Practice Start Test

Indefinite Integration

JEE Mathematics · 69 questions · Page 7 of 7 · Click an option or "Show Solution" to reveal answer

Q61
If    \,\,\, f(3x−43x+4)\left( {{{3x - 4} \over {3x + 4}}} \right) = x + 2, x ≠ \ne −- 43{4 \over 3}, and ∫\int {} f(x) dx = A log∣\left| {} \right.1 −- x ∣\left| {} \right. + Bx + C, then the ordered pair (A, B) is equal to : (where C is a constant of integration)
A (83,23)\left( {{8 \over 3},{2 \over 3}} \right)
B (−83,23)\left( { - {8 \over 3},{2 \over 3}} \right)
C (−83,−23)\left( { - {8 \over 3}, - {2 \over 3}} \right)
D (83,−23)\left( { {8 \over 3}, - {2 \over 3}} \right)
Correct Answer
Option B
Solution

Given, f

(3x−43x+4)\left( {{{3x - 4} \over {3x + 4}}} \right)

= x + 2, x ≠\ne −-

43{4 \over 3}

Let,

3x−43x+4{{3x - 4} \over {3x + 4}}

= t ⇒\Rightarrow

   \,\,\,

3x −- 4 = 3tx + 4t ⇒\Rightarrow

   \,\,\,

3x −- 3tx = 4t + 4 ⇒\Rightarrow

   \,\,\,

x =

4t+43−3t{{4t + 4} \over {3 - 3t}}

So, f(t) =

4t+43−3t{{4t + 4} \over {3 - 3t}}

+ 2 =

10−2t3−3t{{10 - 2t} \over {3 - 3t}}
∴   \therefore\,\,\,

f (x) =

10−2x3−3x{{10 - 2x} \over {3 - 3x}}
∴   \therefore\,\,\,
∫f(x) dx\int {f(x)\,dx}

=

∫2x−103x−3 dx\int {{{2x - 10} \over {3x - 3}}} \,dx

=

∫2x3x−3−10∫dx3x−3\int {{{2x} \over {3x - 3}} - 10\int {{{dx} \over {3x - 3}}} }

=

23∫x−1x+1dx+23∫dxx−1−103∫dxx−1{2 \over 3}\int {{{x - 1} \over {x + 1}}} dx + {2 \over 3}\int {{{dx} \over {x - 1}} - {{10} \over 3}} \int {{{dx} \over {x - 1}}}

=

23{2 \over 3}

x +

23{2 \over 3}

log

∣x−1∣\left| {x - 1} \right|

−-

103{{10} \over 3}

log

∣x−1∣\left| {x - 1} \right|

+ C =

23{2 \over 3}

x −-

83{8 \over 3}

log

∣x−1∣\left| {x - 1} \right|

+ C

∴   \therefore\,\,\,

A = −-

83{8 \over 3}

and B =

23{2 \over 3}
Q62
If ∫tan⁡x1+tan⁡x+tan⁡2xdx=x−KAtan⁡−1\int {{{\tan x} \over {1 + \tan x + {{\tan }^2}x}}dx = x - {K \over {\sqrt A }}{{\tan }^{ - 1}}} (K tan⁡x+1A)+C,(C  \left( {{{K\,\tan x + 1} \over {\sqrt A }}} \right) + C,(C\,\, is a constant of integration) then the ordered pair (K, A) is equal to :
A (2, 1)
B (−-2, 3)
C (2, 3)
D (−-2, 1)
Correct Answer
Option C
Solution

Given,

∫tan⁡x1+tan⁡x+tan⁡2x  dx\int {{{\tan x} \over {1 + \tan x + {{\tan }^2}x}}} \,\,dx

Let tanx = t ⇒\Rightarrow

   \,\,\,

sec2x dx = dt ⇒\Rightarrow

   \,\,\,

dx =

dtsec⁡2x{{dt} \over {{{\sec }^2}x}}

⇒\Rightarrow

   \,\,\,
dx=dt1+tan⁡2xdx = {{dt} \over {1 + {{\tan }^2}x}}

⇒\Rightarrow

   \,\,\,

dx =

dt1+t2{{dt} \over {1 + {t^2}}}
∴   \therefore\,\,\,
∫tan⁡x1+tan⁡x+tan⁡2x  dx\int {{{\tan x} \over {1 + \tan x + {{\tan }^2}x}}} \,\,dx

=

∫t1+t+t2×dt1+t2\int {{t \over {1 + t + {t^2}}}} \times {{dt} \over {1 + {t^2}}}

=

∫t(1+t+t2)(1+t2)  dt\int {{t \over {\left( {1 + t + {t^2}} \right)\left( {1 + {t^2}} \right)}}} \,\,dt

=

∫(11+t2−11+t+t2)  dt\int {\left( {{1 \over {1 + {t^2}}} - {1 \over {1 + t + {t^2}}}} \right)} \,\,dt

=

tan⁡−1(t)−∫11+t+t2+14−14  dt{\tan ^{ - 1}}\left( t \right) - \int {{1 \over {1 + t + {t^2} + {1 \over 4} - {1 \over 4}}}} \,\,dt

=

x−∫1t2+t+14+1−14  dtx - \int {{1 \over {{t^2} + t + {1 \over 4} + 1 - {1 \over 4}}}} \,\,dt

=

x−∫1(t+12)2+(32)2x - \int {{1 \over {{{\left( {t + {1 \over 2}} \right)}^2} + {{\left( {{{\sqrt 3 } \over 2}} \right)}^2}}}}

=

x−132+tan⁡−1(2t+13)+cx - {1 \over {{{\sqrt 3 } \over 2}}} + {\tan ^{ - 1}}\left( {{{2t + 1} \over {\sqrt 3 }}} \right) + c

=

x−23  tan⁡−1(2tan⁡x+13)+cx - {2 \over {\sqrt 3 }}\,\,{\tan ^{ - 1}}\left( {{{2\tan x + 1} \over {\sqrt 3 }}} \right) + c
∴   \therefore\,\,\,

By comparing A = 3 and K = 2

Q63
For x2 ≠ \ne nπ\pi + 1, n ∈ \in N (the set of natural numbers), the integral ∫x2sin⁡(x2−1)−sin⁡2(x2−1)2sin⁡(x2−1)+sin⁡2(x2−1)dx\int {x\sqrt {{{2\sin ({x^2} - 1) - \sin 2({x^2} - 1)} \over {2\sin ({x^2} - 1) + \sin 2({x^2} - 1)}}} dx} is equal to : (where c is a constant of integration)
A log⁡e∣12sec⁡2(x2−1)∣+c{\log _e}\left| {{1 \over 2}{{\sec }^2}\left( {{x^2} - 1} \right)} \right| + c
B 12log⁡e∣sec⁡(x2−1)∣+c{1 \over 2}{\log _e}\left| {\sec \left( {{x^2} - 1} \right)} \right| + c
C 12log⁡e∣sec⁡2(x2−12)∣+c{1 \over 2}{\log _e}\left| {{{\sec }^2}\left( {{{{x^2} - 1} \over 2}} \right)} \right| + c
D log⁡e∣sec⁡(x2−12)∣+c{\log _e}\left| {\sec \left( {{{{x^2} - 1} \over 2}} \right)} \right| + c
Correct Answer
Option D
Solution
∫x2sin⁡(x2−)−sin⁡2(x2−1)2sin⁡(x2−1)+sin⁡2(x2−1)  dx\int {x\sqrt {{{2\sin \left( {{x^2} - } \right) - \sin 2\left( {{x^2} - 1} \right)} \over {2\sin \left( {{x^2} - 1} \right) + \sin 2\left( {{x^2} - 1} \right)}}} } \,\,dx
=∫x2sin⁡(x2−1)−2sin(x2−1)cos⁡(x2−1)2sin⁡(x2−1)+2sin⁡(x2−1)cos⁡(x2−1)  dx= \int {x\sqrt {{{2\sin \left( {{x^2} - 1} \right) - 2sin\left( {{x^2} - 1} \right)\cos \left( {{x^2} - 1} \right)} \over {2\sin \left( {{x^2} - 1} \right) + 2\sin \left( {{x^2} - 1} \right)\cos \left( {{x^2} - 1} \right)}}} } \,\,dx
=∫x1−cos⁡(x2−1)1+cos⁡(x2−1) dx= \int {x\sqrt {{{1 - \cos \left( {{x^2} - 1} \right)} \over {1 + \cos \left( {{x^2} - 1} \right)}}} } \,dx
=∫x2sin⁡2(x2−12)2cos⁡2(x2−12) dx= \int {x\sqrt {{{2{{\sin }^2}\left( {{{{x^2} - 1} \over 2}} \right)} \over {2{{\cos }^2}\left( {{{{x^2} - 1} \over 2}} \right)}}} } \,dx
=∫xtan⁡(x2−12)dx= \int {x\tan } \left( {{{{x^2} - 1} \over 2}} \right)dx

put

x2−12=t{{{x^2} - 1} \over 2} = t
⇒x2−1=2t\Rightarrow {x^2} - 1 = 2t
⇒2xdx=2dt\Rightarrow 2xdx = 2dt
⇒xdx=dt\Rightarrow xdx = dt

∴\therefore

∫tan⁡t dt\int {\tan t\,dt}
=ln⁡∣sec⁡t∣+C= \ln \left| {\sec t} \right| + C
=ln⁡∣sec⁡(x2−12)∣+C= \ln \left| {\sec \left( {{{{x^2} - 1} \over 2}} \right)} \right| + C
Q64
The integral ∫3x13+2x11(2x4+3x2+1)4 dx\int {{{3{x^{13}} + 2{x^{11}}} \over {{{\left( {2{x^4} + 3{x^2} + 1} \right)}^4}}}} \,dx is equal to : (where C is a constant of integration)
A x126(2x4+3x2+1)3{{{x^{12}}} \over {6{{\left( {2{x^4} + 3{x^2} + 1} \right)}^3}}} + CC
B x46(2x4+3x2+1)3+C{{{x^4}} \over {6{{\left( {2{x^4} + 3{x^2} + 1} \right)}^3}}} + C
C x12(2x4+3x2+1)3+C{{{x^{12}}} \over {{{\left( {2{x^4} + 3{x^2} + 1} \right)}^3}}} + C
D x4(2x4+3x2+1)3+C{{{x^4}} \over {{{\left( {2{x^4} + 3{x^2} + 1} \right)}^3}}} + C
Correct Answer
Option A
Solution
∫3x13+2x11(2x4+3x2+1)4dx\int {{{3{x^{13}} + 2{x^{11}}} \over {{{\left( {2{x^4} + 3{x^2} + 1} \right)}^4}}}} dx
∫(3x3+2x5)dx(2+3x2+1x4)4\int {{{\left( {{3 \over {{x^3}}} + {2 \over {{x^5}}}} \right)dx} \over {{{\left( {2 + {3 \over {{x^2}}} + {1 \over {{x^4}}}} \right)}^4}}}}

Let

(2+3x2+1x4)=t\left( {2 + {3 \over {{x^2}}} + {1 \over {{x^4}}}} \right) = t
−12∫dtt4=16t3+C- {1 \over 2}\int {{{dt} \over {{t^4}}}} = {1 \over {6{t^3}}} + C
⇒x126(2x4+3x2+1)3+C\Rightarrow {{{x^{12}}} \over {6{{\left( {2{x^4} + 3{x^2} + 1} \right)}^3}}} + C
Q65
Let n ≥ \ge 2 be a natural number and 0<θ<π2.0 < \theta < {\pi \over 2}. Then ∫(sin⁡nθ−sin⁡θ)1/ncos⁡θsin⁡n+1θ dθ\int {{{{{\left( {{{\sin }^n}\theta - \sin \theta } \right)}^{1/n}}\cos \theta } \over {{{\sin }^{n + 1}}\theta }}} \,d\theta is equal to - (where C is a constant of integration)
A nn2−1(1+1sin⁡n−1θ)n+1n+C{n \over {{n^2} - 1}}{\left( {1 + {1 \over {{{\sin }^{n - 1}}\theta }}} \right)^{{{n + 1} \over n}}} + C
B nn2−1(1−1sin⁡n+1θ)n+1n+C{n \over {{n^2} - 1}}{\left( {1 - {1 \over {{{\sin }^{n + 1}}\theta }}} \right)^{{{n + 1} \over n}}} + C
C nn2−1(1−1sin⁡n−1θ)n+1n+C{n \over {{n^2} - 1}}{\left( {1 - {1 \over {{{\sin }^{n - 1}}\theta }}} \right)^{{{n + 1} \over n}}} + C
D nn2+1(1−1sin⁡n−1θ)n+1n+C{n \over {{n^2} + 1}}{\left( {1 - {1 \over {{{\sin }^{n - 1}}\theta }}} \right)^{{{n + 1} \over n}}} + C
Correct Answer
Option C
Solution
∫(sin⁡nθ−sin⁡θ)1/ncos⁡θsin⁡n+1θ dθ\int {{{{{\left( {{{\sin }^n}\theta - \sin \theta } \right)}^{1/n}}\cos \theta } \over {{{\sin }^{n + 1}}\theta }}} \,d\theta
=∫sin⁡θ(1−1sin⁡n−1θ)1/nsin⁡n+1θ dθ= \int {{{\sin \theta {{\left( {1 - {1 \over {{{\sin }^{n - 1}}\theta }}} \right)}^{1/n}}} \over {{{\sin }^{n + 1}}\theta }}} \,d\theta

Put

1−1sin⁡n−1θ=t1 - {1 \over {{{\sin }^{n - 1}}\theta }} = t

So

(n−1)sin⁡nθcos⁡θdθ=dt{{\left( {n - 1} \right)} \over {{{\sin }^n}\theta }}\cos \theta d\theta = dt

Now

1n−1∫(t)1n+1dt{1 \over {n - 1}}\int {{{\left( t \right)}^{{1 \over n} + 1}}dt}
=1(n−1)(t)1n+11n+1+C= {1 \over {\left( {n - 1} \right)}}{{{{\left( t \right)}^{{1 \over n} + 1}}} \over {{1 \over n} + 1}} + C
=n(n2−1)(1−1sin⁡n−1θ)1n+1+C= {n \over {\left( {n^2 - 1} \right)}}{\left( {1 - {1 \over {{{\sin }^{n - 1}}\theta }}} \right)^{{1 \over n} + 1}} + C
Q66
If ∫dxcos⁡3x2sin⁡2x=(tan⁡x)A+C(tan⁡x)B+k,\int {{{dx} \over {{{\cos }^3}x\sqrt {2\sin 2x} }}} = {\left( {\tan x} \right)^A} + C{\left( {\tan x} \right)^B} + k, where k is a constant of integration, then A + B +C equals :
A 215{{21} \over 5}
B 165{{16} \over 5}
C 710{{7} \over 10}
D 2710{{27} \over 10}
Correct Answer
Option B
Solution
∫dxcos⁡3x2sin⁡2x\int {{{dx} \over {{{\cos }^3}x\sqrt {2\sin 2x} }}}

=

∫dxcos⁡3x4sin⁡xcos⁡x\int {{{dx} \over {{{\cos }^3}x\sqrt {4\sin x\cos x} }}}

=

∫dx2cos⁡4xtan⁡x\int {{{dx} \over {2{{\cos }^4}x\sqrt {\tan x} }}}

Let tan x = t2 ⇒\Rightarrow

   \,\,\,

sec2xdx = 2t dt as sec2x = 1 + tan2x = 1 + t4 =

∫sec⁡4x dx2tan⁡x\int {{{{{\sec }^4}x\,dx} \over {2\sqrt {\tan x} }}}

=

∫sec⁡2x(sec⁡2x dx)2tan⁡x\int {{{{{\sec }^2}x\left( {{{\sec }^2}x\,dx} \right)} \over {2\sqrt {\tan x} }}}

=

∫(1+t4)2t dt2t\int {{{\left( {1 + {t^4}} \right)2t\,dt} \over {2t}}}

=

∫(1+t4) dt\int {\left( {1 + {t^4}} \right)} \,dt

= t +

t55{{{t^5}} \over 5}

+ k =

tan⁡x\sqrt {\tan x}

+

15{1 \over 5}

tan

52^{{5 \over 2}}

x + k By comparing with the given equation, we get A =

12{1 \over 2}

, B =

52{5 \over 2}

, C =

15{1 \over 5}
∴   \therefore\,\,\,

A + B + C =

165{{16} \over 5}
Q67
If ∫1−x2x4\int {{{\sqrt {1 - {x^2}} } \over {{x^4}}}} dx = A(x)(1−x2)m{\left( {\sqrt {1 - {x^2}} } \right)^m} + C, for a suitable chosen integer m and a function A(x), where C is a constant of integration, then (A(x))m equals :
A 127x6{1 \over {27{x^6}}}
B −127x9{{ - 1} \over {27{x^9}}}
C 19x4{1 \over {9{x^4}}}
D 13x3{1 \over {3{x^3}}}
Correct Answer
Option B
Solution
∫1−x2x4\int {{{\sqrt {1 - {x^2}} } \over {{x^4}}}}

dx = A(x)

(1−x2)m{\left( {\sqrt {1 - {x^2}} } \right)^m}

+ C

∫∣x∣1x2−1x4 dx,\int {{{\left| x \right|\sqrt {{1 \over {{x^2}}} - 1} } \over {{x^4}}}} \,dx,

Put

1x2−1=t⇒dtdx=−2x3{1 \over {{x^2}}} - 1 = t \Rightarrow {{dt} \over {dx}} = {{ - 2} \over {{x^3}}}

Case-I

x≥0x \ge 0
−12∫t dt ⇒t3/23+C- {1 \over 2}\int {\sqrt t \,dt\, \Rightarrow {{{t^{3/2}}} \over 3}} + C
⇒−13(1x2−1)3/2\Rightarrow - {1 \over 3}{\left( {{1 \over {{x^2}}} - 1} \right)^{3/2}}
⇒(1−x2)3−3x2+C\Rightarrow {{{{\left( {\sqrt {1 - {x^2}} } \right)}^3}} \over { - 3{x^2}}} + C
A(x)=−13x3  A(x) = - {1 \over {3{x^3}}}\,\,

and

m=3m = 3
(A(x))m=(−13x3)3=−127x9{(A(x))^m} = {\left( { - {1 \over {3{x^3}}}} \right)^3} = - {1 \over {27{x^9}}}

Case-II

x≤0x \le 0

We get

(1−x2)3−3x3+C{{{{\left( {\sqrt {1 - {x^2}} } \right)}^3}} \over { - 3{x^3}}} + C
A(x)=1−3x3,   m=3A(x) = {1 \over { - 3{x^3}}},\,\,\,m = 3
(A(x))m=−127x9{(A(x))^m} = {{ - 1} \over {27{x^9}}}
Q68
If f(x)=∫5x8+7x6(x2+1+2x7)2 dx, (x≥0),f\left( x \right) = \int {{{5{x^8} + 7{x^6}} \over {{{\left( {{x^2} + 1 + 2{x^7}} \right)}^2}}}} \,dx,\,\left( {x \ge 0} \right), f(0)=0,f\left( 0 \right) = 0, then the value of f(1)f(1) is :
A − - 12{1 \over 2}
B − - 14{1 \over 4}
C 12{1 \over 2}
D 14{1 \over 4}
Correct Answer
Option D
Solution
f(x)=∫5x8+7x6(x2+1+2x7)2 dxf\left( x \right) = \int {{{5{x^8} + 7{x^6}} \over {{{\left( {{x^2} + 1 + 2{x^7}} \right)}^2}}}} \,dx
f(x)=∫5x8+7x6x14(x−5+x−7+2)2 dxf\left( x \right) = \int {{{5{x^8} + 7{x^6}} \over {{x^{14}}{{\left( {{x^{ - 5}} + {x^{ - 7}} + 2} \right)}^2}}}} \,dx
f(x)=∫5x−6+7x−8(x−5+x−7+2)2 dxf\left( x \right) = \int {{{5{x^{ - 6}} + 7{x^{ - 8}}} \over {{{\left( {{x^{ - 5}} + {x^{ - 7}} + 2} \right)}^2}}}} \,dx

Let

x−5+x−7+2=t{x^{ - 5}} + {x^{ - 7}} + 2 = t
(−5x−6−7x−8)dx=dt\left( { - 5{x^{ - 6}} - 7{x^{ - 8}}} \right)dx = dt
(5x−6+7x−8)dx=−dt\left( {5{x^{ - 6}} + 7{x^{ - 8}}} \right)dx = - dt
f(x)=∫−dtt2=1t+cf(x) = \int {{{ - dt} \over {{t^2}}}} = {1 \over t} + c
f(x)=1x−5+x−7+2+cf\left( x \right) = {1 \over {{x^{ - 5}} + {x^{ - 7}} + 2}} + c
f(x)=x7x2+1+2x7+cf\left( x \right) = {{{x^7}} \over {{x^2} + 1 + 2{x^7}}} + c
f(0)=0f\left( 0 \right) = 0

∴\therefore

c=0c = 0
f(x)=x7(x2+1+2x7)f\left( x \right) = {{{x^7}} \over {\left( {{x^2} + 1 + 2{x^7}} \right)}}
f(1)=11+1+2=14f(1) = {1 \over {1 + 1 + 2}} = {1 \over 4}
Q69
If ∫2x+57−6x−x2  dx=A7−6x−x2+Bsin⁡−1(x+34)+C\int {{{2x + 5} \over {\sqrt {7 - 6x - {x^2}} }}} \,\,dx = A\sqrt {7 - 6x - {x^2}} + B{\sin ^{ - 1}}\left( {{{x + 3} \over 4}} \right) + C (where C is a constant of integration), then the ordered pair (A, B) is equal to :
A (2, 1)
B (−- 2, −-1)
C (−- 2, 1)
D (2, −-1)
Correct Answer
Option B
Solution

We can write, 7 - 6x - x2 = 16 - (x + 3)2 and

ddx(7−6x−x2)=−(2x+6){d \over {dx}}\left( {7 - 6x - {x^2}} \right) = - (2x + 6)

So,

∫2x+57−6x−x2dx\int {{{2x + 5} \over {\sqrt {7 - 6x - {x^2}} }}} dx

=

∫2x+6−17−6x−x2dx\int {{{2x + 6 - 1} \over {\sqrt {7 - 6x - {x^2}} }}} dx

=

∫2x+67−6x−x2dx\int {{{2x + 6} \over {\sqrt {7 - 6x - {x^2}} }}} dx

-

∫116−(x+3)2dx\int {{1 \over {\sqrt {16 - {{(x + 3)}^2}} }}} dx

=

∫2x+67−6x−x2dx\int {{{2x + 6} \over {\sqrt {7 - 6x - {x^2}} }}} dx

-

∫1(4)2−(x+3)2dx\int {{1 \over {\sqrt {{{\left( 4 \right)}^2} - {{(x + 3)}^2}} }}} dx

=

−27−6x−x2- 2\sqrt {7 - 6x - {x^2}}

-

sin⁡−1(x+34){\sin ^{ - 1}}\left( {{{x + 3} \over 4}} \right)

+ C By comparing with the given equation in the question we get, A = - 2 and B = - 1

Ready for a full JEE mock test? Timed · full syllabus · instant results
Take a Mock Test →