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Maths M.C.Q (1 Marks)

7.1 inderfinite integral · Gujarat Board (GSEB) STD 12 Science (GSEB - English Medium). 565 questions.

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M.C.Q (1 Marks)

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50 questions · auto-graded multiple-choice test.

MCQ 11 Mark
$\int_{}^{} {\frac{{dx}}{{\tan x + \cot x}}} = $
  • A
    $\frac{{\cos 2x}}{4} + c$
  • B
    $\frac{{\sin 2x}}{4} + c$
  • C
    $ - \frac{{\sin 2x}}{4} + c$
  • ✓
    $ - \frac{{\cos 2x}}{4} + c$
Answer
Correct option: D.
$ - \frac{{\cos 2x}}{4} + c$
d
(d)$\int_{}^{} {\frac{{dx}}{{\tan x + \cot x}}} = \int_{}^{} {\frac{{dx}}{{\frac{{\sin x}}{{\cos x}} + \frac{{\cos x}}{{\sin x}}}}} $
$ = \frac{1}{2}\int_{}^{} {2\sin x\cos x\,dx} = \frac{{ - \cos 2x}}{4} + c$.
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MCQ 21 Mark
$\int_{}^{} {\frac{{dx}}{{\sqrt x + \sqrt {x - 2} }} = } $
  • ✓
    $\frac{1}{3}[{x^{3/2}} - {(x - 2)^{3/2}}] + c$
  • B
    $\frac{2}{3}[{x^{3/2}} - {(x - 2)^{3/2}}] + c$
  • C
    $\frac{1}{3}[{(x - 2)^{3/2}} - {x^{3/2}}] + c$
  • D
    $\frac{2}{3}[{(x - 2)^{3/2}} - {x^{3/2}}] + c$
Answer
Correct option: A.
$\frac{1}{3}[{x^{3/2}} - {(x - 2)^{3/2}}] + c$
a
(a)$\int_{}^{} {\frac{{dx}}{{\sqrt x + \sqrt {x - 2} }} = \frac{1}{2}\int_{}^{} {\frac{{x - (x - 2)}}{{\sqrt x + \sqrt {x - 2} }}\,dx} } $
$ = \frac{1}{2}\int_{}^{} {(\sqrt x - \sqrt {x - 2} )\,dx} = \frac{1}{2}\left[ {\frac{{{x^{32}}}}{{32}} - \frac{{{{(x - 2)}^{32}}}}{{32}}} \right] + c$
$ = \frac{1}{3}\left\{ {{x^{32}} - {{(x - 2)}^{32}}} \right\} + c.$
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MCQ 31 Mark
If $\int_{}^{} {(\sin 2x - \cos 2x)} \;dx = \frac{1}{{\sqrt 2 }}\sin (2x - a) + b$, then
  • A
    $a = \frac{\pi }{4},\;b = 0$
  • B
    $a = - \frac{\pi }{4},\;b = 0$
  • C
    $a = \frac{{5\pi }}{4},\;b = $any constant
  • ✓
    $a = - \frac{{5\pi }}{4},\;b = $any constant
Answer
Correct option: D.
$a = - \frac{{5\pi }}{4},\;b = $any constant
d
(d)$\int_{}^{} {(\sin 2x - \cos 2x)\,dx = \frac{1}{{\sqrt 2 }}\sin (2x - a) + b} $
$ \Rightarrow - \frac{1}{2}(\sin 2x + \cos 2x) = \frac{1}{{\sqrt 2 }}\sin (2x - a) + b$
$ \Rightarrow - \left[ {\frac{1}{{\sqrt 2 }}\sin 2x + \frac{1}{{\sqrt 2 }}\cos 2x} \right] = \sin (2x - a) + b\sqrt 2 $
$ \Rightarrow \sin \left( {2x + \frac{{5\pi }}{4}} \right) = \sin (2x - a) + b\sqrt 2 $
$ \Rightarrow b$ is any constant and $a = \frac{{ - 5\pi }}{4}$.
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MCQ 41 Mark
$\int_{}^{} {\left( {1 + x + \frac{{{x^2}}}{{2\;!}} + \frac{{{x^3}}}{{3\;!}} + ..........} \right)\;dx = } $
  • A
    $ - {e^x} + c$
  • ✓
    ${e^x} + c$
  • C
    ${e^{ - x}} + c$
  • D
    $ - {e^{ - x}} + c$
Answer
Correct option: B.
${e^x} + c$
b
(b)$\int_{}^{} {\left( {1 + x + \frac{{{x^2}}}{{2\,!}} + \frac{{{x^3}}}{{3\,!}} + .......} \right){\rm{ }}dx = \int_{}^{} {{e^x}dx = {e^x} + c.} } $
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MCQ 51 Mark
$\int_{}^{} {{{(\sec x + \tan x)}^2}dx = } $
  • ✓
    $2(\sec x + \tan x) - x + c$
  • B
    $1/3{(\sec x + \tan x)^3} + c$
  • C
    $\sec x(\sec x + \tan x) + c$
  • D
    $2(\sec x + \tan x) + c$
Answer
Correct option: A.
$2(\sec x + \tan x) - x + c$
a
(a)$\int_{}^{} {{{(\sec x + \tan x)}^2}dx} $
$ = \int_{}^{} {({{\sec }^2}x + {{\tan }^2}x + 2\sec x\tan x)\,dx} $
$ = \int_{}^{} {(2{{\sec }^2}x - 1 + 2\sec x\tan x)\,dx} $
$ = 2\tan x + 2\sec x - x + c = 2(\sec x + \tan x) - x + c.$
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MCQ 61 Mark
$\int_{}^{} {\frac{{\tan x}}{{\sec x + \tan x}}\;dx = } $
  • A
    $\sec x + \tan x - x + c$
  • ✓
    $\sec x - \tan x + x + c$
  • C
    $\sec x + \tan x + x + c$
  • D
    $ - \sec x - \tan x + x + c$
Answer
Correct option: B.
$\sec x - \tan x + x + c$
b
(b)$\int_{}^{} {\frac{{\tan x}}{{(\sec x + \tan x)}}\,dx} = \int_{}^{} {\frac{{\tan x(\sec x - \tan x)}}{{(\sec x + \tan x)(\sec x - \tan x)}}\,dx} $
Multiplying ${N^r}$ and $D'$ by $(\sec x - \tan x),$ we get
$ = \int_{}^{} {\frac{{\tan x(\sec x - \tan x)}}{{({{\sec }^2}x - {{\tan }^2}x)}}\,dx} = \int_{}^{} {(\sec x\tan x - {{\tan }^2}x)\,dx} $
$ = \int_{}^{} {\sec x\tan x\,dx} - \int_{}^{} {({{\sec }^2}x - 1)\,dx} $
$ = \int_{}^{} {\sec x\tan x\,dx} - \int_{}^{} {{{\sec }^2}x\,dx} + \int_{}^{} {1\,dx} $
$ = \sec x - \tan x + x + c$.
Trick : By inspection,
$\frac{d}{{dx}}\left\{ {\sec x + \tan x} \right\} = \sec x\tan x + {\sec ^2}x$
$ = \sec x(\sec x + \tan x) = \frac{{\sec x}}{{\sec x - \tan x}}$
$ \Rightarrow \frac{d}{{dx}}\left\{ {\sec x - \tan x + x + c} \right\} = \sec x\tan x - {\sec ^2}x + 1$
$ = - \sec x(\sec x - \tan x) + 1 = \frac{{ - \sec x}}{{\sec x + \tan x}} + 1 = \frac{{\tan x}}{{\sec x + \tan x}}$.
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MCQ 71 Mark
$\int_{}^{} {\frac{{dx}}{{{{\sin }^2}x{{\cos }^2}x}} = } $
  • A
    $\tan x + \cot x + c$
  • B
    $\cot x - \tan x + c$
  • ✓
    $\tan x - \cot x + c$
  • D
    None of these
Answer
Correct option: C.
$\tan x - \cot x + c$
c
(c)$\int_{}^{} {\frac{{dx}}{{{{\sin }^2}x{{\cos }^2}x}}} = \int_{}^{} {\frac{{({{\cos }^2}x + {{\sin }^2}x)\,dx}}{{{{\cos }^2}x{{\sin }^2}x}}} $
$ = \int_{}^{} {{\rm{cose}}{{\rm{c}}^2}x\,dx} + \int_{}^{} {{{\sec }^2}x\,dx} = - \cot x + \tan x + c$.
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MCQ 81 Mark
$\int_{}^{} {\frac{{x - 1}}{{{{(x + 1)}^2}}}\;dx = } $
  • ✓
    $\log (x + 1) + \frac{2}{{x + 1}} + c$
  • B
    $\log (x + 1) - \frac{2}{{x + 1}} + c$
  • C
    $\frac{2}{{x + 1}} - \log (x + 1) + c$
  • D
    None of these
Answer
Correct option: A.
$\log (x + 1) + \frac{2}{{x + 1}} + c$
a
(a) $\int_{}^{} {\frac{{x - 1}}{{{{(x + 1)}^2}}}\,dx = \int_{}^{} {\frac{{x + 1 - 2}}{{{{(x + 1)}^2}}}} \,dx} $
$ = \int_{}^{} {\frac{1}{{x + 1}}\,dx} - \int_{}^{} {\frac{2}{{{{(x + 1)}^2}}}\,dx = \log (x + 1) + \frac{2}{{(x + 1)}} + c} $.
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MCQ 91 Mark
$\int_{}^{} {\frac{{{\rm{cosec}}\theta - \cot \theta }}{{{\rm{cosec}}\theta + \cot \theta }}} \;d\theta = $
  • ✓
    $2{\rm{cosec}}\theta - 2\cot \theta - \theta + c$
  • B
    $2\,{\rm{cosec}}\theta - 2\cot \theta + \theta + c$
  • C
    $2\,{\rm{cosec}}\theta + 2\cot \theta - \theta + c$
  • D
    None of these
Answer
Correct option: A.
$2{\rm{cosec}}\theta - 2\cot \theta - \theta + c$
a
(a)$\int_{}^{} {\frac{{{\rm{cosec}}\theta - \cot \theta }}{{{\rm{cosec}}\theta + \cot \theta }}\,d\theta } = \int_{}^{} {{{({\rm{cosec}}\theta - \cot \theta )}^2}d\theta } $
$ = \int_{}^{} {{\rm{cose}}{{\rm{c}}^2}\theta \,d\theta } + \int_{}^{} {{{\cot }^2}\theta \,d\theta } - 2\int_{}^{} {{\rm{cosec}}\theta \cot \theta \,d\theta } $
$ = \int_{}^{} {(2{\rm{cose}}{{\rm{c}}^2}\theta - 1)\,d\theta } - 2\int_{}^{} {{\rm{cosec}}\theta \cot \theta \,d\theta } $
$ = 2{\rm{cosec}}\theta - 2\cot \theta - \theta + c.$
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MCQ 101 Mark
If $\int_{}^{} {(\cos x - \sin x)\;dx = \sqrt 2 \sin (x + \alpha ) + c} $, then $\alpha = $
  • A
    $\frac{\pi }{3}$
  • B
    $ - \frac{\pi }{3}$
  • ✓
    $\frac{\pi }{4}$
  • D
    $ - \frac{\pi }{4}$
Answer
Correct option: C.
$\frac{\pi }{4}$
c
(c) Given that $\int_{}^{} {(\cos x - \sin x)\,dx} = \sqrt 2 \sin (x + \alpha ) + c$
$ \Rightarrow \sin x + \cos x + c = \sqrt 2 \sin (x + \alpha ) + c$
$ \Rightarrow \sqrt 2 \left( {\frac{{\sin x}}{{\sqrt 2 }} + \frac{{\cos x}}{{\sqrt 2 }}} \right) + c = \sqrt 2 \sin (x + \alpha ) + c$
$ \Rightarrow \sqrt 2 \sin \left( {x + \frac{\pi }{4}} \right) + c = \sqrt 2 \sin (x + \alpha ) + c$
$ \Rightarrow \sin \left( {x + \frac{\pi }{4}} \right) = \sin (x + \alpha ) \Rightarrow \alpha = \frac{\pi }{4}.$
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MCQ 111 Mark
If $\int_{}^{} {\frac{{dx}}{{1 + \sin x}} = \tan \left( {\frac{x}{2} + a} \right) + b} $, then
  • A
    $a = \frac{\pi }{4},\;b = 3$
  • B
    $a = - \frac{\pi }{4},\;b = 3$
  • C
    $a = \frac{\pi }{4},\;b = $arbitrary constant
  • ✓
    $a = - \frac{\pi }{4},\;b = $arbitrary constant
Answer
Correct option: D.
$a = - \frac{\pi }{4},\;b = $arbitrary constant
d
(d)$\int_{}^{} {\frac{{dx}}{{1 + \sin x}}} = \tan x - \sec x + c = - \frac{{1 - \sin x}}{{\cos x}}$
$ = - \frac{{{{\left( {\cos \frac{x}{2} - \sin \frac{x}{2}} \right)}^2}}}{{{{\cos }^2}\frac{x}{2} - {{\sin }^2}\frac{x}{2}}} + c = - \frac{{1 - \tan \frac{x}{2}}}{{1 + \tan \frac{x}{2}}} + c$
$ = \frac{{\tan \frac{x}{2} - 1}}{{1 + \tan \frac{x}{2}}} + c = \tan \left( {\frac{x}{2} - \frac{\pi }{4}} \right) + c$
$ \Rightarrow a = - \frac{\pi }{4},\,\,b = $ arbitrary constant.
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MCQ 121 Mark
$\int_{}^{} {\frac{{dx}}{{\sin x + \cos x}}} = $
  • A
    $\log \tan \left( {\frac{\pi }{8} + \frac{x}{2}} \right) + c$
  • B
    $\log \tan \left( {\frac{\pi }{8} - \frac{x}{2}} \right) + c$
  • ✓
    $\frac{1}{{\sqrt 2 }}\log \tan \left( {\frac{\pi }{8} + \frac{x}{2}} \right) + c$
  • D
    None of these
Answer
Correct option: C.
$\frac{1}{{\sqrt 2 }}\log \tan \left( {\frac{\pi }{8} + \frac{x}{2}} \right) + c$
c
(c)$\int_{}^{} {\frac{{dx}}{{\sin x + \cos x}}} = \frac{1}{{\sqrt 2 }}\int_{}^{} {\frac{{dx}}{{\sin x\cos \frac{\pi }{4} + \cos x\sin \frac{\pi }{4}}}} $
$ = \frac{1}{{\sqrt 2 }}\int_{}^{} {{\rm{cosec }}\left( {x + \frac{\pi }{4}} \right)\,dx = \frac{1}{{\sqrt 2 }}\log \tan \left( {\frac{\pi }{8} + \frac{x}{2}} \right)} + c.$
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MCQ 131 Mark
$\int_{}^{} {\frac{{\cos 2x + 2{{\sin }^2}x}}{{{{\cos }^2}x}}dx = } $
  • A
    $2\sec x + c$
  • B
    $2\tan x + c$
  • ✓
    $\tan x + c$
  • D
    None of these
Answer
Correct option: C.
$\tan x + c$
c
(c)$\int_{}^{} {\frac{{\cos 2x + 2{{\sin }^2}x}}{{{{\cos }^2}x}}\,dx} = \int_{}^{} {\frac{{2({{\cos }^2}x + {{\sin }^2}x) - 1}}{{{{\cos }^2}x}}\,dx} $
$ = \int_{}^{} {{{\sec }^2}x\,dx} = \tan x + c$.
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MCQ 141 Mark
$\int_{}^{} {\frac{{5({x^6} + 1)}}{{{x^2} + 1}}dx = } $
  • A
    $5({x^7} + x){\tan ^{ - 1}}x + c$
  • ✓
    ${x^5} - \frac{5}{3}{x^3} + 5x + c$
  • C
    $3{x^4} - 5{x^2} + 15x + c$
  • D
    $5{\tan ^{ - 1}}({x^2} + 1) + \log ({x^2} + 1) + c$
Answer
Correct option: B.
${x^5} - \frac{5}{3}{x^3} + 5x + c$
b
(b)$\int_{}^{} {\frac{{5({x^6} + 1)}}{{{x^2} + 1}}\,dx = \int_{}^{} {\frac{{5({x^2} + 1)({x^4} - {x^2} + 1)}}{{({x^2} + 1)}}\,dx} } $
$ = \int_{}^{} {5({x^4} - {x^2} + 1)\,dx = {x^5} - \frac{5}{3}{x^3} + 5x + c.} $
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MCQ 151 Mark
$\int_{}^{} {\frac{{dx}}{{\sin x + \sqrt 3 \cos x}}} = $
  • A
    $\log \tan \left( {\frac{x}{2} + \frac{\pi }{2}} \right) + c$
  • ✓
    $\frac{1}{2}\log \tan \left( {\frac{x}{2} + \frac{\pi }{6}} \right) + c$
  • C
    $\log \cot \left( {\frac{x}{2} + \frac{\pi }{6}} \right) + c$
  • D
    $\frac{1}{2}\log \cot \left( {\frac{x}{2} + \frac{\pi }{6}} \right) + c$
Answer
Correct option: B.
$\frac{1}{2}\log \tan \left( {\frac{x}{2} + \frac{\pi }{6}} \right) + c$
b
(b)$\int_{}^{} {\frac{{dx}}{{\sin x + \sqrt 3 \cos x}}} = \frac{1}{2}\int_{}^{} {\frac{{dx}}{{\frac{{\sin x}}{2} + \frac{{\sqrt 3 }}{2}\cos x}}} $
$ = \frac{1}{2}\int_{}^{} {\frac{{dx}}{{\sin \left( {x + \frac{\pi }{3}} \right)}}} = \frac{1}{2}\int_{}^{} {{\rm{cosec}}\left( {x + \frac{\pi }{3}} \right)} $
$ = \frac{1}{2}\log \tan \left( {\frac{x}{2} + \frac{\pi }{6}} \right) + c.$
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MCQ 161 Mark
If $\int_{}^{} {\frac{{f(x)\;dx}}{{\log \sin x}} = \log \log \sin x} $, then $f(x) = $
  • A
    $\sin x$
  • B
    $\cos x$
  • C
    $\log \sin x$
  • ✓
    $\cot x$
Answer
Correct option: D.
$\cot x$
d
(d)$\int_{}^{} {\frac{{f(x)\,dx}}{{\log \sin x}}} = \log \log \sin x$
Differentiating both sides, we get
$\frac{{f(x)}}{{\log \sin x}} = \frac{{\cot x}}{{\log \sin x}} \Rightarrow f(x) = \cot x.$
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MCQ 171 Mark
$\int_{}^{} {\frac{1}{{\sqrt {1 + \sin x} }}dx} = $
  • A
    $2\sqrt 2 \log \tan \left( {\frac{\pi }{8} + \frac{x}{4}} \right) + c$
  • B
    $\frac{1}{{\sqrt 2 }}\log \tan \left( {\frac{\pi }{8} + \frac{x}{4}} \right) + c$
  • ✓
    $\sqrt 2 \log \tan \left( {\frac{\pi }{8} + \frac{x}{4}} \right) + c$
  • D
    $\frac{1}{{2\sqrt 2 }}\log \tan \left( {\frac{\pi }{8} + \frac{x}{4}} \right) + c$
Answer
Correct option: C.
$\sqrt 2 \log \tan \left( {\frac{\pi }{8} + \frac{x}{4}} \right) + c$
c
(c)$\int_{}^{} {\frac{1}{{\sqrt {1 + \sin x} }}} \,dx = \int_{}^{} {\frac{1}{{\sqrt 2 \sin \left( {\frac{\pi }{4} + \frac{x}{2}} \right)}}} \,dx$
$ = \frac{1}{{\sqrt 2 }}\int_{}^{} {{\rm{cosec}}\,\left( {\frac{x}{{\rm{2}}} + \frac{\pi }{4}} \right)} \,dx = \sqrt 2 \log \tan \left( {\frac{\pi }{8} + \frac{x}{4}} \right) + c.$
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MCQ 181 Mark
$\int_{}^{} {{a^x}\;da = } $
  • A
    $\frac{{{a^x}}}{{{{\log }_e}a}} + c$
  • B
    ${a^x}{\log _e}a + c$
  • ✓
    $\frac{{{a^{x + 1}}}}{{x + 1}} + c$
  • D
    $x{a^{x - 1}} + c$
Answer
Correct option: C.
$\frac{{{a^{x + 1}}}}{{x + 1}} + c$
c
(c)$\int_{}^{} {{a^x}da} = \frac{{{a^{x + 1}}}}{{x + 1}} + c.$
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MCQ 191 Mark
$\int_{}^{} {\frac{{{e^{5\log x}} - {e^{4\log x}}}}{{{e^{3\log x}} - {e^{2\log x}}}}\;dx = } $
  • A
    $e\;.\;{3^{ - 3x}} + c$
  • B
    ${e^3}\log x + c$
  • ✓
    $\frac{{{x^3}}}{3} + c$
  • D
    None of these
Answer
Correct option: C.
$\frac{{{x^3}}}{3} + c$
c
(c)$\int_{}^{} {\frac{{{e^{5\log x}} - {e^{4\log x}}}}{{{e^{3\log x}} - {e^{2\log x}}}}\,dx} = \int_{}^{} {\frac{{{x^5} - {x^4}}}{{{x^3} - {x^2}}}\,dx} $
$ = \int_{}^{} {\frac{{{x^4}(x - 1)}}{{{x^2}(x - 1)}}\,dx} = \int_{}^{} {{x^2}dx} = \frac{{{x^3}}}{3} + c$.
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MCQ 201 Mark
$\int_{}^{} {\sqrt {1 + \sin x} \;dx = } $
  • A
    $\frac{1}{2}\left( {\sin \frac{x}{2} + \cos \frac{x}{2}} \right) + c$
  • B
    $\frac{1}{2}\left( {\sin \frac{x}{2} - \cos \frac{x}{2}} \right) + c$
  • C
    $2\sqrt {1 + \sin x} + c$
  • ✓
    $ - 2\sqrt {1 - \sin x} + c$
Answer
Correct option: D.
$ - 2\sqrt {1 - \sin x} + c$
d
(d)$\int_{}^{} {\sqrt {1 + \sin x} \,dx} = \int_{}^{} {\sqrt {{{\left( {\sin \frac{x}{2} + \cos \frac{x}{2}} \right)}^2}} } dx$
$ = \int_{}^{} {\sin \frac{x}{2}\,dx} + \int_{}^{} {\cos \frac{x}{2}\,dx} = - 2\cos \frac{x}{2} + 2\sin \frac{x}{2} + c$
$ = - 2\left( {\cos \frac{x}{2} - \sin \frac{x}{2}} \right) + c = - 2\sqrt {(1 - \sin x)} + c.$
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MCQ 211 Mark
$\int_{}^{} {{e^{x\log a}}.\;{e^x}\;dx} $is equal to
  • A
    ${(ae)^x} + c$
  • ✓
    $\frac{{{{(ae)}^x}}}{{\log (ae)}} + c$
  • C
    $\frac{{{e^x}}}{{1 + \log a}} + c$
  • D
    None of these
Answer
Correct option: B.
$\frac{{{{(ae)}^x}}}{{\log (ae)}} + c$
b
(b)$\int_{}^{} {{e^{x\log a}}{e^x}dx} = \int_{}^{} {{e^{\log {a^x}}}.\,{e^x}dx} = \int_{}^{} {{a^x}{e^x}dx} $
$ = \int_{}^{} {{{(ae)}^x}dx} = \frac{{{{(ae)}^x}}}{{\log (ae)}} + C.$
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MCQ 221 Mark
$\int_{}^{} {\frac{1}{{\sqrt {1 + \cos x} }}\;dx = } $
  • ✓
    $\sqrt 2 \log \left( {\sec \frac{x}{2} + \tan \frac{x}{2}} \right) + K$
  • B
    $\frac{1}{{\sqrt 2 }}\log \left( {\sec \frac{x}{2} + \tan \frac{x}{2}} \right) + K$
  • C
    $\log \left( {\sec \frac{x}{2} + \tan \frac{x}{2}} \right) + K$
  • D
    None of these
Answer
Correct option: A.
$\sqrt 2 \log \left( {\sec \frac{x}{2} + \tan \frac{x}{2}} \right) + K$
a
(a)$\int_{}^{} {\frac{1}{{\sqrt {1 + \cos x} }}} \,dx = \int_{}^{} {\frac{{dx}}{{\sqrt {2{{\cos }^2}(x2)} }}} = \frac{1}{{\sqrt 2 }}\int_{}^{} {\sec \frac{x}{2}\,dx} $
$ = \frac{1}{{\sqrt 2 }}\left\{ {\log \left( {\sec \frac{x}{2} + \tan \frac{x}{2}} \right)} \right\}.\frac{1}{{12}} = \sqrt 2 \log \left( {\sec \frac{x}{2} + \tan \frac{x}{2}} \right) + K$.
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MCQ 231 Mark
If $\int {\sqrt 2 \sqrt {1 + \sin x} } \,\,dx = - \,4\cos (ax + b) + c$ then the value of $(a, b) $ is
  • ✓
    $\frac{1}{2},\,\frac{\pi }{4}$
  • B
    $1,\,\frac{\pi }{2}$
  • C
    $1, 1$
  • D
    None of these
Answer
Correct option: A.
$\frac{1}{2},\,\frac{\pi }{4}$
a
(a) $I = \int {\sqrt 2 \sqrt {1 + \sin x} } \,dx$$ = \sqrt 2 \int {\left( {\sin \frac{x}{2} + \cos \frac{x}{2}} \right)\,dx} $
$ = 2\int {\sin \left( {\frac{\pi }{4} + \frac{x}{2}} \right)\,dx = - \,4\cos \,\left( {\frac{x}{2} + \frac{\pi }{4}} \right) + c} $
On comparing, $a = \frac{1}{2},\,b = \frac{\pi }{4}.$
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MCQ 241 Mark
$\int_{}^{} {{{\tan }^4}x\;dx = } $
  • A
    ${\tan ^3}x - \tan x + x + c$
  • ✓
    $\frac{1}{3}{\tan ^3}x - \tan x + x + c$
  • C
    $\frac{1}{3}{\tan ^3}x + \tan x + x + c$
  • D
    $\frac{1}{3}{\tan ^3}x + \tan x + 2x + c$
Answer
Correct option: B.
$\frac{1}{3}{\tan ^3}x - \tan x + x + c$
b
(b)$\int_{}^{} {{{\tan }^4}x\,dx} = \int_{}^{} {{{\tan }^2}x({{\sec }^2}x - 1)\,dx} $
$ = \int_{}^{} {{{\tan }^2}x{{\sec }^2}x\,dx} - \int_{}^{} {{{\tan }^2}x\,dx} = \frac{{{{\tan }^3}x}}{3} - \tan x + x + c$.
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MCQ 251 Mark
$\int_{}^{} {\frac{{\sin 2x}}{{\sin 5x\sin 3x}}} \;dx = $
  • A
    $\log \sin 3x - \log \sin 5x + c$
  • B
    $\frac{1}{3}\log \sin 3x + \frac{1}{5}\log \sin 5x + c$
  • ✓
    $\frac{1}{3}\log \sin 3x - \frac{1}{5}\log \sin 5x + c$
  • D
    $3\log \sin 3x - 5\log \sin 5x + c$
Answer
Correct option: C.
$\frac{1}{3}\log \sin 3x - \frac{1}{5}\log \sin 5x + c$
c
(c)$\int_{}^{} {\frac{{\sin 2x}}{{\sin 5x\sin 3x}}\,dx} = \int_{}^{} {\frac{{\sin (5x - 3x)}}{{\sin 5x\sin 3x}}\,dx} $
$ = \int_{}^{} {\frac{{\sin 5x\cos 3x - \cos 5x\sin 3x}}{{\sin 5x\sin 3x}}\,dx} $
$ = \frac{1}{3}\log \sin 3x - \frac{1}{5}\log \sin 5x + c.$
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MCQ 261 Mark
$\int {{\rm{cose}}{{\rm{c}}^4}x\,dx} = $
  • A
    $\cot x + \frac{{{{\cot }^3}x}}{3} + c$
  • B
    $\tan x + \frac{{{{\tan }^3}x}}{3} + c$
  • ✓
    $ - \cot x - \frac{{{{\cot }^3}x}}{3} + c$
  • D
    $ - \tan x - \frac{{{{\tan }^3}x}}{3} + c$
Answer
Correct option: C.
$ - \cot x - \frac{{{{\cot }^3}x}}{3} + c$
c
(c) $ I = \int {cosec^4}x dx = \int {{\rm{cose}}{{\rm{c}}^2}x} .\,{\rm{cose}}{{\rm{c}}^2}xdx$
$ = \int {{\rm{cose}}{{\rm{c}}^2}x(1 + {{\cot }^2}x)\,dx} $
$ = \int {{\rm{cose}}{{\rm{c}}^2}x\,\,dx} \,\, + \int {{{\cot }^2}x.\,{\rm{cose}}{{\rm{c}}^2}x\,dx} $
$ = - \cot x - \frac{{{{\cot }^3}x}}{3} + c$.
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MCQ 271 Mark
If $\int_{}^{} {\sin 5x\cos 3x\;dx = - \frac{{\cos 8x}}{{16}}} + A$, then $A = $
  • A
    $\frac{{\sin 2x}}{{16}} + $constant
  • ✓
    $ - \frac{{\cos 2x}}{4} + $constant
  • C
    Constant
  • D
    None of these
Answer
Correct option: B.
$ - \frac{{\cos 2x}}{4} + $constant
b
(b)$\int_{}^{} {\sin 5x\cos 3x\;dx} = \frac{1}{2}\int_{}^{} {(\sin 8x + \sin 2x)} dx$
$ = \frac{{ - \cos 8x}}{{16}} - \frac{{\cos 2x}}{4} + c$
Equating to the given value, we get $A = \frac{{ - \cos 2x}}{4} + c$.
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MCQ 281 Mark
$\int_{}^{} {\frac{{dx}}{{\cos (x - a)\cos (x - b)}} = } $
  • A
    ${\rm{cosec}}\,\,(a - b)\log \frac{{\sin (x - a)}}{{\sin (x - b)}} + c$
  • ✓
    ${\rm{cosec}}(a - b)\log \frac{{\cos (x - a)}}{{\cos (x - b)}} + c$
  • C
    ${\rm{cosec}}(a - b)\log \frac{{\sin (x - b)}}{{\sin (x - a)}} + c$
  • D
    ${\rm{cosec}}(a - b)\log \frac{{\cos (x - b)}}{{\cos (x - a)}} + c$
Answer
Correct option: B.
${\rm{cosec}}(a - b)\log \frac{{\cos (x - a)}}{{\cos (x - b)}} + c$
b
(b)$\int_{}^{} {\frac{{dx}}{{\cos (x - a)\cos (x - b)}}} $
$ = \frac{1}{{\sin (a - b)}}\int_{}^{} {\frac{{\sin \left\{ {(x - b) - (x - a)} \right\}}}{{\cos (x - a)\,.\,\cos (x - b)}}\,dx} $
$ = \frac{1}{{\sin (a - b)}}\int_{}^{} {\left\{ {\frac{{\sin (x - b)}}{{\cos (x - b)}} - \frac{{\sin (x - a)}}{{\cos (x - a)}}} \right\}dx} $
$ = {\rm{cosec}}\,(a - b)\log \frac{{\cos (x - a)}}{{\cos (x - b)}} + c$.
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MCQ 291 Mark
If $\int_{}^{} {(\sin 2x + \cos 2x)\;dx = \frac{1}{{\sqrt 2 }}\sin (2x - c) + a} $, then the value of  $a$  and  $c$  is
  • ✓
    $c = \pi /4$ and $a = k$ (an arbitrary constant)
  • B
    $c = - \pi /4$ and $a = \pi /2$
  • C
    $c = \pi /2$ and $a$ is an arbitrary constant
  • D
    None of these
Answer
Correct option: A.
$c = \pi /4$ and $a = k$ (an arbitrary constant)
a
(a)$\int_{}^{} {(\sin 2x + \cos 2x)\,dx} = - \frac{{\cos 2x}}{2} + \frac{{\sin 2x}}{2} + k$
$ = \frac{1}{{\sqrt 2 }}\left( {\sin 2x\cos \frac{\pi }{4} - \cos 2x\sin \frac{\pi }{4}} \right) + k$
$ = \frac{1}{{\sqrt 2 }}\sin \left( {2x - \frac{\pi }{4}} \right) + k$
$ \Rightarrow c = \frac{\pi }{4}$ and $a = k,$ an arbitrary constant.
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MCQ 301 Mark
$\int {\sqrt {1 + \sin \left( {\frac{x}{4}} \right)\,} } dx$
  • ✓
    $8\left( {\sin \frac{x}{8} - \cos \frac{x}{8}} \right) + c$
  • B
    $\left( {\sin \frac{x}{8} + \cos \frac{x}{8}} \right) + c$
  • C
    $\frac{1}{8}\left( {\sin \frac{x}{8} - \cos \frac{x}{8}} \right) + c$
  • D
    $8\left( {\cos \frac{x}{8} - \sin \frac{x}{8}} \right) + c$
Answer
Correct option: A.
$8\left( {\sin \frac{x}{8} - \cos \frac{x}{8}} \right) + c$
a
(a) $\int {\sqrt {1 + \sin \left( {\frac{x}{4}} \right)\,} } dx$
$ = \int {\sqrt {\left( {{{\sin }^2}\frac{x}{8} + {{\cos }^2}\frac{x}{8}} \right) + \left( {2\sin \frac{x}{8}\cos \frac{x}{8}} \right)} \,dx} $
$ = \int {\sqrt {{{\left( {\sin \frac{x}{8} + \cos \frac{x}{8}} \right)}^2}} dx} = \int {\left( {\sin \frac{x}{8} + \cos \frac{x}{8}} \right)} \,dx$
$ = \frac{{ - \cos \frac{x}{8}}}{{\left( {\frac{1}{8}} \right)}} + \frac{{\sin \frac{x}{8}}}{{\left( {\frac{1}{8}} \right)}} + c$ $ = 8\left( {\sin \frac{x}{8} - \cos \frac{x}{8}} \right) + c$.
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MCQ 311 Mark
If $\int {\frac{{2{x^2} + 3}}{{({x^2} - 1)\,\,({x^2} + 4)}}dx = a\log \left( {\frac{{x - 1}}{{x + 1}}} \right) + } \,b{\tan ^{ - 1}}\frac{x}{2} + c$, then values of $a$ and $b$ are
  • A
    $(1, -1)$
  • B
    $(-1, 1)$
  • C
    $\left( {\frac{1}{2},\, - \frac{1}{2}} \right)$
  • ✓
    $\left( {\frac{1}{2},\,\frac{1}{2}} \right)$
Answer
Correct option: D.
$\left( {\frac{1}{2},\,\frac{1}{2}} \right)$
d
(d) $I = \int {\frac{{2{x^2} + 3}}{{({x^2} - 1)({x^2} + 4)}}dx} $
$\because$ $\frac{{2{x^2} + 3}}{{({x^2} - 1)({x^2} + 4)}} = \frac{1}{{({x^2} - 1)}} + \frac{1}{{({x^2} + 4)}}$
$\therefore \,\,I = \int {\frac{{dx}}{{({x^2} - 1)}} + \int {\frac{{dx}}{{{x^2} + 4}}} } $
$I = \frac{1}{{2 \times 1}}\log \left| {\frac{{x - 1}}{{x + 1}}} \right| + \frac{1}{2}{\tan ^{ - 1}}\frac{x}{2} + c$
$ \Rightarrow I = \frac{1}{2}\log \left( {\frac{{x - 1}}{{x + 1}}} \right) + \frac{1}{2}{\tan ^{ - 1}}\frac{x}{2} + c$
==> $a = \frac{1}{2}$, $b = \frac{1}{2}$.
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MCQ 321 Mark
$\int_{}^{} {\frac{{{e^{\sqrt x }}\cos {e^{\sqrt x }}}}{{\sqrt x }}dx} = $
  • ✓
    $2\sin {e^{\sqrt x }}$
  • B
    $\sin {e^{\sqrt x }}$
  • C
    $2\cos {e^{\sqrt x }}$
  • D
    $ - 2\sin {e^{\sqrt x }}$
Answer
Correct option: A.
$2\sin {e^{\sqrt x }}$
a
(a) Put ${e^{\sqrt x }} = t \Rightarrow \frac{{{e^{\sqrt x }}}}{{\sqrt x }} = \,2dt$, (Now proceed yourself).
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MCQ 331 Mark
$\int_{}^{} {\frac{{x\;dx}}{{1 - x\cot x}}} = $
  • A
    $\log (\cos x - x\sin x) + c$
  • B
    $\log (x\sin x - \cos x) + c$
  • ✓
    $\log (\sin x - x\cos x) + c$
  • D
    None of these
Answer
Correct option: C.
$\log (\sin x - x\cos x) + c$
c
(c)$\int_{}^{} {\frac{{x\,dx}}{{1 - x\cot x}}} = \int_{}^{} {\frac{{x\,dx}}{{1 - x\frac{{\cos x}}{{\sin x}}}}} = \int_{}^{} {\frac{{x\sin x}}{{\sin x - x\cos x}}\,dx} $
$ = \int_{}^{} {\frac{{dt}}{t}} = \log t = \log (\sin x - x\cos x) + c.$
$\{$Putting $\sin x - x\cos x = t$,$\}$
$⇒$  $[\cos x - ( - x\sin x + \cos x)]\,dx = dt \Rightarrow x\sin x\,dx = dt\} $

$ = \int_{}^{} {\frac{{dt}}{t}}  = \log t = \log (\sin x - x\cos x) + c.$

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MCQ 341 Mark
$\int_{}^{} {\frac{{{x^3}}}{{\sqrt {{x^2} + 2} }}dx = } $
  • A
    $\frac{1}{3}{({x^2} + 2)^{3/2}} + 2{({x^2} + 2)^{1/2}} + c$
  • ✓
    $\frac{1}{3}{({x^2} + 2)^{3/2}} - 2{({x^2} + 2)^{1/2}} + c$
  • C
    $\frac{1}{3}{({x^2} + 2)^{3/2}} + {({x^2} + 2)^{1/2}} + c$
  • D
    $\frac{1}{3}{({x^2} + 2)^{3/2}} - {({x^2} + 2)^{1/2}} + c$
Answer
Correct option: B.
$\frac{1}{3}{({x^2} + 2)^{3/2}} - 2{({x^2} + 2)^{1/2}} + c$
b
(b)$\int_{}^{} {\frac{{{x^3}}}{{\sqrt {{x^2} + 2} }}\,dx} = \int_{}^{} {\frac{{{x^2}.\,x}}{{\sqrt {{x^2} + 2} }}\,dx} $
Put ${x^2} + 2 = {t^2} \Rightarrow x\,dx = t\,dt$ and ${x^2} = {t^2} - 2,$ then it reduces to $\int_{}^{} {\frac{{({t^2} - 2)}}{t}\,t\,dt} = \int_{}^{} {({t^2} - 2)dt} $
$ = \frac{{{t^3}}}{3} - 2t + c = \frac{{{{({x^2} + 2)}^{32}}}}{3} - 2{({x^2} + 2)^{1/2}} + c.$
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MCQ 351 Mark
$\int_{}^{} {\sec x\log (\sec x + \tan x)\;dx = } $
  • A
    ${[\log (\sec x + \tan x)]^2} + c$
  • ✓
    $\frac{1}{2}{[\log (\sec x + \tan x)]^2} + c$
  • C
    ${\sec ^2}x + \tan x\sec x + c$
  • D
    None of these
Answer
Correct option: B.
$\frac{1}{2}{[\log (\sec x + \tan x)]^2} + c$
b
(b) Let $\log (\sec x + \tan x) = t \Rightarrow \sec x\,dx = dt$
Therefore $\int_{}^{} {\sec x\,\log (\sec x + \tan x)\,dx} = \int_{}^{} {t\,dt} $
$ = \frac{{{t^2}}}{2} + c = \frac{{{{[\log (\sec x + \tan x)]}^2}}}{2} + c.$
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MCQ 361 Mark
$\int_{}^{} {\frac{{dx}}{{{e^x} - 1}} = } $
  • ✓
    $\ln (1 - {e^{ - x}}) + c$
  • B
    $ - \ln (1 - {e^{ - x}}) + c$
  • C
    $\ln ({e^x} - 1) + c$
  • D
    None of these
Answer
Correct option: A.
$\ln (1 - {e^{ - x}}) + c$
a
(a)$\int_{}^{} {\frac{{dx}}{{{e^x} - 1}} = \int_{}^{} {\frac{{{e^{ - x}}}}{{1 - {e^{ - x}}}}\,dx} } $
Put $1 - {e^{ - x}} = t \Rightarrow {e^{ - x}}dx = dt,$ then it reduces to
$\int_{}^{} {\frac{{dt}}{t} = \log t + c} = \log (1 - {e^{ - x}}) + c.$
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MCQ 371 Mark
$\int_{}^{} {\frac{{\sin 2x}}{{{{\sin }^4}x + {{\cos }^4}x}}dx = } $
  • A
    ${\cot ^{ - 1}}({\tan ^2}x) + c$
  • ✓
    ${\tan ^{ - 1}}({\tan ^2}x) + c$
  • C
    ${\cot ^{ - 1}}({\cot ^2}x) + c$
  • D
    ${\tan ^{ - 1}}({\cot ^2}x) + c$
Answer
Correct option: B.
${\tan ^{ - 1}}({\tan ^2}x) + c$
b
(b)$\int_{}^{} {\frac{{\sin 2x}}{{{{\sin }^4}x + {{\cos }^4}x}}\,dx} $
$ = \int_{}^{} {\frac{{2\sin x\cos x}}{{{{\sin }^4}x + {{\cos }^4}x}}\,dx = \int_{}^{} {\frac{{2\tan x{{\sec }^2}x}}{{1 + {{\tan }^4}x}}\,dx} } $
Put ${\tan ^2}x = t \Rightarrow 2\tan x{\sec ^2}x\,dx = dt,$ then it reduced to $ \int {dt\over{1 + t^2}} = tan^{-1} t + c = tan^{-1}(tan^2 x) + c$.
Trick : By inspection,
$\frac{d}{{dx}}\left\{ {{{\cot }^{ - 1}}({{\tan }^2}x)} \right\} = - \frac{{1(2\tan x\,.\,{{\sec }^2}x)}}{{1 + {{\tan }^4}x}} = - \frac{{\sin 2x}}{{{{\cos }^4}x + {{\sin }^4}x}}$
$ \Rightarrow \frac{d}{{dx}}\left\{ {{{\tan }^{ - 1}}({{\tan }^2}x)} \right\} = \frac{{\sin 2x}}{{{{\sin }^4}x + {{\cos }^4}x}}$.
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MCQ 381 Mark
$\int_{}^{} {\frac{{\sin x\cos x}}{{a{{\cos }^2}x + b{{\sin }^2}x}}dx = } $
  • ✓
    $\frac{1}{{2(b - a)}}\log (a{\cos ^2}x + b{\sin ^2}x) + c$
  • B
    $\frac{1}{{b - a}}\log (a{\cos ^2}x + b{\sin ^2}x) + c$
  • C
    $\frac{1}{2}\log (a{\cos ^2}x + b{\sin ^2}x) + c$
  • D
    None of these
Answer
Correct option: A.
$\frac{1}{{2(b - a)}}\log (a{\cos ^2}x + b{\sin ^2}x) + c$
a
(a) Put $a{\cos ^2}x + b{\sin ^2}x = t$$ \Rightarrow 2(b - a)\sin x\cos x = dt,$
then $\int_{}^{} {\frac{{\sin x\cos x\,dx}}{{a{{\cos }^2}x + b{{\sin }^2}x}} = \frac{1}{{2(b - a)}}\int_{}^{} {\frac{1}{t}\,dt} } $
$ = \frac{1}{{2(b - a)}}\log (a{\cos ^2}x + b{\sin ^2}x) + c$.
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MCQ 391 Mark
$\int_{}^{} {\frac{1}{{\sqrt x }}{{\tan }^4}\sqrt x } {\sec ^2}\sqrt x \;dx = $
  • A
    $2{\tan ^5}\sqrt x + c$
  • B
    $\frac{1}{5}{\tan ^5}\sqrt x + c$
  • ✓
    $\frac{2}{5}{\tan ^5}\sqrt x + c$
  • D
    None of these
Answer
Correct option: C.
$\frac{2}{5}{\tan ^5}\sqrt x + c$
c
(c)$\int_{}^{} {\frac{1}{{\sqrt x }}{{\tan }^4}\sqrt x \,.\,{{\sec }^2}\sqrt x \,dx} $
Put $\tan \sqrt x = t \Rightarrow \frac{{{{\sec }^2}\sqrt x }}{{2\sqrt x }}\,dx = dt,$ then it reduces to
$2\int_{}^{} {{t^4}dt} = \frac{2}{5}{(\tan \sqrt x )^5} + c = \frac{2}{5}{\tan ^5}\sqrt x + c$.
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MCQ 401 Mark
$\int_{}^{} {\frac{{\cos 2x}}{{{{(\cos x + \sin x)}^2}}}\;dx = } $
  • A
    $\log \sqrt {\cos x + \sin x} + c$
  • B
    $\log (\cos x - \sin x) + c$
  • ✓
    $\log (\cos x + \sin x) + c$
  • D
    $ - \frac{1}{{\cos x + \sin x}} + c$
Answer
Correct option: C.
$\log (\cos x + \sin x) + c$
c
(c)$\int_{}^{} {\frac{{\cos 2x}}{{{{(\cos x + \sin x)}^2}}}dx} = \int_{}^{} {\frac{{(\cos x - \sin x)(\cos x + \sin x)}}{{{{(\cos x + \sin x)}^2}}}{\rm{ }}} dx$
$ = \int_{}^{} {\frac{{\cos x - \sin x}}{{\cos x + \sin x}}dx} $
Put $t = \sin x + \cos x \Rightarrow dt = (\cos x - \sin x)dx$,

then it reduces to $\int_{}^{} {\frac{1}{t}} \,{\rm{ }}dt = \log t + c = \log (\sin x + \cos x) + c$.

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MCQ 411 Mark
$\int_{}^{} {\frac{{{e^{ - x}}}}{{1 + {e^x}}}\;dx = } $
  • ✓
    $\log (1 + {e^x}) - x - {e^{ - x}} + c$
  • B
    $\log (1 + {e^x}) + x - {e^{ - x}} + c$
  • C
    $\log (1 + {e^x}) - x + {e^{ - x}} + c$
  • D
    $\log (1 + {e^x}) + x + {e^{ - x}} + c$
Answer
Correct option: A.
$\log (1 + {e^x}) - x - {e^{ - x}} + c$
a
(a)$\int_{}^{} {\frac{{{e^{ - x}}}}{{1 + {e^x}}}\,dx} = \int_{}^{} {\frac{{{e^{ - x}}{e^{ - x}}}}{{{e^{ - x}} + 1}}\,dx} $
Put ${e^{ - x}} + 1 = t \Rightarrow - {e^{ - x}}dx = dt,$ then it reduces to
$ - \int_{}^{} {\frac{{(t - 1)}}{t}\,dt} = \int_{}^{} {\left( {\frac{1}{t} - 1} \right)\,dt} $
$ = \log t - t + c = \log ({e^{ - x}} + 1) - ({e^{ - x}} + 1) + c$
$ = \log ({e^x} + 1) - x - {e^{ - x}} - 1 + c$
$ = \log ({e^x} + 1) - x - {e^{ - x}} + c$, $(\because \,\,\,1 = $ constant).
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MCQ 421 Mark
$\int_{}^{} {{{\cos }^5}x\;dx = } $
  • ✓
    $\sin x - \frac{2}{3}{\sin ^3}x + \frac{1}{5}{\sin ^5}x + c$
  • B
    $\sin x + \frac{2}{3}{\sin ^3}x + \frac{1}{5}{\sin ^5}x + c$
  • C
    $\sin x - \frac{2}{3}{\sin ^3}x - \frac{1}{5}{\sin ^5}x + c$
  • D
    None of these
Answer
Correct option: A.
$\sin x - \frac{2}{3}{\sin ^3}x + \frac{1}{5}{\sin ^5}x + c$
a
(a)$\int_{}^{} {{{\cos }^5}x\,dx} = \int_{}^{} {{{\cos }^4}x\cos x\,dx} = \int_{}^{} {{{(1 - {{\sin }^2}x)}^2}\cos xdx} $
Put $\sin x = t \Rightarrow \cos x\;dx = dt$, then it reduces to
$\int_{}^{} {{{(1 - {t^2})}^2}dt = \int_{}^{} {(1 + {t^4} - 2{t^2})\;dt = \frac{{{t^5}}}{5} - \frac{{2{t^3}}}{3} + t + c} } $
$ = \frac{{{{\sin }^5}x}}{5} - \frac{{2{{\sin }^3}x}}{3} + \sin x + c$.
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MCQ 431 Mark
$\int_{}^{} {\sec x{{\tan }^3}x\;dx = } $
  • ✓
    $\frac{1}{3}{\sec ^3}x - \sec x + c$
  • B
    ${\sec ^3}x - \sec x + c$
  • C
    $\frac{1}{3}{\sec ^3}x + \sec x + c$
  • D
    None of these
Answer
Correct option: A.
$\frac{1}{3}{\sec ^3}x - \sec x + c$
a
(a)$\int_{}^{} {\sec x{{\tan }^3}x\;dx = \int_{}^{} {\sec x({{\sec }^2}x - 1)\tan x\;dx} } $
$ = \int_{}^{} {\sec x\tan x{{\sec }^2}x\;dx} - \int_{}^{} {\sec x\tan x\;dx} $
$ = \frac{{{{\sec }^3}x}}{3} - \sec x + c$, (Putting $\sec x = t$ in first part).
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MCQ 441 Mark
$\int_{}^{} {\tan x} {\sec ^2}x\sqrt {1 - {{\tan }^2}x} \;dx = $
  • ✓
    $ - \frac{1}{3}{(1 - {\tan ^2}x)^{3/2}} + c$
  • B
    $\frac{1}{3}{(1 - {\tan ^2}x)^{3/2}} + c$
  • C
    $ - \frac{2}{3}{(1 - {\tan ^2}x)^{2/3}} + c$
  • D
    None of these
Answer
Correct option: A.
$ - \frac{1}{3}{(1 - {\tan ^2}x)^{3/2}} + c$
a
(a)$\int_{}^{} {\tan x\,.\,{{\sec }^2}x\sqrt {1 - {{\tan }^2}x} \,dx} $
Put $\tan x = t \Rightarrow {\sec ^2}x\,dx = dt,$ then it reduces to $\int_{}^{} {t\sqrt {1 - {t^2}} \,dt} $
Now again, put $1 - {t^2} = u,$ then its reduced form is $ - 2tdt = du$
$ - \frac{1}{2}\int_{}^{} {\sqrt u \,du} = - \frac{1}{3}{u^{3/2}} + c = - \frac{1}{3}{(1 - {\tan ^2}x)^{3/2}} + c.$
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MCQ 451 Mark
$\int_{}^{} {\frac{1}{{\log a}}({a^x}\cos {a^x})dx = } $
  • A
    $\sin {a^x} + c$
  • B
    ${a^x}\sin {a^x} + c$
  • ✓
    $\frac{1}{{{{(\log a)}^2}}}\sin {a^x} + c$
  • D
    $\log \sin {a^x} + c$
Answer
Correct option: C.
$\frac{1}{{{{(\log a)}^2}}}\sin {a^x} + c$
c
(c)$\int_{}^{} {\frac{1}{{\log a}}({a^x}\cos {a^x})\,dx} $
Put ${a^x} = t \Rightarrow {a^x}dx = \frac{{dt}}{{\log a}},$ then it reduces to
$\int_{}^{} {\frac{1}{{{{(\log a)}^2}}}\cos t\,dt} = \frac{1}{{{{(\log a)}^2}}}\sin t + c = \frac{1}{{{{(\log a)}^2}}}\sin {a^x} + c$.
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MCQ 461 Mark
$\int_{}^{} {\frac{1}{{{x^3}}}{{[\log {x^x}]}^2}\;dx = } $
  • A
    $\frac{{{x^3}}}{3}(\log x) + x + c$
  • ✓
    $\frac{1}{3}{(\log x)^3} + c$
  • C
    $3\log (\log x) + c$
  • D
    None of these
Answer
Correct option: B.
$\frac{1}{3}{(\log x)^3} + c$
b
(b)$\int_{}^{} {\frac{1}{{{x^3}}}{{[\log {x^x}]}^2}\,dx} = \int_{}^{} {\frac{1}{{{x^3}}}{{[x\log x]}^2}dx} $
$ = \int_{}^{} {\frac{1}{x}{{(\log x)}^2}dx} = \frac{1}{3}{(\log x)^3} + c$, $\{$Putting $\log x = t\} $.
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MCQ 471 Mark
$\int_{}^{} {\frac{{dx}}{{x\log x\log (\log x)}} = } $
  • A
    $2\log (\log x) + c$
  • ✓
    $\log [\log (\log x)] + c$
  • C
    $\log (x\log x) + c$
  • D
    None of these
Answer
Correct option: B.
$\log [\log (\log x)] + c$
b
(b)$\int_{}^{} {\frac{{dx}}{{x\log x\,.\,\log (\log x)}}} $
Put $\log x = t,$ then it reduces to $\int_{}^{} {\frac{{dt}}{{t\,.\,\log (t)}}} $
Again put $\log t = z,$ then reduces form is
$\int_{}^{} {\frac{{dz}}{z}} = \log z = \log [\log (\log x)] + c$.
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MCQ 481 Mark
$\int_{}^{} {\frac{{2x{{\tan }^{ - 1}}{x^2}}}{{1 + {x^4}}}} \;dx = $
  • A
    ${[{\tan ^{ - 1}}{x^2}]^2} + c$
  • ✓
    $\frac{1}{2}{[{\tan ^{ - 1}}{x^2}]^2} + c$
  • C
    $2{[{\tan ^{ - 1}}{x^2}]^2} + c$
  • D
    None of these
Answer
Correct option: B.
$\frac{1}{2}{[{\tan ^{ - 1}}{x^2}]^2} + c$
b
(b) Put $t = {\tan ^{ - 1}}{x^2} \Rightarrow dt = \frac{1}{{1 + {x^4}}}\,2x\,dx,$ then
$\int_{}^{} {\frac{{2x{{\tan }^{ - 1}}{x^2}}}{{1 + {x^4}}}} \,dx = \int_{}^{} {t\,dt} = \frac{{{t^2}}}{2} + c = \frac{1}{2}{({\tan ^{ - 1}}{x^2})^2} + c.$
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MCQ 491 Mark
$\int_{}^{} {2x{{\cos }^3}{x^2}\sin {x^2}dx = } $
  • ✓
    $ - \frac{1}{4}{\cos ^4}{x^2} + c$
  • B
    $\frac{1}{4}{\cos ^4}{x^2} + c$
  • C
    ${\cos ^4}{x^2} + c$
  • D
    None of these
Answer
Correct option: A.
$ - \frac{1}{4}{\cos ^4}{x^2} + c$
a
(a) Put $t = \cos {x^2} \Rightarrow dt = - 2x\sin {x^2}dx,$ then
$\int_{}^{} {2x{{\cos }^3}{x^2}\sin {x^2}dx} = - \int_{}^{} {{t^3}dt} = - \frac{{{t^4}}}{4} + c$
$ = - \frac{1}{4}{\cos ^4}{x^2} + c.$
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MCQ 501 Mark
$\int_{}^{} {\frac{{(x + 1){{(x + \log x)}^2}}}{x}dx = } $
  • A
    $\frac{1}{3}(x + \log x) + c$
  • B
    $\frac{1}{3}{(x + \log x)^2} + c$
  • ✓
    $\frac{1}{3}{(x + \log x)^3} + c$
  • D
    None of these
Answer
Correct option: C.
$\frac{1}{3}{(x + \log x)^3} + c$
c
(c) Put $t = x + \log x \Rightarrow dt = \left( {1 + \frac{1}{x}} \right)\,dx,$ then
$\int_{}^{} {\frac{{(x + 1){{(x + \log x)}^2}}}{x}\,dx} = \int_{}^{} {{t^2}dt} = \frac{{{t^3}}}{3} + c$
$ = \frac{1}{3}{(x + \log x)^3} + c.$
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M.C.Q (1 Marks) - Maths STD 12 Science Questions - Vidyadip