JEE Main Mathematics - Indefinite Integration
Exam Duration: 60 Mins Total Questions : 30
If \(\int{{1-x^7}\over x(1+x^7)}dx=\alpha\ ln\ |x|+b\ ln\ |x^7+1|+C,\) then
- (a)
\(a=1, b={2\over 7}\)
- (b)
\(a=-1, b={2\over 7}\)
- (c)
\(a=1, b=-{2\over 7}\)
- (d)
\(a=-1, b=-{2\over 7}\)
If \({d\over dx}[f(x)]=xcos\ x+sin\ x\ and\ f(0)=2,\) then f(x) is equal to
- (a)
x sin x
- (b)
x cos x + sin x + 2
- (c)
x sin x+2
- (d)
x cos x+2
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
\(\int{dx\over \sqrt{sin^3\ x.sin(x+\alpha)}}\) is equal to
- (a)
\(2\ cosec\ \alpha \sqrt{cos\ \alpha+sin\ \alpha.cot x}+C\)
- (b)
\(-2\ cosec\ \alpha \sqrt{cos\ \alpha+sin\ \alpha.cot x}+C\)
- (c)
\(cosec\ \alpha \sqrt{cos\ \alpha+sin\ \alpha.cot x}+C\)
- (d)
None of the above
\(\int{dx\over cos^3\ x.\sqrt{sin2x}}\) is equal to
- (a)
\(\sqrt2(\sqrt{cos\ x}+{1\over 5}tan^{5/2}x)+C\)
- (b)
\(\sqrt2(\sqrt{tan\ x}+{1\over 5}tan^{5/2}x)+C\)
- (c)
\(\sqrt2(\sqrt{tan\ x}+{1\over 5}tan^{5/2}x)+C\)
- (d)
\(\sqrt2(\sqrt{tan\ x}+{1\over 5}tan^{5/2}x)+C\)
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
\(If \int{dx\over 3\sqrt{sin^{11}xcos\ x}}=-[{{3\over 8}f(x)+{3\over 2}g(x)}]+C,\) 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)
\(f(x)=tan^{4/3}x, g(x)=tan^{-4/3}x\)
\(\int{x^2-1\over x^4+x^2+1}dx\) 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)
\({ 2}log ({{x^2-x+1}\over x^2+x+1})+C\)
\(\int{tan x\over \sqrt{sin^4x+cos^4\ x}}dx\) is equal to
- (a)
\(ln(tan^2x+\sqrt{1+tan^2x})+C\)
- (b)
\(sec\ x+C\)
- (c)
\(\sqrt{1-tan^2x+C}\)
- (d)
None of the above
The equation of a curve passing through origin is given by \(y=\int x^3.cos\ x^4dx.\) If the equation of the curve is written in the form \(x=g(y),\) then
- (a)
\(g(y)=3\sqrt{sin^{-1}4y}\)
- (b)
\(g(y)=\sqrt{sin^{-1}4y}\)
- (c)
\(g(y)=4\sqrt{sin^{-1}4y}\)
- (d)
\(g(y)=5\sqrt{sin^{-1}4y}\)
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
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)
\({1\over 3}[x^3\Psi(x^3)-\int x^3\Psi(x^3)dx]+C\)
\(\int \left\{ {(ln\ x-1)\over 1+(ln\ x)^2} \right\}^2 dx\) is equal to
- (a)
\({x\over (ln\ x)^2+1}+C\)
- (b)
\({xe^x\over 1+x^2}+C\)
- (c)
\({x\over x^2+1}+C\)
- (d)
\({ln\ x\over (ln\ x)^2+1}+C\)
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
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)
\(ln|sec({ x }^{ 2 }+1|+c\)
\(\int { sec^{ 4/9 } } \theta \cos { ec } ^{ 14/9 }\theta \) is equal to
- (a)
\(\frac { 5 }{ 9 } \cos { { \theta }^{ -5/9 } } +c\)
- (b)
\(-\frac { 9 }{ 5 } (tan\theta )^{ -5/9 }+c\)
- (c)
\(\frac { 9 }{ 5 } (tan\theta )^{ -9/5 }+c\)
- (d)
\(-\frac { 5 }{ 9 } (tan\theta )^{ -9/5 }+c\)
\(\int { \frac { \left( { \sqrt { x } }^{ 5 } \right) }{ \left( { \sqrt { x } }^{ 7 } \right) +{ x }^{ 6 } } } dx=\lambda \quad ln\left( \frac { { x }^{ a } }{ { x }^{ a }+1 } \right) +c\) 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)
\(\frac { 1 }{ 2 } x|x|ln|x|-\frac { 1 }{ 4 } x|x|+c\)
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
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)
\(\left\{ \frac { 1 }{ 4 } ,\frac { 3 }{ 8 } ,\frac { 1 }{ 32 } \right\} \)
\(\int\) 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
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 \(\int { \left( \frac { { 4e }^{ x }+{ 6e }^{ -x } }{ { 9 }e^{ x }-{ 4e }^{ -x } } \right) } \) 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)
\(\frac { 1 }{ 3 } { sin }\left( 3x+\frac { \pi }{ 7 } \right) sin\left( 3x-\frac { \pi }{ 7 } +c \right) \)
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 \(\int { sec^{ 6 }xdx=\frac { 1 }{ 5 } tan^{ 5 }x } +Btan^{ 3 }x+Ctanx+D\) 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 \(\int { { cos }^{ n } } xdx=-\frac { cos^{ n-1 }xsinx }{ n } +f(n)\int { cos^{ n-2 }xdx } \), then f(5) is equal to
- (a)
1/5
- (b)
2/5
- (c)
3/5
- (d)
4/5