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Indefinite Integration

JEE Mathematics · 69 questions · Page 6 of 7 · Click an option or "Show Solution" to reveal answer

Q51
If ∫1a2sin⁡2x+b2cos⁡2x dx=112tan⁡−1(3tan⁡x)+\int \dfrac{1}{\mathrm{a}^2 \sin ^2 x+\mathrm{b}^2 \cos ^2 x} \mathrm{~d} x=\dfrac{1}{12} \tan ^{-1}(3 \tan x)+ constant, then the maximum value of asin⁡x+bcos⁡x\mathrm{a} \sin x+\mathrm{b} \cos x, is :
A 41\sqrt{41}
B 39\sqrt{39}
C 40\sqrt{40}
D 42\sqrt{42}
Correct Answer
Option C
Solution
∫1a2sin⁡2x+b2cos⁡2xdx=112tan⁡−1(3tan⁡x)+cI=∫sec⁡2xb2+a2tan⁡2xdxtan⁡x=t⇒sec⁡2xdx=dtI=∫dtb2+a2t2=1batan⁡−1(atb)+cI=1abtan⁡−1(abtan⁡x)+c⇒ab=12 and ab=3⇒a2=36 and b2=4⇒ Maximum value of asin⁡x+bcos⁡x is a2+b2\begin{aligned} & \int \frac{1}{a^2 \sin ^2 x+b^2 \cos ^2 x} d x=\frac{1}{12} \tan ^{-1}(3 \tan x)+c \\ & I=\int \frac{\sec ^2 x}{b^2+a^2 \tan ^2 x} d x \\ & \tan x=t \\ & \Rightarrow \sec ^2 x d x=d t \\ & I=\int \frac{d t}{b^2+a^2 t^2} \\ & =\frac{1}{b a} \tan ^{-1}\left(\frac{a t}{b}\right)+c \\ & I=\frac{1}{a b} \tan ^{-1}\left(\frac{a}{b} \tan x\right)+c \\ & \Rightarrow a b=12 \text{ and } \frac{a}{b}=3\\ & \Rightarrow a^2=36 \text{ and } b^2=4\\ & \Rightarrow \text{ Maximum value of } a \sin x+b \cos x \text{ is } \sqrt{a^2+b^2} \end{aligned}
=36+4\quad=\sqrt{36+4}
=40\quad=\sqrt{40}
Q52
If ∫ex(xsin⁡−1x1−x2+sin⁡−1x(1−x2)3/2+x1−x2)dx=g(x)+C\int \mathrm{e}^x\left(\dfrac{x \sin ^{-1} x}{\sqrt{1-x^2}}+\dfrac{\sin ^{-1} x}{\left(1-x^2\right)^{3 / 2}}+\dfrac{x}{1-x^2}\right) \mathrm{d} x=\mathrm{g}(x)+\mathrm{C}, where C is the constant of integration, then g(12)g\left(\dfrac{1}{2}\right) equals :
A π6e3\dfrac{\pi}{6} \sqrt{\dfrac{\mathrm{e}}{3}}
B π6e2\dfrac{\pi}{6} \sqrt{\dfrac{\mathrm{e}}{2}}
C π4e3\dfrac{\pi}{4} \sqrt{\dfrac{\mathrm{e}}{3}}
D π4e2\dfrac{\pi}{4} \sqrt{\dfrac{\mathrm{e}}{2}}
Correct Answer
Option A
Solution
∵ddx(xsin⁡−1x1−x2)=sin⁡−1x(1−x2)3/2+x1−x2⇒∫ex(xsin⁡−1x1−x2+sin⁡−1x(1−x2)3/2+x1−x2)dx=ex⋅xsin⁡−1x1−x2+c=g(x)+C Note : assuming g(x)=xexsin⁡−1x1−x2g(1/2)=e1/22⋅π6×23=π6e3\begin{aligned} &\begin{aligned} & \because \frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{x} \sin ^{-1} \mathrm{x}}{\sqrt{1-\mathrm{x}^2}}\right)=\frac{\sin ^{-1} \mathrm{x}}{\left(1-\mathrm{x}^2\right)^{3 / 2}}+\frac{\mathrm{x}}{1-\mathrm{x}^2} \\ & \Rightarrow \int \mathrm{e}^{\mathrm{x}}\left(\frac{\mathrm{x} \sin ^{-1} \mathrm{x}}{\sqrt{1-\mathrm{x}^2}}+\frac{\sin ^{-1} \mathrm{x}}{\left(1-\mathrm{x}^2\right)^{3 / 2}}+\frac{\mathrm{x}}{1-\mathrm{x}^2}\right) \mathrm{dx} \\ & =\mathrm{e}^{\mathrm{x}} \cdot \frac{\mathrm{x} \sin ^{-1} \mathrm{x}}{\sqrt{1-\mathrm{x}^2}}+\mathrm{c}=\mathrm{g}(\mathrm{x})+\mathrm{C} \end{aligned}\\ &\text{ Note : assuming } g(x)=\frac{\mathrm{xe}^{\mathrm{x}} \sin ^{-1} x}{\sqrt{1-\mathrm{x}^2}}\\ &g(1 / 2)=\frac{\mathrm{e}^{1 / 2}}{2} \cdot \frac{\frac{\pi}{6} \times 2}{\sqrt{3}}=\frac{\pi}{6} \sqrt{\frac{\mathrm{e}}{3}} \end{aligned}

Comment : In this question we will not get a unique function g(x)\mathrm{g}(\mathrm{x}), but in order to match the answer we will have to assume g(x)=xexsin⁡−1x1−x2g(x)=\dfrac{\mathrm{xe}^{\mathrm{x}} \sin ^{-1} \mathrm{x}}{\sqrt{1-\mathrm{x}^2}}.

Q53
If f(x)=∫1x1/4(1+x1/4)dx,f(0)=−6f(x)=\int \dfrac{1}{x^{1 / 4}\left(1+x^{1 / 4}\right)} \mathrm{d} x, f(0)=-6, then f(1)f(1) is equal to :
A 4(log⁡e2−2)4\left(\log _{\mathrm{e}} 2-2\right)
B log⁡e22+2\log _{e^2} 2+2
C 2−log⁡e22-\log \mathrm{e}^2
D 4(log⁡e2+2)4\left(\log _e 2+2\right)
Correct Answer
Option A
Solution

Substitution: Start by substituting x=t4 x = t^4 which implies dx=4t3 dt dx = 4t^3 \, dt .

Integral Transformation: ∫1x1/4(1+x1/4) dx⇒∫4t3 dtt(1+t)=∫4t1+t dt \int \dfrac{1}{x^{1/4}(1 + x^{1/4})} \, dx \Rightarrow \int \dfrac{4t^3 \, dt}{t(1 + t)} = \int \dfrac{4t}{1 + t} \, dt Rewriting the Integral: The expression 4t1+t \dfrac{4t}{1 + t} can be decomposed as: 4t1+t=4(t2−1+11+t)=4((t−1)+1t+1) \dfrac{4t}{1 + t} = 4\left(\dfrac{t^2 - 1 + 1}{1 + t}\right) = 4\left((t - 1) + \dfrac{1}{t + 1}\right) Separate and Integrate: 4∫(t−1) dt+4∫1t+1 dt 4 \int (t - 1) \, dt + 4 \int \dfrac{1}{t + 1} \, dt Solving the Integrals: The integral of t−1 t - 1 is (t−1)22 \dfrac{(t - 1)^2}{2} .

The integral of 1t+1 \dfrac{1}{t + 1} is ln⁡(t+1) \ln(t + 1) .

Therefore: f(x)=4{(t−1)22+ln⁡(t+1)}+C f(x) = 4 \left\{ \dfrac{(t - 1)^2}{2} + \ln(t + 1) \right\} + C Substitute t=x1/4 t = x^{1/4} : Replace t t back with x1/4 x^{1/4} : f(x)=2(x1/4−1)2+4ln⁡(1+x1/4)+C f(x) = 2(x^{1/4} - 1)^2 + 4 \ln(1 + x^{1/4}) + C Using the Condition f(0)=−6 f(0) = -6 : f(0)=2(0−1)2+4ln⁡(1+0)+C=2×1−0+C=−6 f(0) = 2(0 - 1)^2 + 4 \ln(1 + 0) + C = 2 \times 1 - 0 + C = -6 Solving gives C=−8 C = -8 .

Find f(1) f(1) : f(1)=2(11/4−1)2+4ln⁡(1+11/4)−8 f(1) = 2(1^{1/4} - 1)^2 + 4 \ln(1 + 1^{1/4}) - 8 Simplifies to: f(1)=0+4ln⁡2−8=4(ln⁡2−2) f(1) = 0 + 4 \ln 2 - 8 = 4(\ln 2 - 2)

Q54
Let I(x)=∫dx(x−11)1113(x+15)1513\mathrm{I}(x)=\int \dfrac{d x}{(x-11)^{\dfrac{11}{13}}(x+15)^{\dfrac{15}{13}}}. If I(37)−I(24)=14(1 b113−1c113),b,c∈N\mathrm{I}(37)-\mathrm{I}(24)=\dfrac{1}{4}\left(\dfrac{1}{\mathrm{~b}^{\dfrac{1}{13}}}-\dfrac{1}{\mathrm{c}^{\dfrac{1}{13}}}\right), \mathrm{b}, \mathrm{c} \in \mathcal{N}, then 3( b+c)3(\mathrm{~b}+\mathrm{c}) is equal to
A 39
B 22
C 40
D 26
Correct Answer
Option A
Solution
I(x)=∫dx(x−11)1113(x+15)1513 Put x−11x+15=t⇒26(x+5)2dx=dtI(x)=126∫dtt11/13=126⋅t2/132/13\begin{aligned} & I(x)=\int \frac{d x}{(x-11)^{\frac{11}{13}}(x+15)^{\frac{15}{13}}} \\ & \text{ Put } \frac{x-11}{x+15}=t \Rightarrow \frac{26}{(x+5)^2} d x=d t \\ & I(x)=\frac{1}{26} \int \frac{d t}{t^{11 / 13}}=\frac{1}{26} \cdot \frac{t^{2 / 13}}{2 / 13} \end{aligned}
I(x)=14(x−11x+15)2/13+CI(37)−I(24)=14(2652)2/13−14(1339)2/13=14(122/13−132/13)=14(141/13−191/13)∴b=4,c=93( b+c)=39\begin{aligned} & \mathrm{I}(\mathrm{x})=\frac{1}{4}\left(\frac{\mathrm{x}-11}{\mathrm{x}+15}\right)^{2 / 13}+\mathrm{C} \\ & \mathrm{I}(37)-\mathrm{I}(24)=\frac{1}{4}\left(\frac{26}{52}\right)^{2 / 13}-\frac{1}{4}\left(\frac{13}{39}\right)^{2 / 13} \\ & =\frac{1}{4}\left(\frac{1}{2^{2 / 13}}-\frac{1}{3^{2 / 13}}\right) \\ & =\frac{1}{4}\left(\frac{1}{4^{1 / 13}}-\frac{1}{9^{1 / 13}}\right) \\ & \therefore \mathrm{b}=4, \mathrm{c}=9 \\ & 3(\mathrm{~b}+\mathrm{c})=39 \end{aligned}
Q55
Let ∫x3sin⁡x dx=g(x)+C\int x^3 \sin x \mathrm{~d} x=g(x)+C, where CC is the constant of integration. If 8(g(π2)+g′(π2))=απ3+βπ2+γ,α,β,γ∈Z8\left(g\left(\dfrac{\pi}{2}\right)+g^{\prime}\left(\dfrac{\pi}{2}\right)\right)=\alpha \pi^3+\beta \pi^2+\gamma, \alpha, \beta, \gamma \in Z, then α+β−γ\alpha+\beta-\gamma equals :
A 47
B 55
C 62
D 48
Correct Answer
Option B
Solution
∫x3sin⁡xdx=−x3cos⁡x+∫3x2cos⁡xdx=−x3cos⁡x+3x2sin⁡x−∫6xsin⁡xdx=−x3cos⁡x+3x2sin⁡x+6xcos⁡x−6sin⁡x+c\begin{aligned} & \int x^3 \sin x d x=-x^3 \cos x+\int 3 x^2 \cos x d x \\ & =-x^3 \cos x+3 x^2 \sin x-\int 6 x \sin x d x \\ & =-x^3 \cos x+3 x^2 \sin x+6 x \cos x-6 \sin x+c \end{aligned}

So g(x)=−x3cos⁡x+3x2sin⁡x+6xcos⁡x−6sin⁡xg(x)=-x^3 \cos x+3 x^2 \sin x+6 x \cos x-6 \sin x

g(π2)=3π24−6g′(x)=x3sin⁡xg′(π2)=π388(g(π2)+g′(π2))=π3+6π2−48\begin{aligned} & g\left(\frac{\pi}{2}\right)=\frac{3 \pi^2}{4}-6 \\ & g^{\prime}(x)=x^3 \sin x \\ & g^{\prime}\left(\frac{\pi}{2}\right)=\frac{\pi^3}{8} \\ & 8\left(g\left(\frac{\pi}{2}\right)+g^{\prime}\left(\frac{\pi}{2}\right)\right)=\pi^3+6 \pi^2-48 \end{aligned}

So α+β−γ=55\alpha+\beta-\gamma=55

Q56
 Let f(x)=∫x33−x2dx. If 5f(2)=−4, then f(1) is equal to  \text{ Let } f(x)=\int x^3 \sqrt{3-x^2} d x \text{. If } 5 f(\sqrt{2})=-4 \text{, then } f(1) \text{ is equal to }
A −625-\dfrac{6 \sqrt{2}}{5}
B −825-\dfrac{8 \sqrt{2}}{5}
C −225-\dfrac{2 \sqrt{2}}{5}
D −425-\dfrac{4 \sqrt{2}}{5}
Correct Answer
Option A
Solution
∫x33−x2dx3−x2=t2−2xdx=2tdt=−∫t2(3−t2)dt=∫t4−3t2dt=t55−t3+c\begin{aligned} & \int x^3 \sqrt{3-x^2} d x \\ & 3-x^2=t^2 \\ & -2 x d x=2 t d t \\ & =-\int t^2\left(3-t^2\right) d t=\int t^4-3 t^2 d t \\ & =\frac{t^5}{5}-t^3+c \end{aligned}
f(x)=(3−x2)525−(3−x2)32+c∵f(2)=−45⇒−45=15−1+c⇒c=0∴f(x)=(3−x2)525−(3−x2)32 Now f(1)=2525−232=232[25−1]=22(−35)=−652\begin{aligned} & f(x)=\frac{\left(3-x^2\right)^{\frac{5}{2}}}{5}-\left(3-x^2\right)^{\frac{3}{2}}+c \\ & \because f(\sqrt{2})=-\frac{4}{5} \\ & \Rightarrow \quad-\frac{4}{5}=\frac{1}{5}-1+c \Rightarrow c=0 \\ & \begin{aligned} \therefore \quad f(x) & =\frac{\left(3-x^2\right)^{\frac{5}{2}}}{5}-\left(3-x^2\right)^{\frac{3}{2}} \\ \text{ Now } f(1) & =\frac{2^{\frac{5}{2}}}{5}-2^{\frac{3}{2}}=2^{\frac{3}{2}}\left[\frac{2}{5}-1\right] \\ & =2 \sqrt{2}\left(-\frac{3}{5}\right)=-\frac{6}{5} \sqrt{2} \end{aligned} \end{aligned}
Q57
If ∫ \int \, x5.e−-4x3 dx = 148{1 \over {48}}e−-4x3 f(x) + C, where C is a constant of inegration, then f(x) is equal to -
A −-2x3 −- 1
B −- 2x3 + 1
C 4x3 + 1
D −-4x3 −- 1
Correct Answer
Option D
Solution
∫x5.e−4x3 dx=148e−4x3f(x)+c\int {{x^5}} .{e^{ - 4{x^3}}}\,dx = {1 \over {48}}{e^{ - 4{x^3}}}f\left( x \right) + c

Put

x3=t{x^3} = t
3x2 dx=dt3{x^2}\,dx = dt
∫x3.e−4x3. x2dx\int {{x^3}.{e^{ - 4{x^3}}}.\,{x^2}} dx
13∫t.e−4tdt{1 \over 3}\int {t.{e^{ - 4t}}dt}
13[t.e−4t−4−∫e−4t−4dt]{1 \over 3}\left[ {t.{{{e^{ - 4t}}} \over { - 4}} - \int {{{{e^{ - 4t}}} \over { - 4}}dt} } \right]
−e−4t48[4t+1]+c- {{{e^{ - 4t}}} \over {48}}\left[ {4t + 1} \right] + c
−e−4x348[4x3+1]+c{{ - {e^{ - 4{x^3}}}} \over {48}}\left[ {4{x^3} + 1} \right] + c

∴\therefore

f(x)=−1−4x3f(x) = - 1 - 4{x^3}
Q58
If ∫x+12x−1 dx\int {{{x + 1} \over {\sqrt {2x - 1} }}} \,dx = f(x) 2x−1\sqrt {2x - 1} + C, where C is a constant of integration, then f(x) is equal to :
A 23{2 \over 3} (x −- 4)
B 13{1 \over 3} (x + 4)
C 13{1 \over 3} (x + 1)
D 23{2 \over 3} (x + 2)
Correct Answer
Option B
Solution
2x−1=t⇒2x−1=t2⇒2dx=2t.dt\sqrt {2x - 1} = t \Rightarrow 2x - 1 = {t^2} \Rightarrow 2dx = 2t.dt
∫x+12x−1dx=∫t2+12+1ttdt=∫t2+32dt\int {{{x + 1} \over {\sqrt {2x - 1} }}dx = \int {{{{{{t^2} + 1} \over 2} + 1} \over t}tdt = \int {{{{t^2} + 3} \over 2}dt} } }
=12(t33+3t)=t6(t2+9)+c= {1 \over 2}\left( {{{{t^3}} \over 3} + 3t} \right) = {t \over 6}\left( {{t^2} + 9} \right) + c
=2x−1(2x−1+96)+c=2x−1(x+43)+c= \sqrt {2x - 1} \left( {{{2x - 1 + 9} \over 6}} \right) + c = \sqrt {2x - 1} \left( {{{x + 4} \over 3}} \right) + c
⇒f(x)=x+43\Rightarrow f\left( x \right) = {{x + 4} \over 3}
Q59
The integral ∫dx(1+x)x−x2\int {{{dx} \over {\left( {1 + \sqrt x } \right)\sqrt {x - {x^2}} }}} is equal to : (where C is a constant of integration.)
A −21+x1−x+C - 2\sqrt {{{1 + \sqrt x } \over {1 - \sqrt x }}} + C
B −21−x1+x+C - 2\sqrt {{{1 - \sqrt x } \over {1 + \sqrt x }}} + C
C −1−x1+x+C - \sqrt {{{1 - \sqrt x } \over {1 + \sqrt x }}} + C
D 21+x1−x+C2\sqrt {{{1 + \sqrt x } \over {1 - \sqrt x }}} + C
Correct Answer
Option B
Solution

I =

∫dx(1+x)x−x2\int {{{dx} \over {\left( {1 + \sqrt x } \right)\sqrt {x - {x^2}} }}}

=

∫dx(1+x)x1−x\int {{{dx} \over {\left( {1 + \sqrt x } \right)\sqrt x \sqrt {1 - x} }}}

Let 1 +

x\sqrt x

= t ⇒\Rightarrow

   \,\,\,
12x dx{1 \over {2\sqrt x }}\,dx

= dt I =

∫2dtt2t−t2\int {{{2dt} \over {t\sqrt {2t - {t^2}} }}}

Again let t =

1z{1 \over z}

⇒\Rightarrow

   \,\,\,

dt = −-

1z2dz{1 \over {{z^2}}}dz
∴   \therefore\,\,\,

I = 2

∫−1z2dz1z2z−1z2\int {{{ - {1 \over {{z^2}}}dz} \over {{1 \over z}\sqrt {{2 \over z} - {1 \over {{z^2}}}} }}}

= 2

∫−dz2z−1\int {{{ - dz} \over {\sqrt {2z - 1} }}}

=

−22z−1+C- 2\sqrt {2z - 1} + C

=

−22t−1+C- 2\sqrt {{2 \over t} - 1} + C

=

−22−tt+C- 2\sqrt {{{2 - t} \over t}} + C

=

−21−x1+x+C- 2\sqrt {{{1 - \sqrt x } \over {1 + \sqrt x }}} + C
Q60
Let In=∫tan⁡nx dx, (n>1).{I_n} = \int {{{\tan }^n}x\,dx} ,\,\left( {n > 1} \right). If I4+I6{I_4} + {I_6} = atan⁡5x+bx5+Ca{\tan ^5}x + b{x^5} + C, where C is a constant of integration, then the ordered pair (a,b)\left( {a,b} \right) is equal to
A (15,0)\left( {{1 \over 5},0} \right)
B (15,−1)\left( {{1 \over 5}, - 1} \right)
C (−15,0)\left( { - {1 \over 5},0} \right)
D (−15,1)\left( { - {1 \over 5},1} \right)
Correct Answer
Option A
Solution

Given, In =

∫tan⁡nx dx,   n>1\int {{{\tan }^n}x\,dx,\,\,\,n > 1}
∴   \therefore\,\,\,

I4 =

∫tan⁡4x dx\int {{{\tan }^4}x\,dx}

and I6 =

∫tan⁡6x dx\int {{{\tan }^6}} x\,dx
∴   \therefore\,\,\,

I = I4 + I6 =

∫(tan⁡4x+tan⁡6x)dx\int {\left( {{{\tan }^4}x + {{\tan }^6}x} \right)} dx

=

∫tan⁡4x(1+tan⁡2x)dx\int {{{\tan }^4}} x\left( {1 + {{\tan }^2}x} \right)dx

=

∫tan⁡4x.sec⁡2x dx\int {{{\tan }^4}} x.{\sec ^2}x\,dx

Let, tanx = t ⇒\Rightarrow

   \,\,\,

sec2x dx = dt

∴   \therefore\,\,\,

I =

∫t4 dt\int {{t^4}\,dt}

=

15{1 \over 5}

t5 + C =

15{1 \over 5}

tan5x + C

∴   \therefore\,\,\,

By comparing with the question, we get A =

15{1 \over 5}

, B = 0

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