Definite Integration

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

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

$ = \int\limits_0^1 {{x^4}\left[ {1 + {x^2} - 4x + {{4\{ 1 + {x^2} - 1\} } \over {1 + {x^2}}}} \right]dx} $

$ = \int\limits_0^1 {{x^4}\left[ {1 + {x^2} - 4x + 4 - {4 \over {1 + {x^2}}}} \right]dx} $

$ = \int\limits_0^1 {\left[ {{x^6} - 4{x^5} + 5{x^4} - 4{{{x^4} - 1 + 1} \over {1 + {x^2}}}} \right]dx} $

$ = \int\limits_0^1 {({x^6} - 4{x^5} + 5{x^4} - 4{x^2} + 4)dx - 4\int\limits_0^1 {{{dx} \over {1 + {x^2}}}} } $

$ = \left[ {{{{x^7}} \over 7} - {{2{x^6}} \over 3} + {x^5} - {{4{x^3}} \over 3} + 4x} \right]_0^1 - 4[{\tan ^{ - 1}}x]_0^1$

$ = \left[ {{1 \over 7} - {2 \over 3} + 1 - {4 \over 3} + 4} \right] - \pi = {{22} \over 7} - \pi $

We have,

${e^{ - x}}f(x) = 2 + \int\limits_0^x {\sqrt {{t^4} + 1} \,dt\,x \in ( - 1,1)} $

On differentiating w.r.t. x, we get

${e^{ - x}}(f'(x) - f(x)) = \sqrt {{x^4} + 1} $

$ \Rightarrow f'(x) = f(x) + \sqrt {{x^4} + 1} \,{e^x}$

$\because$ ${f^{ - 1}}$ is the inverse of f

$\therefore$ ${f^{ - 1}}(f(x)) = x$

$ \Rightarrow {f^{ - 1'}}(f(x))f'(x) = 1$

$ \Rightarrow {f^{ - 1'}}(f(x)) = {1 \over {f'(x)}}$

$ \Rightarrow {f^{ - 1'}}(f(x)) = {1 \over {f(x) + \sqrt {{x^4} + 1} \,{e^x}}}$

At $x = 0$, $f(x) = 2$

${f^{ - 1'}}(2) = {1 \over {2 + 1}} = {1 \over 3}$

Let $I = \int_0^\pi {\left[ {\cot x} \right]dx\,\,\,\,\,\,...\left( 1 \right)} $

$ = \int_0^\pi {\left[ {\cot \left( {\pi - x} \right)} \right]} dx$

$ = \int_0^\pi {\left[ { - \cot x} \right]dx\,\,\,\,\,\,...\left( 2 \right)} $

Adding two values of $I$ in $e{q^n}s\left( 1 \right)\,\,\& \,\,\left( 2 \right),$

We get $2I = \int_0^\pi {\left( {\left[ {\cot x} \right] + \left[ { - \cot x} \right]} \right)dx} $

$ = \int_0^\pi {\left( { - 1} \right)dx} $

$\left[ {} \right.$ As $\left[ x \right] + \left[ { - x} \right] = - 1,\,\,if\,\,$ $x \notin z$

and $\left[ x \right] + \left[ { - x} \right] = 0,\,\,if\,\,x \in z$ $\left. {} \right]$

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

(A) We have the integral

${I_n} = \int\limits_{ - \pi }^\pi {{{\sin nx} \over {(1 + {\pi ^x})\sin x}}dx} $

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

Now, ${I_{n + 2}} - {I_n} = \int\limits_0^\pi {{{\sin (n + 2)x - \sin nx} \over {\sin x}}dx} $

(B) Since ${I_3} = {I_5} = \,...\, = {I_{21}}$, we have

$\sum\limits_{m = 1}^{10} {{I_{2m + 1}} = 10{I_3} = 10\int\limits_0^\pi {{{\sin 3x} \over {\sin x}}dx = 10\int\limits_0^\pi {(3 - 4{{\sin }^2}x)dx} } } $

$ = 10[3x - 2x + \sin 2x]_0^\pi = 2\pi $

(C) Since ${I_2} = {I_4} = \,...\, = {I_{20}}$, we have

$\sum\limits_{m = 1}^{10} {{I_{2m}} = 10\int\limits_0^\pi {{{\sin 2x} \over {\sin x}}dx = 20[\sin x]_0^\pi = 0} } $

Given $g'(x) = \int\limits_0^{{e^x}} {{{f'(t)dt} \over {1 + {t^2}}}} $

$g''(x) = {{dg(x)} \over {d({e^x})}}.\,{{d({e^x})} \over {dx}}$

$ = {{f'({e^x})} \over {1 + {e^{2x}}}}\,.\,{e^x}$

$ = {{{e^x}} \over {1 + {e^{2x}}}}\,.\,{{2a({e^{2x}} - 1)} \over {{{({e^{2x}} + a{e^x} + 1)}^2}}} > 0$

when $x \in (0,\infty )$

$g'(x)$ < 0 when $x \in ( - \infty ,0)$

$y' = {1 \over {3[1 - f{{(x)}^2}]}}$

Clearly $f(x)$ is an odd function then $g'(x)$ is an even function

So,

$\int\limits_{ - 1}^1 {g'(x) = 2\int\limits_0^1 {g'(x)dx} } $

$ = 2[g(x)]_0^1 = 2[g(1) - g(0)]$

$ = 2g(1)$

We know that ${{\sin x} \over x} < 1,$ for $x \in \left( {0,1} \right)$

$ \Rightarrow {{\sin x} \over {\sqrt x }} < \sqrt x $ on $x \in \left( {0,1} \right)$

$ \Rightarrow \int\limits_0^1 {{{\sin x} \over {\sqrt x }}dx < \int\limits_0^1 {\sqrt x dx} = \left[ {{{2{x^{3/2}}} \over 3}} \right]} _0^1$

$ \Rightarrow \int\limits_0^1 {{{\sin x} \over {\sqrt x }}} dx < {2 \over 3} \Rightarrow I < {2 \over 3}$

Also ${{\cos x} \over {\sqrt x }} < {1 \over {\sqrt x }}$ for $x \in \left( {0,1} \right)$

$ \Rightarrow \int\limits_0^1 {{{\cos x} \over {\sqrt x }}dx < \int\limits_0^1 {{x^{ - 1/2}}dx} } $

$\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left[ {2\sqrt x } \right]_0^1 = 2$

$ \Rightarrow \int\limits_0^1 {{{\cos x} \over {\sqrt x }}dx < 2} $

$ \Rightarrow J < 2$

Given $f\left( x \right) = f\left( x \right) + f\left( {{1 \over x}} \right),$

where $f\left( x \right) = \int_1^x {{{\log \,t} \over {1 + t}}} \,dt$

$\therefore$ $F\left( e \right) = f\left( e \right) + f\left( {{1 \over e}} \right)$

$ \Rightarrow F\left( e \right)$

$ = \int_1^e {{{\log \,t} \over {1 + t}}dt + \int_1^{1/e} {{{\log t} \over {1 + t}}} } dt\,\,\,....\left( A \right)$

Now for solving, $I = \int_1^{1/e} {{{\log t} \over {1 + t}}dt} $

$\therefore$ Put ${1 \over t} = z \Rightarrow - {1 \over {{t^2}}}dt = dz$

$ \Rightarrow dt = - {{dz} \over {{z^2}}}$

and limit for $t = 1 \Rightarrow z = 1$ and for

$t = 1/e \Rightarrow z = e$

$\therefore$ $I = \int_1^e {{{\log \left( {{1 \over z}} \right)} \over {1 + {1 \over z}}}} \left( { - {{dz} \over {{z^2}}}} \right)$

$ = \int_1^e {{{\left( {\log 1 - \log z} \right).z} \over {z + 1}}\left( { - {{dz} \over {{z^2}}}} \right)} $

$ = \int_1^e { - {{\log z} \over {\left( {z + 1} \right)}}\left( { - {{dz} \over z}} \right)} $

[ as $\log 1 = 0$ ]

$ = \int_1^e {{{\log z} \over {z\left( {z + 1} \right)}}} dz$

$\therefore$ $I = \int_1^e {{{\log \,t} \over {t\left( {t + 1} \right)}}dt} $

$\left[ {} \right.$ By property $\int_a^b {f\left( t \right)dt} $

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$ $ = \int_a^b {f\left( x \right)dx} $ $\left. {} \right]$

Equation $(A)$ becomes

$F\left( e \right) = \int_1^e {{{\log t} \over {1 + t}}dt + \int_1^e {{{\log t} \over {t\left( {1 + t} \right)}}} } dt$

$ = \int_1^e {{{t.\log t + \log t} \over {t\left( {1 + t} \right)}}} dt$

$ = \int_1^e {{{\left( {\log t} \right)\left( {t + 1} \right)} \over {t\left( {1 + t} \right)}}} $

$ \Rightarrow F\left( e \right) = \int_1^e {{{\log t} \over t}dt} $

Let $\log t = x$

$\therefore$ ${1 \over t}dt = dx$

$\left[ {} \right.$ for limit $t = 1,x = 0$

and $t = e,x = \log \,e = 1$ $\left. {} \right]$

$\therefore$ $F\left( e \right) = \int_0^1 {x\,dx} $

$ \Rightarrow F\left( e \right) = \left[ {{{{x^2}} \over 2}} \right]_0^1$

$ \Rightarrow F\left( e \right) = {1 \over 2}$

$\int_{\sqrt 2 }^x {{{dt} \over {t\sqrt {{t^2} - 1} }}} = {\pi \over 2}$

$\therefore$ $\left[ {{{\sec }^{ - 1}}t} \right]_{\sqrt 2 }^x = {\pi \over 2}$

$\left[ {} \right.$ As $\int {{{dx} \over {x\sqrt {{x^2} - 1} }}} = {\sec ^{ - 1}}x$ $\left. {} \right]$

$ \Rightarrow {\sec ^{ - 1}}x - {\sec ^{ - 1}}\sqrt 2 = {\pi \over 2}$

$ \Rightarrow {\sec ^{ - 1}}x - {\pi \over 4} = {\pi \over 2}$

$ \Rightarrow {\sec ^{ - 1}}x = {\pi \over 2} + {\pi \over 4}$

$ \Rightarrow {\sec ^{ - 1}}x = {{3\pi } \over 4}$

$ \Rightarrow x = \sec {{3\pi } \over 4}$

$ \Rightarrow x = - \sqrt 2 $

$\mathop {\lim }\limits_{x \to {\pi \over 4}} {{\int\limits_2^{{{\sec }^2}x} {f(t)\,dt} } \over {{x^2} - {{{\pi ^2}} \over {16}}}}$ (${0 \over 0}$ form)

$ = \mathop {\lim }\limits_{x \to {\pi \over 4}} {{f({{\sec }^2}x)2\sec x\sec x\tan x} \over {2x}}$

(Applying L' Hospital Rule)

$ = {{2f(2)} \over {{\pi \over 4}}} = {{8f(2)} \over \pi }$

(A) $\int\limits_{ - 1}^1 {{{dx} \over {1 + {x^2}}} = {{\tan }^{ - 1}}x]_{ - 1}^1} $

$ = {\tan ^{ - 1}}(1) - {\tan ^{ - 1}}( - 1)$

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

$ = {{2\pi } \over 4} = {\pi \over 2}$

(B) $\int\limits_0^1 {{{dx} \over {\sqrt {1 - {x^2}} }} = {{\sin }^{ - 1}}x]_0^1} $

$ = {\sin ^{ - 1}}(1) - {\sin ^{ - 1}}(0)$

$ = {\pi \over 2} - 0$

$ = {\pi \over 2}$

(C) $\int\limits_2^3 {{{dx} \over {1 - {x^2}}} = {1 \over 2}\log \left| {{{1 + x} \over {1 - x}}} \right|} $

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

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

(D) $\int\limits_1^2 {{{dx} \over {x\sqrt {{x^2} - 1} }}[{{\sec }^{ - 1}}x]_1^2} $

$ = {\sec ^{ - 1}}2 - {\sec ^{ - 1}}1$

$ = {\pi \over 3} - 0$

$ = {\pi \over 3}$

Let $a = k + h$ where $k$ is an integer such that

$\left[ a \right] = k$ and $0 \le h < 1$

$\therefore$ $\int\limits_1^a {\left[ x \right]f'\left( x \right)dx = \int\limits_1^2 {1f'\left( x \right)} } \,dx$

$\,\,\,\,\,\,\,\,\,\,\,\, + \int\limits_2^3 {2f'\left( x \right)dx + } ....$

$\,\,\,\,\,\,\,\,\,\,\,\,\int\limits_{k - 1}^k {\left( {k - 1} \right)dx + \int\limits_k^{k + h} {kf'\left( x \right)} } dx$

$\left\{ {f\left( 2 \right) - f\left( 1 \right)} \right\} + 2\left\{ {f\left( 3 \right) - f\left( 2 \right)} \right\}$

$\,\,\,\,\,\,\,\,\,\,\,\,\,$ $ + 3\left\{ {f\left( 4 \right) - f\left( 3 \right)} \right\} + \,........\,$

$\,\,\,\,\,\,\,\,\,\,\,\,\,$ $ + \left( {k - 1} \right)\left\{ {f\left( k \right) - f\left( {k - 1} \right)} \right\}$

$\,\,\,\,\,\,\,\,\,\,\,\,\,$ $ + k\left\{ {f\left( {k + h} \right) - f\left( k \right)} \right\}$

$ = - f\left( 1 \right) - f\left( 2 \right) - f\left( 3 \right).......$

$\,\,\,\,\,\,\,\,\,\,\,\,\,$ $ - f\left( k \right) + kf\left( {k + h} \right)$

$ = \left[ a \right]f\left( a \right) - \left\{ {f\left( 1 \right) + f\left( 2 \right)} \right.$

$\,\,\,\,\,\,\,\,\,\,\,\,\,$ $\left. { + f\left( 3 \right) + ........f\left( {\left[ a \right]} \right)} \right\}$

$I = \int\limits_0^\pi {xf\left( {\sin \,x} \right)dx} $

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

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

$ \Rightarrow 2I = \pi {\pi \over 0}f\left( {\sin x} \right)dx$

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

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

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

$I = \int\limits_{ - {{3\pi } \over 2}}^{ - {\pi \over 2}} {\left[ {{{\left( {x + \pi } \right)}^3} + {{\cos }^2}\left( {x + 3\pi } \right)} \right]} \,dx$

Put $x + \pi = t$

$I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {\left( {{t^3} + {{\cos }^2}t} \right)dt} $

$ = 2\int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{\cos }^2}} tdt$

$\left[ {} \right.$ using the property of even and odd function $\left. {} \right]$

$ = \int\limits_0^{{\pi \over 2}} {\left( {1 + \cos 2t} \right)} dt = {\pi \over 2} + 0$

${I_1} = \int\limits_0^1 {{2^{{x^2}}}} dx,\,{I_2} = \int\limits_0^1 {{2^{{x^3}}}} dx,$

$ = {I_3} = \int\limits_0^1 {{2^{{x^2}}}} dx,\,$

${I_4} = \int\limits_0^1 {{2^{{x^3}}}} dx\,\,$

$\forall 0 < x < 1,\,{x^2} > {x^3}$

$ \Rightarrow \int\limits_0^1 {{2^{{x^2}}}} \,dx > \int\limits_0^1 {{2^{{x^3}}}} dx$

and $\int\limits_1^2 {{2^{{x^3}}}dx} > \int\limits_1^2 {{2^{{x^2}}}dx} $

$ \Rightarrow {I_1} > {I_2}$ and ${I_4} > {I_3}$

$\mathop {\lim }\limits_{n \to \infty } \left[ {{1 \over {{n^2}}}{{\sec }^2}{1 \over {{n^2}}} + {2 \over {{n^2}}}{{\sec }^2}{4 \over {{n^2}}} + {3 \over {{n^2}}}se{c^2}{9 \over {{n^2}}} + ... + {1 \over n}{{\sec }^2}1} \right]$

$ = \mathop {\lim }\limits_{n \to \infty } {r \over {{n^2}}}{\sec ^2}{{{r^2}} \over {{n^2}}} = \mathop {\lim }\limits_{n \to \infty } {1 \over n}.{\pi \over n}{\sec ^2}{{{r^2}} \over {{n^2}}}$

$ \Rightarrow $ Given limit is equal to value of integral

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\int\limits_0^1 {x{{\sec }^2}} \,{x^2}\,dx$

or ${1 \over 2}\int\limits_0^1 {2x\,\sec } \,\,{x^2}dx = {1 \over 2}\int\limits_0^1 {{{\sec }^2}} tdl$

$\left[ {} \right.$ put $\left. {{x^2} = t} \right]$

$ = {1 \over 2}\left( {\tan t} \right)_0^1 = {1 \over 2}\tan \,1.$

$I = \int\limits_3^6 {{{\sqrt x } \over {\sqrt {9 - x} + \sqrt x }}} dx\,\,\,\,\,\,\,\,\,...\left( 1 \right)$

$I = \int\limits_3^6 {{{\sqrt {9 - x} } \over {\sqrt {9 - x} + \sqrt x }}} dx\,\,\,\,\,\,\,\,\,...\left( 2 \right)$

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

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

$2I = \int\limits_3^6 {dx} = 3 \Rightarrow I = {3 \over 2}$

$\mathop {\lim }\limits_{x \to 2} \int\limits_6^{f\left( x \right)} {{{4{t^3}} \over {x - 2}}} dt$

$ = \mathop {\lim }\limits_{x \to 0} {{\int\limits_6^{f\left( x \right)} {4{t^3}dt} } \over {x - 2}}$

This limit resembles a derivative because the fraction has the form $0/0$ as $x \to 2$ since both the numerator (integral from $6$ to $f(x)$) and the denominator ($x - 2$) are zero when $x = 2$.

Applying $L'$ Hospital rule

$\mathop {\lim }\limits_{x \to 2} {{\left[ {4f{{\left( x \right)}^3}f'\left( x \right)} \right]} \over 1}$

$ = 4{\left( {f\left( 2 \right)} \right)^3}f'\left( 2 \right)$

$ = 4 \times {6^3} \times {1 \over {48}} = 18$

Let $I = \int\limits_{ - \pi }^\pi {{{{{\cos }^2}x} \over {1 + {a^x}}}} dx\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$

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

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

$ = \int\limits_{ - \pi }^\pi {{{{{\cos }^2}x} \over {1 + {\alpha ^x}}}} dx\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)$

Adding equations $(1)$ and $(2)$ we get

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

$ = \int\limits_{ - \pi }^\pi {{{\cos }^2}} x\,dx$

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

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

$ = 4\int\limits_0^{{\pi \over 2}} {{{\sin }^2}} x\,dx$

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

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

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

$ \Rightarrow I + I = 2\left( {{\pi \over 2}} \right) = \pi $

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

$\begin{aligned} & \int_\limits{0}^{\pi} e^{|\cos x|}\left[2 \sin \left(\frac{1}{2} \cos x\right)+3 \cos \left(\frac{1}{2} \cos x\right)\right] \sin x d x \\ & =\int_\limits{0}^{\pi} e^{|\cos x|} 2 \sin x \sin \left(\frac{1}{2} \cos x\right) d x \\ & +\int_\limits{0}^{\pi} e^{|\cos x|} 3 \cos \left(\frac{1}{2} \cos x\right) \sin x d x \end{aligned}$

We know that

$\begin{aligned} & \int_{0}^{2 a} f(x) d x=0 \text { if } f(2 a-x)=-f(x) \\ & \quad=2 \int_{0}^{a} f(x) d x \text { if } f(2 a-x)=f(x) \\ & \quad=0+2 \int_{0}^{\frac{\pi}{2}} e^{|\cos x|} \cdot 3 \cdot \cos \left(\frac{1}{2} \cos x\right) \sin x d x \\ & \quad=6 \int_{0}^{\frac{\pi}{2}} e^{|\cos x|} \cdot \cos \left(\frac{\cos x}{2}\right) \sin x d x \end{aligned}$

Now put $\quad \cos x=t$

$-\sin x d x=d t$

when,

$\begin{aligned} & x=0 t=1 \\ & x=\frac{\pi}{2} t=0 \end{aligned}$

$\begin{aligned} & =-6 \int_\limits{1}^{0} e^{t} \cos \left(\frac{t}{2}\right) d t \\ & =6 \int_\limits{0}^{1} e^{t} \cos \left(\frac{t}{2}\right) d t \end{aligned}$

Let, $\mathrm{I}=\int e^{t} \cos \left(\frac{t}{2}\right) d t$

$=e^{t} 2 \sin \left(\frac{t}{2}\right)-2 \int e^{t} \sin \left(\frac{t}{2}\right) d t$

$ =2 e^{t} \sin \left(\frac{t}{2}\right)-2\left[e^{t}\left(\frac{-\cos \frac{t}{2}}{\frac{1}{2}}\right)+2 \int e^{t} \cos \frac{t}{2} d t\right]$

$=2 e^{t} \sin \left(\frac{t}{2}\right)+4 e^{t} \cos \left(\frac{t}{2}\right)-4 \mathrm{I}$

$=2 e^{t} \sin \left(\frac{t}{2}\right)+4 e^{t} \cos \left(\frac{t}{2}\right)-4 \mathrm{I}$

$=2 e^{t} \sin \left(\frac{t}{2}\right)+4 e^{t} \cos \left(\frac{t}{2}\right)-4 \mathrm{I}$

$5 \mathrm{I}=2 e^{t} \sin \left(\frac{t}{2}\right)+4 e^{t} \cos \left(\frac{t}{2}\right)$

$I=\frac{1}{5}\left[2 e^{t} \sin \left(\frac{t}{2}\right)+4 e^{t} \cos \left(\frac{t}{2}\right)\right]$

Required Integration

$\begin{aligned} & =\frac{6}{5}\left[2 e \sin \left(\frac{1}{2}\right)+4 e\left(\cos \frac{1}{2}\right)-4 e\right] \\ & =24 \frac{e}{5}\left[\cos \left(\frac{1}{2}\right)+\left(\frac{1}{2}\right) \sin \left(\frac{1}{2}\right)-1\right] \end{aligned}$

$\mathop {Lim}\limits_{n \to \infty } \sum\limits_{r = 1}^n {{1 \over n}} {e^{{r \over n}}}\,\,$

$\left[ {} \right.$ Using definite integrals as limit of sum $\left. {} \right]$

$ = \int\limits_\theta ^1 {{e^x}} dx = e - 1$

Let $I = \int\limits_0^\pi {xf\left( {\sin x} \right)} dx$

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

$\therefore$ $2I = \pi \int\limits_2^\pi {f\left( {\sin x} \right)} dx$

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

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

$ \Rightarrow A = \pi $

$I = \int\limits_0^{{\pi \over 2}} {{{{{\left( {\sin \,x + \cos x} \right)}^2}} \over {\sqrt {1 + \sin 2x} }}} dx$

We know $\left[ {{{\left( {\sin x + \cos x} \right)}^2} = 1 + \sin 2x} \right],\,$ So

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

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

$\left[ {} \right.$ $\sin x + \cos x > 0$ $\,\,if\,\,0 < x < {\pi \over 2}$ $\left. {} \right]$

or $I = \left[ { - \cos x + \sin x} \right]_0^{{\pi \over 2}} $

= - cos ${\pi \over 2}$ + sin ${\pi \over 2}$ + cos 0 - sin 0

= - 0 + 1 + 1 - 0

= 2

$\int\limits_{ - 2}^3 {\left| {1 - {x^2}} \right|} dx = \int\limits_{ - 2}^3 {\left| {{x^2} - 1} \right|} dx$

Now $\left| {{x^2} - 1} \right| = \left\{ {\matrix{ {{x^2} - 1} & {if} & {x \le - 1} \cr {1 - {x^2}} & {if} & { - 1 \le x \le 1} \cr {{x^2} - 1} & {if} & {x \ge 1} \cr } } \right.$

$\therefore$ Integral is

$\int\limits_{ - 2}^{ - 1} {\left( {{x^2} - 1} \right)dx + \int\limits_{ - 1}^1 {\left( {1 - {x^2}} \right)} } dx + \int\limits_1^3 {\left( {{x^2} - 1} \right)dx} $

$\left[ {{{{x^3}} \over 3} - x} \right]_{ - 2}^{ - 1} + \left[ {x - {{{x^3}} \over 3}} \right]_{ - 1}^1 + \left[ {{{{x^3}} \over 3} - x} \right]_1^3$

$ = \left( { - {1 \over 3} + 1} \right) - \left( { - {8 \over 3} + 2} \right) + \left( {2 - {2 \over 3}} \right) + \left( {{{27} \over 3} - 3} \right) - \left( {{1 \over 3} - 1} \right)$

$ = {2 \over 3} + {2 \over 3} + {4 \over 3} + 6 + {2 \over 3} = {{28} \over 3}$

$f\left( x \right) = {{{e^x}} \over {1 + {e^x}}}$

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

$\therefore$ $f\left( x \right) + f\left( { - x} \right) = 1\forall x$

Now ${I_1} = \int\limits_{f\left( { - a} \right)}^{f\left( a \right)} {xg\left\{ {x\left( {1 - x} \right)} \right\}} dx$

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

$\left[ {} \right.$ using $\int\limits_a^b {f\left( x \right)} dx\,a$

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

$ = {I_2} - {I_1} \Rightarrow 2{I_1} = {I_2}$

$\therefore$ ${{{I_2}} \over {{I_1}}} = 2$

$F\left( t \right) = \int\limits_0^t {f\left( {t - y} \right)g\left( y \right)} dy$

$ = \int\limits_0^t {{e^{t - y}}} ydy = {e^t}\int\limits_0^t {{e^{ - y}}} \,ydy$

$ = {e^t}\left[ { - y{e^{ - y}} - {e^{ - y}}} \right]_0^t$

$ = - {e^t}\left[ {y{e^{ - y}} + {e^{ - y}}} \right]_0^t$

$ = - {e^t}\left[ {t\,{e^{ - t}} + {e^{ - t}} - 0 - 1} \right]$

$ = - {e^t}\left[ {{{t + 1 - {e^t}} \over {{e^t}}}} \right]$

$ = {e^t} - \left( {1 + t} \right)$

Given $f'\left( x \right) = f\left( x \right) \Rightarrow {{f'\left( x \right)} \over {f\left( x \right)}} = 1$

Integrating log

$f\left( x \right) = x + c \Rightarrow f\left( x \right) = {e^{x + c}}$

$f\left( 0 \right) = 1 \Rightarrow f\left( x \right) = {e^x}$

$\therefore$ $\int\limits_0^1 {f\left( x \right)g\left( x \right)} dx$

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

$ = \int\limits_0^1 {{x^2}} {e^x}dx - \int\limits_0^1 {{e^{2x}}} dx$

$ = \left[ {{x^2}{e^x}} \right]_0^1 - 2\left[ {x{e^x} - {e^x}} \right]_0^1 - {1 \over {20}}\left[ {{e^{2x}}} \right]_0^1$

$ = e - \left[ {{{{e^2}} \over 2} - {1 \over 2}} \right] - 2\left[ {e - e + 1} \right]$

$ = e - {{{e^2}} \over 2} - {3 \over 2}$

$I = \int\limits_0^1 {x{{\left( {1 - x} \right)}^n}} dx$

$ = \int\limits_0^1 {\left( {1 - x} \right){{\left( {1 - 1 + x} \right)}^n}dx} $

$ = \int\limits_0^1 {\left( {1 - x} \right)} {x^n}dx$

$ = \left[ {{{{x^{n + 1}}} \over {n + 1}} - {{{x^{n + 2}}} \over {n + 2}}} \right]_0^1$

$ = {1 \over {n + 1}} - {1 \over {n + 2}}$

The given expression can be written as

$\mathop {\lim }\limits_{n \to \infty } {1 \over n}{\sum\limits_{r = 1}^n {\left( {{r \over n}} \right)} ^4} - \mathop {\lim }\limits_{n \to \infty } {1 \over n}.\mathop {\lim }\limits_{n \to \infty } {1 \over n}{\left( {{r \over n}} \right)^3}$

$ = \int\limits_0^1 {{x^4}} \,\,dx - \mathop {\lim }\limits_{n \to \infty } {1 \over n} \times \int\limits_0^1 {{x^3}} \,\,dx$

$ = \left[ {{{{x^5}} \over 5}} \right]_0^1 - 0 = {1 \over 5}$

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

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

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

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

$\left[ {} \right.$ As given that $f\left( {a + b - x} \right) = f\left( x \right)$ $\left. {} \right]$

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

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

$\mathop {\lim }\limits_{x \to 0} {{{d \over {dx}}\int\limits_0^{{x^2}} {{{\sec }^2}tdt} } \over {{d \over x}\left( {x\sin x} \right)}}$

$ = \mathop {\lim }\limits_{x \to 0} {{{{\sec }^2}{x^2}.2x} \over {\sin \,x + x\,\cos \,x}}$ (by $L'$ Hospital rule)

$\mathop {\lim }\limits_{x \to 0} {{2{{\sec }^2}{x^2}} \over {\left( {{{\sin x} \over x} + \cos \,x} \right)}} = {{2 \times 1} \over {1 + 1}} = 1$

We have $\int\limits_0^2 {f\left( x \right)} dx = {3 \over 4};Now,$

$\int\limits_0^2 {xf'\left( x \right)} dx$

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

$ = \left[ {x\,f\left( x \right)} \right]_0^2 - {3 \over 4}$

$ = 2f\left( 2 \right) - {3 \over 4}$

$ = 0 - {3 \over 4}$

$\left( {} \right.$ As $f\left( 2 \right) = 0$ $\left. {} \right)$ $ = - {3 \over 4}.$

$\int\limits_0^2 {\left[ {{x^2}} \right]} dx = \int\limits_0^1 {\left[ {{x^2}} \right]dx} + \int\limits_1^{\sqrt 2 } {\left[ {{x^2}} \right]} dx + $

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$\int\limits_{\sqrt 2 }^{\sqrt 3 } {\left[ {{x^2}} \right]} + \int\limits_{\sqrt 3 }^2 {\left[ {{x^2}} \right]} dx$

$ = \int\limits_0^1 {0dx} + \int\limits_1^{\sqrt 2 } {1dx} + \int\limits_{\sqrt 2 }^{\sqrt 3 } {2dx} + \int\limits_{\sqrt 3 }^2 {3dx} $

$ = \left[ x \right]_1^{\sqrt n } + \left[ {2x} \right]_{\sqrt 2 }^{\sqrt 3 } + \left[ {3x} \right]_{\sqrt 3 }^2$

$ = \sqrt 2 - 1 + 2\sqrt 3 - 2\sqrt 2 + 6 - 3\sqrt 3 $

$ = 5 - \sqrt 3 - \sqrt 2 $

${I_n} + {I_{n + 2}}$

$ = \int\limits_0^{\pi /4} {{{\tan }^n}} \,x\left( {1 + {{\tan }^2}x} \right)dx$

$ = \int\limits_0^{\pi /4} {{{\tan }^n}} x\,{\sec ^2}x\,dx$

$ = \left[ {{{{{\tan }^{n + 1}}x} \over {n + 1}}} \right]_0^{\pi /4}$

$ = {{1 - 0} \over {n + 1}} = {1 \over {n + 1}}$

$\therefore$ ${I_n} + {I_{n + 2}} = {1 \over {n + 1}}$

$ \Rightarrow \mathop {\lim }\limits_{n \to \infty } n\left[ {{I_n} + {I_{n + 2}}} \right]$

$ = \mathop {\lim }\limits_{n \to \infty } \,n.{1 \over {n + 1}} = \mathop {\lim }\limits_{n \to \infty } {n \over {n + 1}}$

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

$I = \int\limits_0^{10\pi } {\left| {\sin x} \right|} dx$

$ = 10\int\limits_0^\pi {\left| {\sin x\,} \right|} \,dx$

$ = 10\int\limits_0^\pi {\sin \,x\,dx} $

$\left[ \, \right.$ as $\left| {\sin x} \right|$ is periodic with period $\pi $

and $sin$ $x > 0$ if $0 < x$ $ < \pi $ $\left. \, \right]$

$I = 20\int\limits_0^{\pi /2} {\sin x} \,dx$

$ = 20\left[ { - \cos \,x_0^{\pi /2}} \right] = 20$

$\int_{ - \pi }^\pi {{{2x\left( {1 + \sin \,x} \right)} \over {1 + {{\cos }^2}x}}} dx$

$ = \int_{ - \pi }^\pi {{{2x\,dx} \over {1 + {{\cos }^2}x}} + 2\int_{ - \pi }^\pi {{{x\,\sin x} \over {1 + {{\cos }^2}x}}} } dx$

$ = 0 + 4\int_0^\pi {{{x\sin x\,dx} \over {1 + {{\cos }^2}x}}} ;$

$\left[ \, \right.$ as $\int\limits_{ - a}^a {f\left( x \right)} dx = 0$ $\left. \, \right]$

if $f(x)$ is odd

$ = 2\int\limits_0^a {f\left( x \right)} dx$ if $f(x)$ is even.

$I = 4\int_0^\pi {{{\left( {\pi - x} \right)\sin \left( {\pi - x} \right)} \over {1 + {{\cos }^2}\left( {\pi - x} \right)}}} dx$

$I = 4\int_0^\pi {{{\left( {\pi - x} \right)\sin \,x} \over {1 + {{\cos }^2}x}}} $

$ \Rightarrow I = 4\pi \int_0^\pi {{{\sin x\,dx} \over {1 + {{\cos }^2}x}}} - 4\int {{{x\sin x\,dx} \over {1 + {{\cos }^2}x}}} $

$ \Rightarrow 2I = 4\pi \int_0^\pi {{{\sin x} \over {1 + {{\cos }^2}x}}} dx$

put $\cos x = t \Rightarrow - \sin xdx = dt$

$\therefore$ $I = - 2\pi \int\limits_1^{ - 1} {{1 \over {1 + {t^2}}}} dt$

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

$ = 2\pi \left[ {{{\tan }^{ - 1}}t} \right]_{ - 1}^1$

$ = 2\pi \left[ {{{\tan }^{ - 1}}1 - {{\tan }^{ - 1}}\left( { - 1} \right)} \right]$

$ = 2\pi \left[ {{\pi \over 4} - \left( {{{ - \pi } \over 4}} \right)} \right] = 2\pi {\pi \over 2} = {\pi ^2}$

We have $\mathop {\lim }\limits_{x \to \infty } {{{1^p} + {2^p} + .... + {n^p}} \over {{n^{p + 1}}}};$

$\mathop {\lim }\limits_{x \to \infty } \sum\limits_{r = 1}^n {{{{r^p}} \over {{n^p}.n}}} = \int\limits_0^1 {{x^p}} dx$

$ = {\left[ {{{{x^{p + 1}}} \over {p + 1}}} \right]_0} = {1 \over {p + 1}}$