IISER Mathematics - Indefinite Integration
Exam Duration: 45 Mins Total Questions : 30
If \(\int e^x \left\{ {f(x)-f'(x)} \right\} dx=\phi(x),\) then \(\int e^x{f(x)}dx\) is equal to
- (a)
\(\phi(x)=e^xf(x)\)
- (b)
\(\phi(x)-e^xf(x)\)
- (c)
\({1\over 2}\left\{\phi(x)+e^xf(x)\right\}\)
- (d)
\({1\over 2}\left\{\phi(x)+e^xf'(x)\right\}\)
If \(f(x)=\int{x^2+sin^2\ x\over 1+x^2}.sec^2xdx\) and f(0)=0, then f(1) is equal to
- (a)
\(1-{\pi\over 4}\)
- (b)
\({\pi\over 4}-1\)
- (c)
\(tan\ 1-{\pi\over 4}\)
- (d)
None of the above
If \(\int{f(x)\over log\ sin\ x}dx\)=log log sin x, the f(x) is equal to
- (a)
sin x
- (b)
cos x
- (c)
log sin x
- (d)
cot x
\(\int{dx\over (1+\sqrt{x})\sqrt{x-x^2}}\) is equal to
- (a)
\({2(\sqrt x -1)\over \sqrt{1-x}}+C\)
- (b)
\({2(1-\sqrt x)\over \sqrt{1-x}}+C\)
- (c)
\({\sqrt x -1\over \sqrt{1-x}}+C\)
- (d)
\({1-\sqrt x\over 2\sqrt{1-x}}+C\)
\(\int{x^9dx\over (4x^2+1)^6}\) is equal to
- (a)
\({1\over 5x}({4+{1\over x^2}})^{-5}+C\)
- (b)
\({1\over 5}({4+{1\over x^2}})^{-5}+C\)
- (c)
\({1\over 10}({1+{4x^2}})^{-5}+C\)
- (d)
\({1\over10}(4+{1\over x^2})^{-5}+C\)
If \(\int[{ln(ln\ x)+{1\over (ln\ x)^2}}]dx=x[f(x)-g(x)]+C,\) then
- (a)
\(f(x)=ln(ln\ x),\ g(x)={1\over in\ (x)}\)
- (b)
\(f(x)=ln\ x,\ g(x)={1\over in\ (x)}\)
- (c)
\(f(x)={ln\over ln\ x},\ g(x)=ln(ln\ x)\)
- (d)
\(f(x)=ln\ x,\ g(x)=ln(ln\ x )\)
\(\int{dx\over x^4+x^2}={A\over x^2}+{B\over x}+ ln|{x\over x+1}|+C,\) then
- (a)
\(A=-1/2, B=1\)
- (b)
\(A=1, B=-1/2\)
- (c)
\(A=1/2, B=1\)
- (d)
\(A=1/2, B=1/2\)
\(If \int{dx\over 3\sqrt{sin^{11}xcos\ x}}=-[{{3\over 8}f(x)+{3\over 2}g(x)}]+C,\) then
- (a)
\(f(x)=tan^{-8/3}x, g(x)=tan^{-2/3}x\)
- (b)
\(f(x)=tan^{8/3}x, g(x)=tan^{-2/3}x\)
- (c)
\(f(x)=tan^{-8/3}x, g(x)=tan^{2/3}x\)
- (d)
\(f(x)=tan^{4/3}x, g(x)=tan^{-4/3}x\)
\(\int{x^2+cos^2X\over x^2+1}.cosec^2x\ dx\) is equal to
- (a)
\(cot\ x+cot^{-1}x+C\)
- (b)
\(-e^{in\ tan^{-1}}x-cot\ x+C\)
- (c)
\(-cot\ x+cot^{-1}x+C\)
- (d)
\(-tan^{-1}x-{cosec\ x\over sec\ x}+C\)
If \(If I = \int{sin\ x+sin^3\ x\over cos2x}dx=pcos\ x+Q\ log|f(x)|+R,\) then
- (a)
\(P={1\over 2},Q={1\over 4\sqrt{2}}\)
- (b)
\(P={1\over 4},Q=-{1\over \sqrt{2}}\)
- (c)
\(f(x)={cos\ x+1\over \sqrt2cos\ x-1}\)
- (d)
\(f(x)={\sqrt2cos\ x-1\over \sqrt2cos\ x+1}\)
\(\int{{x+3\sqrt{x^2}+6\sqrt x}\over x(1+3\sqrt{x})}dx\) is equal to
- (a)
\({3\over 2}x^{2/3}+6\ tan^{-1}c^{1/6}+C\)
- (b)
\({3\over 2}x^{2/3}-6\ tan^{-1}c^{1/6}+C\)
- (c)
\(-{3\over 2}x^{2/3}+6\ tan^{-1}c^{1/6}+C\)
- (d)
\({1\over 2}x^{2/3}-6\ tan^{-1}c^{1/6}+C\)
\(\int{2\over (2-x)^2}3\sqrt{2-x\over 2+x}dx\) is equal to
- (a)
\({4\over 3}({2+x\over 2-x})^{2/3}+C\)
- (b)
\({3\over 4}({2+x\over 2-x})^{2/3}+C\)
- (c)
\({3\over 4}({2-x\over 2+x})^{2/3}+C\)
- (d)
\({3\over 4}({2+x\over 2-x})^{4/3}+C\)
\(\int{x^3\over (1+x^2)}^{1/3}dx\) is equal to
- (a)
\({20\over 3}(1+x^2)^{2/3}(2x^2-3)+C\)
- (b)
\({3\over 20}(1+x^2)^{2/3}(2x^2-3)+C\)
- (c)
\({3\over 20}(1+x^2)^{2/3}(2x^2+3)+C\)
- (d)
\({3\over 20}(1+x^2)^{3/2}(2x^2-3)+C\)
If \(\int\ sin^{-1}\ x.cos^{-1}\ x\ dx=f^{-1}(x)\) \([Ax-xf^{-1}(x)-2\sqrt{1+x^2}]+{\pi\over 2}\sqrt{1-x^2} +2x+C\), then
- (a)
\(f(x)= sin\ 2x\)
- (b)
\(f(x)=cos\ x\)
- (c)
\(A={\pi\over 4}\)
- (d)
\(A={\pi\over 2}\)
The integral \(\int({1+x-{1\over x}})e^{x+{1\over x}}dx\) is equal to
- (a)
\((x+1)e^{({x+{1\over x}})}+C\)
- (b)
\(-x\ e^{({x+{1\over x}})}+C\)
- (c)
\((x-1)e^{({x+{1\over x}})}+C\)
- (d)
\(xe^{({x+{1\over x}})}+C\)
\(\int { \left( x-{ ^{ 11 }{ C }_{ 1 }{ x }^{ 2 }+^{ 11 }{ C }_{ 2 }{ x }^{ 3 }-^{ 11 }{ C }_{ 2 }{ x }^{ 4 } }+...-^{ 11 }{ C }_{ 11 }{ x }^{ 12 } \right) } \)dx equals
- (a)
\(\frac { \left( 1-{ x }^{ 12 } \right) }{ 12 } -\frac { \left( 1-{ x }^{ 11 } \right) }{ 11 } +c\)
- (b)
\(\frac { \left( 1-{ x }^{ 13 } \right) }{ 13 } -\frac { \left( 1-{ x }^{ 12 } \right) }{ 12 } +c\)
- (c)
\(\frac { \left( 1-{ x }^{ 11 } \right) }{ 11 } -\frac { \left( 1-{ x }^{ 12 } \right) }{ 12 } +c\)
- (d)
\(\frac { \left( 1-{ x }^{ 12 } \right) }{ 12 } -\frac { \left( 1-{ x }^{ 13 } \right) }{ 13 } +c\)
\(\int { \frac { 3+2\cos { x } }{ (2-3\cos { { x) }^{ 2 } } } } \)dx is equal to
- (a)
\(\left( \frac { \sin { x } }{ 2+3\cos { x } } \right) +c\)
- (b)
\(\left( \frac { 2\sin { x } }{ 2+3\sin { x } } \right) +c\)
- (c)
\(\left( \frac { 2\sin { x } }{ 2+3\sin { x } } \right) +c\)
- (d)
\(\left( \frac { 2\sin { x } }{ 2+3\sin { x } } \right) +c\)
\(\int { \frac { \left( { \sqrt { x } }^{ 5 } \right) }{ \left( { \sqrt { x } }^{ 7 } \right) +{ x }^{ 6 } } } dx=\lambda \quad ln\left( \frac { { x }^{ a } }{ { x }^{ a }+1 } \right) +c\) then a is
- (a)
=2
- (b)
>2
- (c)
<2
- (d)
=1
\(\int\)|x| ln|x| dx equals (x = 0)
- (a)
\(\frac { { x }^{ 2 } }{ 2 } ln|x|-\frac { { x }^{ 2 } }{ 4 } +c\)
- (b)
\(\frac { 1 }{ 2 } x|x|ln\quad x+\frac { 1 }{ 4 } x|x|+c\)
- (c)
\(-\frac { { x }^{ 2 } }{ 2 } ln|x|+\frac { { x }^{ 2 } }{ 4 } +c\)
- (d)
\(\frac { 1 }{ 2 } x|x|ln|x|-\frac { 1 }{ 4 } x|x|+c\)
The primitive function of the function f (x) = \(\frac { \sqrt { { a }^{ 2 }-{ x }^{ 2 } } }{ { x }^{ 4 } } \) is
- (a)
\(c+\frac { \sqrt { { a }^{ 2 }-{ x }^{ 2 } } }{ { 3a }^{ 2 }{ x }^{ 3 } } \)
- (b)
\(c-\frac { \left( a^{ 2 }-{ x }^{ 2 } \right) ^{ 3/2 } }{ { 2a }^{ 2 }{ x }^{ 3 } } \)
- (c)
\(c-\frac { \left( a^{ 2 }-{ x }^{ 2 } \right) ^{ 3/2 } }{ { 3a }^{ 2 }{ x }^{ 3 } } \)
- (d)
None of these
If lt means ln ln ln....x, the in being repeated r times, then \(\int { \{ x.l(x).{ l }^{ 2 }(x){ l }^{ 3 }(x)... } { l }^{ r }(x){ \} }^{ -1 }dx\) is equal to
- (a)
lr+1(x)+ c
- (b)
\(\frac { l^{ r+1 }(x) }{ r+1 } +c\)
- (c)
lr(x)+c
- (d)
none of these
If \(\int\) f(x) sin x cox dx = \(\frac { 1 }{ 2\left( { b }^{ 2 }-{ a }^{ 2 } \right) } \) ln f(x)+c then f(x) equal to
- (a)
\(\frac { 1 }{ ({ a }^{ 2 }{ sin }^{ 2 }+{ b }^{ 2 }{ cos }^{ 2 }x) } \)
- (b)
\(\frac { 1 }{ ({ a }^{ 2 }{ sin }^{ 2 }-{ b }^{ 2 }{ cos }^{ 2 }x) } \)
- (c)
\(\frac { 1 }{ ({ a }^{ 2 }{ sin }^{ 2 }-{ b }^{ 2 }{ sin }^{ 2 }x) } \)
- (d)
none of these
If \(\int { \frac { sinx }{ sin\quad (x-a) } } \) dx=Ax + B ln sin (x -a )+C, then
- (a)
A = sin a
- (b)
B = cos a
- (c)
A = cos a
- (d)
B = sin a
If \(\int { \left( \frac { { 4e }^{ x }+{ 6e }^{ -x } }{ { 9 }e^{ x }-{ 4e }^{ -x } } \right) } \) dx=Ax+b loge (9e2x-4)+c, then
- (a)
A=3/2
- (b)
B=35/36
- (c)
C is indefine
- (d)
A+b=\(\frac { 19 }{ 36 } \)
A primitive of sin 6x is
- (a)
\(\frac { 1 }{ 3 } (sin^{ 6 }x-{ sin }^{ 3 }x)+c\)
- (b)
\(-\frac { 1 }{ 3 } { cos }^{ 2 }3x+c\)
- (c)
\(\frac { 1 }{ 3 } { sin }^{ 2 }3x+c\)
- (d)
\(\frac { 1 }{ 3 } { sin }\left( 3x+\frac { \pi }{ 7 } \right) sin\left( 3x-\frac { \pi }{ 7 } +c \right) \)
Repeated application of integration by parts gives us, the reduction formula if the integrand is dependent of n, n ∈ N.
If \(I_{ n }=\int { xsin^{ n }xdx } \) and \(I_{ n }=-\frac { xsin^{ n-1 }xcosx }{ n } +\frac { sin^{ n }x }{ { n }^{ 2 } } +f(n){ I }_{ n-2 }\), then the value of f(n) is equal to
- (a)
\(\frac { n-1 }{ n } \)
- (b)
\(\frac { n-2}{ n-1 } \)
- (c)
\(\frac { n+1 }{ n } \)
- (d)
\(\frac { n+1 }{ n-1 } \)
If the primitive of sin(ln x) is f(x) {sin g(x) - cos h(x) } + c(c being the constant of integration , then
- (a)
lim f(x) = 1
- (b)
lim \(\frac { g(x) }{ h(x) } =1\)
- (c)
g(e3) = 3
- (d)
h(e5) = 5
if \(\int\) (cos-1) x+cos-1 \(\sqrt { \left( 1-{ x }^{ 2 } \right) } \)) dx = Ax + f(x) sin-1 x -2 \(\sqrt { \left( 1-{ x }^{ 2 } \right) } \) +C x [-1,0], then
- (a)
f(x) = x
- (b)
f(x) = - 2x
- (c)
A = \(\frac { \pi }{ 4 } \)
- (d)
A = \(\frac { \pi }{ 2 } \)
If the intergrand a rational function of x and fractional power of a linear fractional function of the form \(\frac { ax+b }{ cx+d } \quad \int { f\left( \frac { ax+b }{ cx+d } \right) } ^{ m/n }..,,,\left( \frac { ax+b }{ cx+d } \right) ^{ r/s }dx\) In this form substitute \(\frac { ax+b }{ cx+d } \)=tm where m is the LCM of the demominators of fractional powers of \(\frac { ax+b }{ cx+d } \)
The Value of \(\int\) \(\frac { dx }{ \left( 1+{ x } \right) ^{ 1/2 }-(1+x)^{ 1/3 } } \) is
- (a)
\(2{ \lambda }^{ 1/2 }+3{ \lambda }^{ 1/3 }+{ 6\lambda }^{ 1/6 }+6ln|{ \lambda }^{ 1/6 }-1|+c\)
- (b)
\(2{ \lambda }^{ 1/2 }-3{ \lambda }^{ 1/3 }+{ 6\lambda }^{ 1/6 }+6ln|{ \lambda }^{ 1/6 }-1|+c\)
- (c)
\(2{ \lambda }^{ 1/2 }+3{ \lambda }^{ 1/3 }-{ 6\lambda }^{ 1/6 }+6ln|{ \lambda }^{ 1/6 }-1|+c\)
- (d)
\(2{ \lambda }^{ 1/2 }+3{ \lambda }^{ 1/3 }+{ 6\lambda }^{ 1/6 }-6ln|{ \lambda }^{ 1/6 }-1|+c\)
If an integral can not be evaluated, then it is connected to another integral of lower degree but of same type. This is called reduction formula, we can derive reduction formulas' for the integral of the form f \(\int\)sinnxdx, \(\int\)cosnxdx, \(\int\) tanntxdx, \(\int\)cotnxdx, \(\int\)secnxdx, \(\int\)cosecnxdx by using integration by parts. In term these reduction formulas can be used to compute integrals of sin x, cos x, tan x etc . If \(\int { tan^{ 6 }xdx=\frac { tan^{ 5 }x }{ 5 } + } Atan^{ 3 }x+Btanx+Cx+D\), then A+B+C is equal to
- (a)
-1/2
- (b)
-1/4
- (c)
-1/3
- (d)
-1/5
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