Mathematics - Indefinite Integration
Exam Duration: 45 Mins Total Questions : 30
If \(\int f(x)dx=f(x)\), then \(\int \left\{ {f(x)}^2dx \right\}\) is equal to
- (a)
\({1\over 2}\int \left\{ {f(x)} \right\}^2\)
- (b)
\({f \left\{(x)\right\}}^3\)
- (c)
\({f \left\{(x)\right\}}^3\over 3\)
- (d)
\({f \left\{(x)\right\}}^2\)
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 \(\int{f(x)\over log\ sin\ x}dx\)=log log sin x, the f(x) is equal to
- (a)
sin x
- (b)
cos x
- (c)
log sin x
- (d)
cot x
\(\int{dx\over (x+1)^2\sqrt{x^2}+2x+2}\) is equal to
- (a)
\(-{\sqrt{x^2+2x+2}\over x+1}+c\)
- (b)
\(-{\sqrt{x^2+2x+1}\over (x+1)^2}+c\)
- (c)
\({-\sqrt{x^2+2x+2}\over x+1}+c\)
- (d)
None of these
\(\int{dx\over x^4+x^2}={A\over x^2}+{B\over x}+ ln|{x\over x+1}|+C,\) 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 \(\int f(x)dx=F(x),\) then \(\int x^3f(x)dx\) is equal to
- (a)
\({1\over 2}[x^2\left\{F(x)\right\}^2-\int\left\{F(x)^2\right\}dx\)
- (b)
\({1\over 2}[x^2\left\{F(x)\right\}^2-\int F(x)^2d(x^2)]\)
- (c)
\({1\over 2}[x^2F(x)-{1\over 2}\int\left\{F(x)\right\}^2dx\)
- (d)
None of the above
\(\int {tan\ x \over \sqrt{a+b.tan^2x}}dx\), \(a>b>0\) is equal to
- (a)
\({1\over \sqrt{a-b}}\ ln\ {cos\ x+\sqrt{cos^2x+{b\over a-b}}}|+C\)
- (b)
\({1\over \sqrt{a-b}}\ ln\ {sin\ x+\sqrt{sin^2x+{b\over a-b}}}|+C\)
- (c)
\(-{1\over \sqrt{a-b}}\ ln\ {cos\ x+\sqrt{cos^2x+{b\over a-b}}}|+C\)
- (d)
None of the above
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)
\({1\over 2}x^{2/3}-6\ tan^{-1}c^{1/6}+C\)
\(\int{\sqrt{1+\sqrt x}\over x}dx\) is equal to
- (a)
\(2\sqrt{1+\sqrt x}-2 \ ln\ |{\sqrt{1+\sqrt x}+1\over \sqrt{1+\sqrt x }-1}|+C\)
- (b)
\(4\sqrt{1+\sqrt x}-2\ ln|{\sqrt{1+\sqrt x}+1\over{\sqrt{1+\sqrt x}}-1}|+C\)
- (c)
\(4\sqrt{1+\sqrt x}+2\ ln|{\sqrt{1+\sqrt x}-1\over{\sqrt{1+\sqrt x}}+1}|+C\)
- (d)
None of the above
\(\int{x^3\over (1+x^2)}^{1/3}dx\) 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)
\(A={\pi\over 2}\)
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)
\({1\over 3}[x^3\Psi(x^3)-\int x^3\Psi(x^3)dx]+C\)
The Integral \(\int { \frac { dx }{ \sqrt { x+\sqrt [ 3 ]{ { x }^{ 2 } } } } } \) represents the function
- (a)
\(6\{ \sqrt [ 3 ]{ { x }^{ 2 } } -\sqrt [ 3 ]{ x } +1n|1+\sqrt [ 3 ]{ x| } \} +c\)
- (b)
\(3\{ \sqrt [ 3 ]{ { x } } -6\sqrt [ 6 ]{ x } +61n|1+\sqrt [ 6 ]{ x| } \} +c\)
- (c)
\(3\{ \sqrt [ 3 ]{ { x } } +6\sqrt [ 6 ]{ x } +61n|1+\sqrt [ 6 ]{ x| } \} +c\)
- (d)
\(6\{ \sqrt [ 3 ]{ { { x }^{ 2 } } } +3\sqrt [ 3 ]{ x } +61n|1+\sqrt [ 3 ]{ x| } \} +c\)
Given f(x) \({ \left| \begin{matrix} 0 & { x }^{ 2 }-sin\quad x & cos\quad x-2 \\ sin\quad x-x2 & 0 & 1-2x \\ 2-cox\quad x & 2x-1 & 0 \end{matrix} \right| }\), then \(\int { f(x) } \) dx is equal to
- (a)
\(\frac { { x }^{ 3 } }{ 3 } -{ x }^{ 2 }\) sin x + sin 2x+ c
- (b)
\(\frac { { x }^{ 3 } }{ 3 } -{ x }^{ 2 }\) sin x - sin 2x+ c
- (c)
\(\frac { { x }^{ 3 } }{ 3 } -{ x }^{ 2 }\) cos x - cos 2x+ c
- (d)
none of the above
\(\int { \left| ln\quad x \right| dx } \) (0< x<1)
- (a)
x + x|ln x|+ c
- (b)
x| ln x|-x + c
- (c)
x +|ln x | + c
- (d)
x - | ln x| + 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
Let f(x) be a function such that f(0)=f'(0), f''(x) = sec4 x + 4, then the function is
- (a)
\(ln\left( \sin { x } \right) |+\frac { 1 }{ 3 } { tan }^{ 3 }x+x\)
- (b)
\(\frac { 2 }{ 3 } ln|\left( \sec { x } \right) |+\frac { 1 }{ 6 } { tan }^{ 2 }x+{ 2x }^{ 2 }\)
- (c)
\(ln|cosx|+\frac { 1 }{ 6 } { cos }^{ 2 }x+\frac { { x }^{ 2 } }{ 5 } \)
- (d)
None of the above
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
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)
\(\left\{ \frac { 1 }{ 4 } ,\frac { 3 }{ 8 } ,\frac { 1 }{ 32 } \right\} \)
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
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)
\(-\frac { 1 }{ (1+n) } \left( 1+\frac { 1 }{ { x }^{ n } } \right) ^{ { 1 }+{ 1/ }_{ n } }+c\)
Repeated application of integration by parts gives us, the reduction formula if the integrand is dependent of n, n ∈ N.
If \(I_{ m,p }=\int { x^{ m }.(a+bx } ^{ n })^{ p }dx\) and \(I_{ m,p }=\frac { x^{ m+1 }(a+bx^{ n })^{ p } }{ (m+1) } -f(m,n,p,b)I_{ m+n,p-1 }\), then the value of f (1,2,3,4) is equal to
- (a)
8
- (b)
10
- (c)
12
- (d)
None of these
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)
\(\frac { n+1 }{ n-1 } \)
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 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 \(\int { \frac { \sqrt [ 4 ]{ x } }{ \sqrt { x-1 } } } \)dx is
- (a)
\(-\frac { 4 }{ 3 } { x }^{ 3/4 }+{ 4x }^{ 1/4 }+21n\left| \frac { { x }^{ 1/4 }-1 }{ { x }^{ 1/4 }+1 } \right| +c\)
- (b)
\(\frac { 4 }{ 3 } { x }^{ 3/4 }+{ 4x }^{ 1/4 }+21n\left| \frac { { x }^{ 1/4 }-1 }{ { x }^{ 1/4 }+1 } \right| +c\)
- (c)
\(-\frac { 4 }{ 3 } { x }^{ 3/4 }-{ 4x }^{ 1/4 }+21n\left| \frac { { x }^{ 1/4 }-1 }{ { x }^{ 1/4 }+1 } \right| +c\)
- (d)
\(\frac { 4 }{ 3 } { x }^{ 3/4 }-{ 4x }^{ 1/4 }+21n\left| \frac { { x }^{ 1/4 }-1 }{ { x }^{ 1/4 }+1 } \right| +c\)
If the integrand 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 \(\int { \frac { x }{ \sqrt [ 3 ]{ (a+bx) } } } dx\) is
- (a)
\(-\frac { 3 }{ 5b^{ 2 } } (a+bx)^{ 5/3 }+\frac { 3a }{ 2b^{ 2 } } (a+bx)^{ 2/3 }+c\)
- (b)
\(-\frac { 3 }{ 5b^{ 2 } } (a+bx)^{ 5/3 }-\frac { 3a }{ 2b^{ 2 } } (a+bx)^{ 2/3 }+c\)
- (c)
\(\frac { 3 }{ 5b^{ 2 } } (a+bx)^{ 5/3 }-\frac { 3a }{ 2b^{ 2 } } (a+bx)^{ 2/3 }+c\)
- (d)
\(\frac { 3 }{ 5b^{ 2 } } (a+bx)^{ 5/3 }+\frac { 3a }{ 2b^{ 2 } } (a+bx)^{ 2/3 }+c\)
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 \(\int { sin^{ 5 }xdx=-\frac { sin^{ 4 }xcosx }{ 5 } } \) + A sin2 x cos x + B cos x + C then A+B is equal to
- (a)
\(-\frac{2}{3}\)
- (b)
\(-\frac{3}{4}\)
- (c)
\(-\frac{4}{5}\)
- (d)
\(-\frac{5}{6}\)
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 \(\int { sec^{ 6 }xdx=\frac { 1 }{ 5 } tan^{ 5 }x } +Btan^{ 3 }x+Ctanx+D\) then B+ C is equal to
- (a)
\(\frac{7}{3}\)
- (b)
\(\frac{5}{3}\)
- (c)
\(\frac{11}{3}\)
- (d)
\(\frac{13}{3}\)
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