Mathematics - Indefinite Integration
Exam Duration: 45 Mins Total Questions : 30
\(\int{x^9dx\over (4x^2+1)^6}\) 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)
\({1\over10}(4+{1\over x^2})^{-5}+C\)
\(\int{\sqrt{x^2+1}[ln(x^2+1)2-2ln\ x]\over x^4}dx\) is equal to
- (a)
\({(x^2+1)^{3/2}\over x^3}{[{2\over 3}-ln({x^2+1\over x^2})]}+C\)
- (b)
\({(x^2+1)^{3/2}\over 3x^3}{[ln({x^2+1\over x^2})-{2\over 3}]}+C\)
- (c)
\({(x^2+1)^{3/2}\over 3x^3}{[{2\over 3}-ln({x^2+1\over x^2})]}+C\)
- (d)
\({\sqrt{x^2+1\over 3x^3}}({ln{x^2+1\over x^2}-{2\over 3}})+C\)
If g(x) be a differentiable function satisfying \({d\over dx}\left\{ {g(x)} \right\}=g(x)\) and \(g(0)=1,\) then \(\int g(x)({2-sin2x\over 1-cos\ 2x})dx\) is equal to
- (a)
\(g(x)\ cot\ x\ +C\)
- (b)
\(-g(x)\ cot\ x\ +C\)
- (c)
\({9(x)\over 1-cos\ 2\ x} +C\)
- (d)
None of the above
\(\int{x^3-1\over x^3+x}dx\) is equal to
- (a)
\(x+1 ln|x|+1/2ln(x^2+1)-tan^{-1}x+C\)
- (b)
\(x- ln|x|+1/2ln(x^2+1)-tan^{-1}x+C\)
- (c)
\(x- ln|x|-1/2ln(x^2+1)-tan^{-1}x+C\)
- (d)
\(x+ ln|x|-1/2ln(x^2+1)+ C\)
\(\int{x^2-1\over x^4+x^2+1}dx\) is equal to
- (a)
\({1\over 2}log ({{x^2+x+1}\over x^2-x+1})+C\)
- (b)
\({1\over 2}log ({{x^2-x-1}\over x^2+x+1})+C\)
- (c)
\({1\over 2}log ({{x^2-x+1}\over x^2+x+1})+C\)
- (d)
\({ 2}log ({{x^2-x+1}\over x^2+x+1})+C\)
\(\int{x^2+cos^2X\over x^2+1}.cosec^2x\ dx\) 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)
\(-tan^{-1}x-{cosec\ x\over sec\ x}+C\)
\(\int{2\over (2-x)^2}3\sqrt{2-x\over 2+x}dx\) 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)
\({3\over 4}({2+x\over 2-x})^{4/3}+C\)
If \(\int\ f(x)dx=\Psi(x),\) then \(\int x^5f(x^3)dx\) is equal to
- (a)
\({1\over 3}[x^3\Psi(x^3)-\int x^2\Psi(x^3)dx]+C\)
- (b)
\({1\over 3}[x^3\Psi(x^3)-3\int x^3\Psi(x^3)dx]+C\)
- (c)
\({1\over 3}x^3\Psi(x^3)-\int x^2\Psi(x^3)dx+C\)
- (d)
\({1\over 3}[x^3\Psi(x^3)-\int x^3\Psi(x^3)dx]+C\)
Let x2 + 1 \(\neq \eta \pi \), n \(\epsilon \) N, then
\(\int { x } { \sqrt { \left\{ \frac { 2\quad sin\quad ({ x }^{ 2 }+1)-sin\quad \{ 2{ x }^{ 2 }+1\} }{ 2\quad sin\quad (\quad { x }^{ 2 }+1)+sin\quad \{ 2{ x }^{ 2 }+1\} } \right\} } }dx\) is
- (a)
\(In\left| \frac { 1 }{ 2 } sec({ x }^{ 2 }+1) \right| +c\)
- (b)
\(In\left| sec\left\{ \frac { 1 }{ 2 } ({ x }^{ 2 }+1) \right\} \right| +c\)
- (c)
\(\frac { 1 }{ 2 } in|sec({ x }^{ r }+1)|+c\)
- (d)
\(ln|sec({ x }^{ 2 }+1|+c\)
\(\int\)|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)
\(\frac { 1 }{ 2 } x|x|ln|x|-\frac { 1 }{ 4 } x|x|+c\)
If a particle is moving with velocity v(t) = cos\(\pi\) along a straight line such that at t = 0, s = 4 its position function is given by
- (a)
\(\frac { 1 }{ \pi } \cos { \pi t+2 } \)
- (b)
\(-\frac { 1 }{ \pi } \sin { \pi t+4 } \)
- (c)
\(\frac { 1 }{ \pi } \sin { \pi t+4 } \)
- (d)
none of these
If \(\int\) f(x) cos x dx = \(\frac { 1 }{ 2 } \){f(x)}2 +c, then f(x) is
- (a)
x + c
- (b)
sin x + c
- (c)
cos x + c
- (d)
c
The aniderivative of f(x) = \(\frac { 1 }{ 3+5\sin { x+3\cos { x } } } \) whose graph passes through the point (0.0) is
- (a)
\(\frac { 1 }{ 5 } \left( ln|1-\frac { 5 }{ 3 } tan(x/2)| \right) \)
- (b)
\(\frac { 1 }{ 5 } \left( ln|1+\frac { 5 }{ 3 } tan(x/2)| \right) \)
- (c)
\(\frac { 1 }{ 5 } \left( ln|1+\frac { 5 }{ 3 } cot(x/2)| \right) \)
- (d)
none of the above
\(\int\) xx (1+ln | x|) dx is equal to
- (a)
xx ln|x| + c
- (b)
ex + c
- (c)
xx + c
- (d)
none of these
if \(\int\) cos4 x dx= Ax + B sin 2x+c sin 4x+ D, then {A,B, C} equals
- (a)
\(\left\{ \frac { 3 }{ 8 } ,\frac { 1 }{ 32 } ,\frac { 1 }{ 4 } \right\} \)
- (b)
\(\left\{ \frac { 3 }{ 8 } ,\frac { 1 }{ 4 } ,\frac { 1 }{ 32 } \right\} \)
- (c)
\(\left\{ \frac { 1 }{ 32 } ,\frac { 1 }{ 4 } ,\frac { 3 }{ 8 } \right\} \)
- (d)
\(\left\{ \frac { 1 }{ 4 } ,\frac { 3 }{ 8 } ,\frac { 1 }{ 32 } \right\} \)
\(\int\) \(\frac { x{ e }^{ x } }{ \left( 1+x \right) ^{ 2 } } \) dx is equal to
- (a)
\(\frac { { e }^{ x } }{ \left( x+1 \right) } +c\)
- (b)
\({ e }^{ x }(x+1)+c\)
- (c)
\(-\frac { { e }^{ x } }{ \left( x+1 \right) ^{ 2 } } +c\)
- (d)
\(\frac { { e }^{ x } }{ 1+{ x }^{ 2 } } +c\)
If the derivative of f(x) w.r.t x is \(\frac { (1/2)-sin^{ 2 }x }{ f(x) } \), then f(x) is periodic function with period
- (a)
\(\pi\)/2
- (b)
\(\pi\)
- (c)
2\(\pi\)
- (d)
not defined
\(\int\) x2/3 (1+x1/2)-5/3 dx is equal to
- (a)
3(1 + x-1/2)-1/3 + c
- (b)
3(1 + x-1/2)-2/3 + c
- (c)
3(1 + x1/2)-2/3 + c
- (d)
none of these
The value of the integral \(\int { \frac { dx }{ { x }^{ n }(1+{ x }^{ n })^{ 1/n } } n\epsilon N\quad } \)is
- (a)
\(\frac { 1 }{ (1-n) } \left( 1+\frac { 1 }{ { x }^{ n } } \right) ^{ 1-1/n }+c\)
- (b)
\(\frac { 1 }{ (1+n) } \left( 1+\frac { 1 }{ { x }^{ n } } \right) ^{ { 1 }+{ 1/ }_{ n } }+c\)
- (c)
\(-\frac { 1 }{ (1-n) } \left( 1-\frac { 1 }{ { x }^{ n } } \right) ^{ { 1 }-{ 1/ }_{ n } }+c\)
- (d)
\(-\frac { 1 }{ (1+n) } \left( 1+\frac { 1 }{ { x }^{ n } } \right) ^{ { 1 }+{ 1/ }_{ n } }+c\)
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 \(\int { \frac { sinx }{ sin\quad (x-a) } } \) 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
Let \(\int { \frac { { x }^{ 1/2 } }{ \sqrt { x-{ x }^{ 3 } } } } dx=\frac { 2 }{ 3 } gof(x)+c\), then
- (a)
\(f(x)=\sqrt { x } \)
- (b)
\(f(x)={ x }^{ 3/2 }\)
- (c)
\(f(x)=x^{ 2/3 }\)
- (d)
\(g(x)=sin^{ -1 }x\)
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)
\(\frac { n+1 }{ n-1 } \)
Repeated application of integration by parts gives us, the reduction formula if the integrand is dependent of n, n ∈ N.
If \(I_{ n }=\int { e^{ ax }.sinxdx } \) and \(I_{ n }=\frac { e^{ ax }sin^{ n-1 }x(asinx-cosx) }{ (n+a^{ 2 }) } -Asin^{ n }x+BI_{ n-2 }\), then A+B is equal to
- (a)
\(\frac { a\left( n^{ 2 }+1 \right) }{ n(n+a^{ 2 }) } \)
- (b)
\(\frac { a\left( n^{ 2 }-1 \right) }{ n(n+a^{ 2 }) } \)
- (c)
\(\frac { a\left( n^{ 2 }+1 \right) }{ n(n^{ 2 }+a^{ 2 }) } \)
- (d)
\(\frac { a\left( n^{ 2 }-1 \right) }{ n(n^{ 2 }+a^{ 2 }) } \)
The Value of \(\int { \frac { (t-|t|)^{ 2 } }{ (1+{ t }^{ 2 }) } dt } \) is equal to
- (a)
4 (x-tan-1 x),if x< 0
- (b)
O, if x > 0
- (c)
In (1+x2), if x > 0
- (d)
none of above
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
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 \(\int { { cosec }^{ n } } xdx=-\frac { cosec^{ n-2 }cotx }{ (n-1) } +f(n)\int { cosec^{ n-2 }xdx } \\ \) then f(n+1) is equal to
- (a)
\(\frac{n-1}{n}\)
- (b)
\(\frac{n-2}{n-1}\)
- (c)
\(\frac{n-3}{n-2}\)
- (d)
none of these