1 \over 2}\int_0^1 {\left[ {{1 \over {1 + x}} - {{(x - 1)} \over {(1 + {x^2})}}} \right]dx} $

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

$ = {1 \over 2}\left[ {{\pi \over 4} + {1 \over 2}\ln 2} \right] = {\pi \over 8} + {1 \over 4}\ln 2$

$I = {\pi ^2}\int_0^2 {\sin \left( {{{\pi x} \over 2}} \right){{(x - [x])}^{[x]}}dx} $

$ = {\pi ^2}\int_0^1 {\sin \left( {{{\pi x} \over 2}} \right){x^0}dx + {\pi ^2}\int_1^2 {\sin \left( {{{\pi x} \over 2}} \right)(x - 1)dx} } $

$ = {\pi ^2}\left[ {{{ - 2} \over \pi }\cos {{\pi x} \over 2}} \right]_0^1 + {\pi ^2}\left[ {(x - 1){2 \over \pi }\left( { - \cos {{\pi x} \over 2}} \right)} \right]_1^2 + {\pi ^2}\int_1^2 {{2 \over \pi }\cos {{\pi x} \over 2}dx} $

$ = \left. {{\pi ^2}\left( {{2 \over \pi }} \right) + {{2{\pi ^2}} \over \pi }(1 - 0) + 2\pi \,.\,{2 \over \pi }\left( {\sin {{\pi x} \over 2}} \right)} \right|_1^2$

$ = 2\pi + 2\pi + 4(0 - 1) = 4\pi - 4 = 4(\pi - 1)$

$I = \int\limits_0^5 {{{x + [x]} \over {{e^{x - [x]}}}}dx = \alpha {e^{ - 1}} + \beta } $

$I = \int\limits_0^1 {{x \over {{e^x}}}dx + \int\limits_1^2 {{{x + 1} \over {{e^{x - 1}}}}dx + \int\limits_2^3 {{{x + 2} \over {{e^{x - 2}}}}dx + \int\limits_3^4 {{{x + 3} \over {{e^{x - 3}}}}dx + \int\limits_4^5 {{{x + 4} \over {{e^{x - 4}}}}dx} } } } } $

Let $I = {I_1} + {I_2} + {I_3} + {I_4} + {I_5}$

Here, ${I_2} = \int\limits_1^2 {{{x + 1} \over {{e^{x - 1}}}}dx} $ Put $x = t + 1 \Rightarrow dx = dt$

$ = \int\limits_0^1 {{{t + 2} \over {{e^t}}}dt = \int\limits_0^1 {{t \over {{e^t}}}dt + \int\limits_0^1 {{2 \over {{e^t}}}dt} } } $

${I_2} = {I_1} + 2\int\limits_0^1 {{e^{ - t}}dt = {I_1} + 2(1 - {e^{ - 1}})} $

Similarly,

${I_3} = {I_1} + 4(1 - {e^{ - 1}})$

${I_4} = {I_1} + 6(1 - {e^{ - 1}})$

${I_5} = {I_1} + 8(1 - {e^{ - 1}})$

$I = {I_1} + {I_2} + {I_3} + {I_4} + {I_5} = 5{I_1} + (2 + 4 + 6 + 8)(1 - {e^{ - 1}})$

$ = 5{I_1} + 20(1 - {e^{ - 1}})$

${I_1} = \int_0^1 {x{e^{ - 1}}dx = - [{e^{ - x}}(x + 1)_0^1 = 1 - 2{e^{ - 1}}} $

$\therefore$ $5{I_1} + 20(1 - {e^{ - 1}}) = 5(1 - 2{e^{ - 1}}) + 20(1 - {e^{ - 1}}) = 25 - 30{e^{ - 1}}$

$\therefore$ $\alpha$ = $-$30, $\beta$ = 25

Also it satisfy $5\alpha + 6\beta = 0$

Now, ${(\alpha + \beta )^2} = {( - 30 + 25)^2} = {( - 5)^2} = 25$

$\sqrt {{{\left( {{{x + 1} \over {x - 1}}} \right)}^2} + {{\left( {{{x - 1} \over {x + 1}}} \right)}^2} - 2} $

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

$ = \left| {{{x + 1} \over {x - 1}} - {{x - 1} \over {x + 1}}} \right|$

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

$ = \left| {{{{x^2} + 2x + 1 - {x^2} + 2x - 1} \over {{x^2} - 1}}} \right|$

$ = \left| {{{4x} \over {{x^2} - 1}}} \right|$

$\therefore$ $I = \int\limits_{ - {1 \over {\sqrt 2 }}}^{{1 \over {\sqrt 2 }}} {{{\left[ {{{\left( {{{x + 1} \over {x - 1}}} \right)}^2} + {{\left( {{{x - 1} \over {x + 1}}} \right)}^2} - 2} \right]}^{{1 \over 2}}}dx} $

$I = \int\limits_{{1 \over {\sqrt 2 }}}^{{1 \over {\sqrt 2 }}} {\left| {{{4x} \over {{x^2} - 1}}} \right|dx} $

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

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

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

$\therefore$ f(x) is a even function.

$\therefore$ $I = 2\int_0^{{1 \over {\sqrt 2 }}} {\left| {{{4x} \over {{x^2} - 1}}} \right|dx} $

Using property, If f(x) is an even function then,

$\int_{ - a}^a {f(x) = 2\int_0^a {f(x)dx} } $

x > 0 when $x \in \left[ {0,{1 \over {\sqrt 2 }}} \right]$

$\Rightarrow$ x² > 0 when $x \in \left[ {0,{1 \over {\sqrt 2 }}} \right]$

$\Rightarrow$ x² $-$ 1 < 0 when $x \in \left[ {0,{1 \over {\sqrt 2 }}} \right]$

$ \Rightarrow {1 \over {{x^2} - 1}} < 0$ when $x \in \left[ {0,{1 \over {\sqrt 2 }}} \right]$

$ \Rightarrow {{4x} \over {{x^2} - 1}} < 0$ when $x \in \left[ {0,{1 \over {\sqrt 2 }}} \right]$

$\therefore$ $\left| {{{4x} \over {{x^2} - 1}}} \right| = - {{4x} \over {{x^2} - 1}}$ when $x \in \left[ {0,{1 \over {\sqrt 2 }}} \right]$

$\therefore$ $I = 2\int_0^{{1 \over {\sqrt 2 }}} { - \left( {{{4x} \over {{x^2} - 1}}} \right)dx} $

$ = - 4\int_0^{{1 \over {\sqrt 2 }}} {{{2x} \over {{x^2} - 1}}dx} $

$ = - 4\left[ {{{\log }_e}\left| {{x^2} - 1} \right|} \right]_0^{{1 \over {\sqrt 2 }}}$

$ = - 4\left[ {\log \left| {{1 \over 2} - 1} \right| - \log \left| { - 1} \right|} \right]$

$ = - 4{\log _e}\left| { - {1 \over 2}} \right|$

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

$ = - 4\log _e^{{2^{ - 1}}}$

$ = 4\log _e^2$

$ = \log _e^{{2^4}}$

$ = \log _e^{16}$