Maths M.C.Q (1 Marks)

$ = \frac{1}{2}\int_0^{\pi /2} {x{{\sec }^2}\frac{x}{2}} dx + \int_0^{\pi /2} {\tan \frac{x}{2}dx} $.

$ = \left| {\,x\tan \frac{x}{2}\,} \right|_0^{\pi /2} = \frac{\pi }{2}\tan \frac{\pi }{4} = \frac{\pi }{2}$.

$=  - [{e^x}\cos x]_0^{\pi /2} + \int_0^{\pi /2} {{e^x}\cos x\,dx} $

$ = - [{e^x}\cos x]_0^{\pi /2} + [{e^x}\sin x]_0^{\pi /2} - \int_0^{\pi /2} {{e^x}\sin x\,dx} $

$\therefore $$2I = [{e^x}(\sin x - \cos x)]_0^{\pi /2} = ({e^{\pi /2}} + 1)$

Hence $\int_0^{\pi /2} {{e^x}\sin xdx = \frac{1}{2}({e^{\pi /2}} + 1)} $.

$=[{e^x}\log x]_1^e - \int_1^e {{e^x}\log x\,dx + \int_1^e {{e^x}\log x\,dx} } $

$= [{e^e}\log e - {e^1}{\log _e}1] = {e^e}$.

Hence depends on $c$.

$= \int_0^{2\pi } {\left| {\sin \frac{x}{4} + \cos \frac{x}{4}} \right|dx = 4\left[ {\sin \frac{x}{4} - \cos \frac{x}{4}} \right]} _0^{2\pi }$

$ = 4[1 - 0 - 0 + 1] = 8$.

$ = \left[ {\frac{{\sin 3x}}{3}(2 + 3{x^2})} \right]_0^{\pi /6} - \int_0^{\pi /6} {\frac{{\sin 3x}}{3}} .6x.dx$

$ = \frac{1}{{36}}({\pi ^2} + 16)$.

$ = \int_{ - 1}^3 {\left\{ {{{\tan }^{ - 1}}\left( {\frac{x}{{{x^2} + 1}}} \right) + {{\cot }^{ - 1}}\left( {\frac{x}{{{x^2} + 1}}} \right)} \right\}} dx$

$ = \int_{ - 1}^3 {\frac{\pi }{2}dx = } 2\pi $.

$ = \left[ {\log \left( {1 + \frac{x}{2}} \right)\frac{{{x^2}}}{2}} \right]_0^2 - \int_0^1 {\frac{1}{{1 + \frac{x}{2}}}\frac{1}{2}\frac{{{x^2}}}{2}} dx$

$ = \frac{1}{2}\log \frac{3}{2} - \frac{1}{2}\int_0^1 {\frac{{{x^2}}}{{x + 2}}dx} $

$ = \frac{1}{2}\log \frac{3}{2} - \frac{1}{2}\left[ {\frac{1}{2} - 2 + 4\log 3 - 4\log 2} \right] $

$= \frac{3}{4} + \frac{3}{2}\log \frac{2}{3}$

On comparing with the given value $a = \frac{3}{4},b = \frac{3}{2}$.

$= \left[ {{{\sin }^{ - 1}}\left( {\frac{{\frac{{2x - 1}}{2}}}{{1/2}}} \right)} \right]_{1/4}^{1/2}$

$ = [{\sin ^{ - 1}}(2x - 1)]_{1/6}^{1/2} = \pi /6$.

==> $I = \int_0^1 {({x^2} - 1)} dx + 2\int_0^1 {\frac{{dx}}{{1 + {x^2}}}} $

==> $I = \left[ {\frac{{{x^3}}}{3} - x} \right]_0^1 + 2\,[{\tan ^{ - 1}}x]_0^1$

$ = - \frac{2}{3} + \frac{\pi }{2} = \frac{{(3\pi - 4)}}{6}$.

$ \Rightarrow \frac{{{a^2}}}{2} \le a + 4$

$ \Rightarrow {a^2} \le 2a + 8$ 

$ \Rightarrow {a^2} - 2a - 8 \le 0$

$ \Rightarrow (a - 4)(a + 2) \le 0$

$ \Rightarrow - 2 \le a \le 4$.

Since $\sin \theta $ is positive in interval $(0,\pi )$

$\therefore I = \int_0^\pi {{{\sin }^3}\theta \,d\theta = \int_0^\pi {\sin \theta (1 - {{\cos }^2}\theta )\,\,d\theta } } $

$ = \int_0^\pi {\sin \theta \,d\theta + \int_0^\pi {( - \sin \theta )\,{{\cos }^2}\theta \,d\theta } } $

$ = [ - \cos \theta ]_0^\pi + \left( {\frac{{{{\cos }^3}\theta }}{3}} \right)_0^\pi = \frac{4}{3}$.

$ = \left[ {\frac{3}{2}\log ({x^2} + 9) + \frac{1}{3}{{\tan }^{ - 1}}\left( {\frac{x}{3}} \right)} \right]_0^3$

$ = \frac{3}{2}(\log 18 - \log 9) + \frac{1}{3}\left( {\frac{\pi }{4}} \right)$

$ = \frac{3}{2}\log 2 + \frac{\pi }{{12}} $

$= \log (2\sqrt 2 ) + \frac{\pi }{{12}}$.

Put $1 + x = {t^2} \Rightarrow dx = 2t\,dt$

When $x = 3 \to 8,$ then $t = 2 \to 3$

$\therefore$  $I = 2\int_2^3 {\frac{{5 - 3{t^2}}}{{{t^2} - 1}}dt} $;

$I = 2\int_2^3 {\left( {\frac{2}{{{t^2} - 1}} - 3} \right)} \,dt$

$I = 2\left[ {\frac{2}{{2.1}}\log \frac{{t - 1}}{{t + 1}} - 3t} \right]_2^3$;

$I = 2\log \left( {\frac{3}{{2{e^3}}}} \right)$.

$ = \left| {\frac{{ - {x^2}}}{2} - 2x} \right|_{ - 4}^{ - 2} + \,\,\left| {\frac{{{x^2}}}{2} + 2x} \right|_{ - 2}^4 = 20$.

$ = \left| {2{x^2} + 3x} \right|_1^2 + \left| {\frac{{3{x^2}}}{2} + 5x} \right|_2^4 = 37$.

$ = [x - x\log x]_{1/e}^1 + [x\log x - x]_1^e$

$ = (1 - 0) - \left\{ {\frac{1}{e} - \frac{1}{e}( - 1)} \right\} + e - e + 1$

$ = 2 - \frac{2}{e} = 2\left( {1 - \frac{1}{e}} \right)$.

$ = \frac{1}{2}\left[ { - 3\cos \theta + \frac{{\cos 3\theta }}{3}} \right]_0^\pi $

$= \frac{1}{2}\left[ { - 3( - 1 - 1) + \frac{{( - 1 - 1)}}{3}} \right] = \frac{8}{3}$.

$ = \int_0^2 {(2 - x)} \,dx + \int_2^3 { - (2 - x)\,dx} $

$ = \int_0^2 {(2 - x)} \,dx - \int_2^3 {\,(2 - x)\,dx} = \left[ {2x - \frac{{{x^2}}}{2}} \right]_0^2 - \left[ {2x - \frac{{{x^2}}}{2}} \right]_2^3$

$\Rightarrow$ $I = [4-2] - \left[ 6- \frac{{9}}{{2}} - (4-2) \right] $

$ = 2 - \left[ {4 - \frac{9}{2}} \right]$$ = \frac{5}{2}$.

$ = [x - {x^3}]_0^{1/\sqrt 3 } + [{x^3} - x]_{1/\sqrt 3 }^1$

$ = \frac{1}{{\sqrt 3 }} - \frac{1}{{3\sqrt 3 }} + \frac{{ - 1}}{{3\sqrt 3 }} + \frac{1}{{\sqrt 3 }}$

$ = \frac{4}{{3\sqrt 3 }}$.

$I = \int_{\,0}^{\,\pi /2} {\cos x\,dx} - \int_{\,\pi /2}^{\,\pi } {\cos x\,dx} $

$= [\sin x]_0^{\pi /2} - [\sin x]_{\pi /2}^\pi $

$I = \left[ {\sin \frac{\pi }{2} - \sin 0} \right] - \left[ {\sin \pi - \sin \frac{\pi }{2}} \right] $

$=1+ 1 = 2.$

$ = - (b - a) = \,|b| - |a|$.

$ = \int_0^{\pi /4} {({{\sec }^2}x - 1){{\tan }^{n - 2}}x\,\,dx} $

$ = \int_0^{\pi /4} {{{\sec }^2}x{{\tan }^{n - 2}}x\,\,dx} - \int_0^{\pi /4} {{{\tan }^{n - 2}}x\,\,dx} $

$ = \left[ {\frac{{{{\tan }^{n - 1}}x}}{{n - 1}}} \right]_0^{\pi /4} - {u_{n - 2}}$

$ \Rightarrow {u_n} + {u_{n - 2}} = \frac{1}{{n - 1}}$.

$ = [f'(x + a)]_0^{b - c} = f'(b - c + a) - f'(a)$.

so that reduced integral is 

$\int_0^1 {\left( {\frac{1}{{1 + t}} - \frac{1}{{2 + t}}} \right)\,\,dt = [\log (1 + t) - \log (2 + t)]_0^1} $

$ = \log \frac{2}{3} - \log \frac{1}{2} = \log \frac{4}{3}$.

$= \int_{\pi /3}^{\pi /2} {\,\,\frac{{\sin x}}{{{{(1 - \cos x)}^3}}}\,dx} $

Now, put $1 - \cos x = t$

Also, when $x = \frac{\pi }{3},t = \frac{1}{2}$ and $x = \frac{\pi }{2}\,,\,\,t = 1$

Therefore, $I = \int_{1/2}^1 {\frac{{dt}}{{{t^3}}} = \left| {\frac{{{t^{ - 2}}}}{{ - 2}}} \right|} _{1/2}^1 = \frac{3}{2}$.

$\therefore $ $dx = {\sec ^2}\theta \,d\theta $

As $x = 1 \Rightarrow \theta = \frac{\pi }{4}$ and

$x = 0 \Rightarrow \theta = 0$, then

$I = 2\int_0^{\pi /4} {\theta {{\sec }^2}\theta \,d\theta = 2[\theta \tan \theta ]_0^{\pi /4} - 2\int_0^{\pi /4} {\tan \theta \,d\theta } } $

$= \frac{\pi }{2} + 2\,[\log \cos x]_0^{\pi /4} = \frac{\pi }{2} - 2\log \sqrt 2 $.

Put $t = \cos \theta \Rightarrow dt = - \sin \theta \,\,d\theta ,$ then

$I =$$ - \int_1^0 {{t^{1/2}}(1 - {t^2})dt = \int_0^1 {({t^{1/2}} - {t^{5/2}})} } $$dt$

$I = $$\left[ {\frac{2}{3}{t^{3/2}} - \frac{2}{7}{t^{7/2}}} \right]_0^1 = \frac{8}{{21}}$.

$=\int_0^{\pi /4} {\frac{{{{\sin }^3}\theta }}{{{{\cos }^3}\theta }}.{{\sec }^4}\theta \,d\theta } $

Putting $\tan \theta = t,$ it reduces to

$\int_0^1 {{t^3}(1 + {t^2})\,dt}  =$$  \left| {\frac{{{t^4}}}{4} + \frac{{{t^6}}}{6}} \right|_0^1 = \frac{5}{{12}}$.

$\Rightarrow dx = {\sec ^2}\theta \,\,d\theta $

Also as $x = 0,\theta = 0$ and $x = 1,\theta = \frac{\pi }{4}$

Therefore, $\int_0^1 {{{\tan }^{ - 1}}x\,dx = \int_0^{\pi /4} {\theta {{\sec }^2}\theta \,d\theta } } $

$ = \frac{\pi }{4}$$ - \log \sqrt 2 = \frac{\pi }{4} - \frac{1}{2}\log 2$.

Put $t = \sin \theta \Rightarrow dt = \cos \theta \,\,d\theta ,$

then we have

$\int_{1/\sqrt 2 }^1 {\frac{1}{{{t^2}}}dt} = \left[ {\frac{{ - 1}}{t}} \right]_{1/\sqrt 2 }^1 $

$= \sqrt 2 - 1$.

Put ${\sin ^{ - 1}}x = t$

$\Rightarrow \frac{1}{{\sqrt {1 - {x^2}} }}dx = dt$ and $x = \sin t$

Also $t = 0$ to $\frac{\pi }{4}$

as $x = 0$ to $\frac{1}{{\sqrt 2 }}$

$ \Rightarrow I = \int_0^{\pi /4} {t.{{\sec }^2}t\,dt = \frac{\pi }{4} - \frac{1}{2}\log 2} $.

$= 2\int_0^{\pi /2} {{{\sin }^2}x\cos xdx} $

Put $t = \sin x \Rightarrow dt = \cos x\,dx$

Now, $I = 2\int_0^1 {{t^2}dt = \frac{2}{3}[{t^3}]_0^1 = \frac{2}{3}} $.

$ = \int_0^{\pi /2} {\frac{{dx}}{{2{{\sin }^2}\frac{x}{2} + 2{{\cos }^2}\frac{x}{2} + {{\cos }^2}\frac{x}{2} - {{\sin }^2}\frac{x}{2}}}} $

$ = \int_0^{\pi /2} {\frac{{dx}}{{{{\sin }^2}\frac{x}{2} + 3{{\cos }^2}\frac{x}{2}}}} $

$= \int_0^{\pi /2} {\frac{{{{\sec }^2}\frac{x}{2}}}{{3 + {{\tan }^2}\frac{x}{2}}}dx} $

Put $t = \tan \frac{x}{2} $

$\Rightarrow dt = \frac{1}{2}{\sec ^2}\frac{x}{2}dx$, then

$I = 2\int_0^1 {\frac{{dt}}{{3 + {t^2}}} = \frac{2}{{\sqrt 3 }}{{\tan }^{ - 1}}\left( {\frac{1}{{\sqrt 3 }}} \right)} $.

As $x = 2 \Rightarrow t = \log 2$

and $x = 1 \Rightarrow t = 0$, we have

$\int_1^2 {\frac{{\cos (\log x)}}{x}} dx = - \int_0^{\log 2} {\cos t\,dt} = [\sin t]_0^{\log 2}$$ = \sin (\log 2)$.

$\Rightarrow dx = - 2\sin \theta \,d\theta ,$ then

$\int_0^2 {\sqrt {\frac{{2 + x}}{{2 - x}}} } dx = - 2\int_{\pi /2}^0 {\sqrt {\frac{{1 + \cos \theta }}{{1 - \cos \theta }}} } \sin \theta \,d\theta $

$ = 4\int_0^{\pi /2} {\frac{{\cos (\theta /2)}}{{\sin (\theta /2)}}\sin \frac{\theta }{2}\cos \frac{\theta }{2}d\theta } $

$ = 2\int_0^{\pi /2} {(1 + \cos \theta )\,d\theta } $

$ = 2[\theta + \sin \theta ]_0^{\pi /2} $

$= 2\left[ {\frac{\pi }{2} + 1} \right] = \pi + 2$.

$ = [\tan x - \sec x]_0^\pi = [\tan \pi - \sec \pi + 1] $

$= [0 + 1 + 1] = 2$.

Now $\int_0^{\pi /2} {\frac{{\sin x\cos x}}{{1 + {{\sin }^4}x}}dx = \frac{1}{2}\int_0^1 {\frac{1}{{1 + {t^2}}}dt = \frac{1}{2}[{{\tan }^{ - 1}}t]_0^1 = \frac{\pi }{8}} } $.

$\int_0^2 {\frac{{{x^3}}}{{{{({x^2} + 1)}^{3/2}}}}dx = \frac{1}{2}} \int_1^5 {\frac{{(t - 1)}}{{{t^{3/2}}}}dt = \frac{1}{2}\int_1^5 {[{t^{ - 1/2}} - {t^{ - 3/2}}]\,dt} } $

$ = \frac{1}{2}\left[ {2\sqrt t + 2\frac{1}{{\sqrt t }}} \right]_1^5 $

$= \frac{1}{2}\left[ {2\sqrt 5 + \frac{2}{{\sqrt 5 }} - 2 - 2} \right]$

$ = \left[ {\sqrt 5 + \frac{1}{{\sqrt 5 }} - 2} \right] $

$= \frac{{6 - 2\sqrt 5 }}{{\sqrt 5 }}$.

We put $\cos x = t \Rightarrow - \sin x\,dx = dt,$ then

$I = \int_0^1 {\frac{{t.dt}}{{{t^2} + 3t + 2}} = \int_0^1 {\left[ {\frac{2}{{t + 2}} - \frac{1}{{t + 1}}} \right]} } \,dt$

$ = [2\log (t + 2) - \log (\,t + 1)]_0^1$

$ = [2\log 3 - \log 2 - 2\log 2]$

$ = [2\log 3 - 3\log 2] = [\log 9 - \log 8] $

$= \log \left( {\frac{9}{8}} \right)$.

$\Rightarrow {e^x}dx = 2t\,dt$

Also as $x = 0$ to $\log 5,t = 0$ to $2$

Therefore, $\int_0^{\log 5} {\frac{{{e^x}\sqrt {{e^x} - 1} }}{{{e^x} + 3}}} dx = \int_0^2 {\frac{{2{t^2}}}{{{t^2} + 4}}dt} $

$ = 2\left[ {\int_0^2 {1dt - 4\int_0^2 {\frac{{dt}}{{{t^2} + 4}}} } } \right] = 4 - \pi $.

${I_n} = \int_0^{\pi /4} {{{\sec }^2}\theta {{\tan }^{n - 2}}\theta \,d\theta } - \int_0^{\pi /2} {{{\tan }^{n - 2}}\theta } \,d\theta $

${I_n} = \left[ {\frac{{{{\tan }^{n - 1}}\theta }}{{n - 1}}} \right]_0^{\pi /4} - {I_{n - 2}} $

$\Rightarrow {I_n} + {I_{n - 2}} = \frac{1}{{n - 1}}$

Hence ${I_8} + {I_6} = \frac{1}{{8 - 1}} = \frac{1}{7}$.

Put $\log (\sec x + \tan x) = t \Rightarrow \sec x\,dx = dt$

$ \Rightarrow I = \int_0^{\log (\sqrt 2 + 1)} {t\,dt = \left[ {\frac{{{t^2}}}{2}} \right]} _0^{\log (\sqrt 2 + 1)} $

$= \frac{{{{[\log (\sqrt 2 + 1)]}^2}}}{2}$.

$\therefore \,\,\,\int_0^{\pi /4} {\frac{{{{\sec }^2}x}}{{(1 + \tan x)(2 + \tan x)}}dx} $

$ = \int_1^2 {\frac{{dt}}{{t(1 + t)}}} = \int_1^2 {\frac{{dt}}{t} - \int_1^2 {\frac{{dt}}{{1 + t}}} } = [\log t - \log (1 + t)]_1^2$

$ = {\log _e}2 - {\log _e}3 + {\log _e}2 = {\log _e}\frac{4}{3}$.

Dividing the numerator and denominator by ${\cos ^2}x,$ we get

$I = \int_0^{\pi /2} {\frac{{\frac{1}{{{{\cos }^2}x}}dx}}{{{a^2} + {b^2}\frac{{{{\sin }^2}x}}{{{{\cos }^2}x}}}} = \int_0^{\pi /2} {\frac{{{{\sec }^2}x}}{{{a^2} + {b^2}{{\tan }^2}x}}dx} } $.

Substituting $b\,\,\tan x = t$ and $b\,\,{\sec ^2}x\,dx = dt$ and limit when $x = 0$, 

then $t = 0$ and when $x = \frac{\pi }{2},$ then $t = \infty ,$

therefore, $I = \int_0^\infty {\frac{{\frac{{dt}}{b}}}{{{a^2} + {t^2}}}} = \frac{1}{b}\left[ {\frac{1}{a}{{\tan }^{ - 1}}\left( {\frac{t}{a}} \right)} \right]_0^\infty $

$ = \frac{1}{{ab}}\left[ {{{\tan }^{ - 1}}\infty - {{\tan }^{ - 1}}0} \right] $

$= \frac{1}{{ab}}\left( {\frac{\pi }{2} - 0} \right) = \frac{\pi }{{2ab}}$.

$I = \int_0^{\pi /8} {\,(1 - {{\sin }^2}4\theta )\cos 4\theta \,d\theta } $

Put $\sin 4\theta = t \Rightarrow \cos 4\theta \,d\theta = \frac{{dt}}{4}$

When $\theta = 0 \to \frac{\pi }{8},$ then $t = 0 \to 1$

$\therefore$  $I = \frac{1}{4}\int_0^1 {(1 - {t^2})dt = \frac{1}{4}} \left[ {t - \frac{{{t^3}}}{3}} \right]_0^1 = \frac{1}{6}$.

Put $\frac{x}{c} = t \Rightarrow dx = c\,dt$ and

$x = bc \Rightarrow t = b$

$x = ac \Rightarrow t = a$ then,

$I = \int_a^b {f(t)dt = \int_a^b {f(x)dx} } $.