Integrals Test

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Integrals Test - 35

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Question 1
1 / -0

$$\int \dfrac {2x + 5}{\sqrt {7 - 6x - x^{2}}} dx = A\sqrt {7 - 6x - x^{2}} + B\sin^{-1} \left (\dfrac {x + 3}{4}\right ) + C$$
(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
$$(2, 1)$$

Solution

Note that $$7-6x-x^2 = 16 - (x+3)^2$$ and $$\dfrac{d}{dx} (7-6x-x^2) = -2x-6$$

So, we have

$$\int \dfrac{2x+5}{\sqrt{7-6x-x^2}} dx = \int \dfrac{2x+6}{\sqrt{7-6x-x^2}} dx - \int \dfrac{1}{\sqrt{16-(x+3)^2}} dx\\ = -2\sqrt{7-6x-x^2} - \sin^{-1} (\dfrac{x+3}{4}) + C$$

So, we have $$A = -2, \ B = -1$$.

Thus option A is the correct answer.
Question 2
1 / -0

$$\displaystyle\int\displaystyle\frac{e^{\cot^{-1}x}}{1+x^2}(x^2-x+1)dx$$

A
$$\displaystyle\frac{e^{\cot^{-1}x}}{1+x^2}$$

B
$$x\cdot e^{\cot^{-1}x}$$

C
$$e^{\cot^{-1}x}$$

D
$$-e^{\cot^{-1}x}$$

Solution

Consider, $$\displaystyle I=\int \dfrac{e^{cot^{-1}x}}{1+x^2}(x^2-x+1)dx$$

$$\Rightarrow$$ $$\displaystyle I=\int e^{cot^{-1}x}-\int \dfrac{xe^{cot^{-1}x}}{1+x^2}dx$$

$$\Rightarrow$$ $$\displaystyle \int e^{cot^{-1}x}=xe^{cot^{-1}x}+\int \dfrac{xe^{cot^{-1}x}}{1+x^2}+C$$

$$\Rightarrow$$ $$\displaystyle \int e^{cot^{-1}x}-\int \dfrac{xe^{cot^{-1}x}}{1+x^2}dx = xe^{cot^{-1}x}+C$$

$$\displaystyle \Rightarrow I = xe^{cot^{-1}x}+C$$

Hence, the answer is option (B).
Question 3
1 / -0

$$\displaystyle\int \displaystyle\frac{\cos \alpha}{\sin x\cos (x-\alpha)}dx=$$___________ $$+$$c where $$0 < x < \alpha < \pi_{/2}$$ and $$\alpha$$-constant.

A
$$-ln|tan x+\cot \alpha|$$

B
$$ln|\cot x+\tan\alpha|$$

C
$$ln|tan x+\cot \alpha|$$

D
$$-ln|\cot x+\tan\alpha|$$

Solution

$$\int \dfrac{cos\alpha}{sinxcos(x-\alpha)}dx$$

$$=\int \dfrac{cos(x-(x-\alpha))}{sinxcos(x-\alpha)}dx$$

$$=\int \dfrac{cosxcos(x-\alpha)+sinxsin(x-\alpha)}{sinxcos(x-\alpha)}dx$$

$$=\int (cotx+tan(x-\alpha))dx$$

$$= ln|sinx|-ln|cos(x-\alpha)|$$

$$= ln|\dfrac{sinx}{cos(x-\alpha)}|$$

$$= ln|\dfrac{sinx}{cosxcos(\alpha)+sinxsin(\alpha)}|$$

$$=ln|\dfrac{sec(\alpha)}{cotx+tan(\alpha)}|$$

$$=ln|(sec(\alpha))-ln(cotx+tan(\alpha))|$$

$$=-ln|(cotx+tan(\alpha))|+C$$

Hence, the answer is option (D).
Question 4
1 / -0

Let $$I=\displaystyle \int _{ \pi /4 }^{ \pi /3 }{ \cfrac { \sin { x } }{ x } } dx$$. Then?

A
$$\cfrac { 1 }{ 2 } \le I\le 1\quad $$

B
$$4\le I\le 2\sqrt { 30 } $$

C
$$\cfrac { \sqrt { 3 } }{ 8 } \le I\le \cfrac { \sqrt { 2 } }{ 6 } $$

D
$$1\le I\le \cfrac { 2\sqrt { 3 } }{ \sqrt { 2 } } $$

Solution

$$I=\displaystyle \int^{\tfrac{\pi}{3}}_{\tfrac{\pi}{4}}\dfrac{sinx}{x}dx$$

$$\dfrac{sinx}{x}$$ is a decreasing function in given interval

difference of limits $$=\dfrac{\pi}{3}-\dfrac{\pi}{4}=\dfrac{\pi}{12}$$

so, $$\dfrac{\pi}{12}\cdot\dfrac{sin\dfrac{\pi}{3}}{\dfrac{\pi}{3}} \le I \le \dfrac{\pi}{12}

\cdot\dfrac{sin\dfrac{\pi}{4}}{\dfrac{\pi}{4}}$$

$$\dfrac{\sqrt3}{8} \leq I \leq \dfrac{\sqrt2}{6}$$
Question 5
1 / -0

$$\int _{ 0 }^{ 1 }{ x{ \left[ 1-x \right] }^{ 11 } } dx=........$$

A
$$-\cfrac { 1 }{ 132 } $$

B
$$-\cfrac { 1 }{ 156 } $$

C
$$-\cfrac { 1 }{ 121 } $$

D
$$-\cfrac { 1 }{ 12 } $$

Solution

$$\rightarrow I=\int _{ 0 }^{ 1 }{ x{ (1-x) }^{ 11 } } dx$$
Take $$1-x$$ is place of $$x$$
$$=\int _{ 0 }^{ 1 }{ (1-x){ \left[ 1-(1-x) \right] }^{ 11 }dx } $$
$$=-\int _{ 0 }^{ 1 }{ \left( 1-x \right) { x }^{ 11 } } dx=-\int _{ 0 }^{ 1 }{ \left( { x }^{ 11 }-{ x }^{ 12 } \right) } dx=-{ \left[ \cfrac { { x }^{ 12 } }{ 12 } -\cfrac { { x }^{ 13 } }{ 13 } \right] }_{ 0 }^{ 1 }=\cfrac { 1 }{ 13 } -\cfrac { 1 }{ 12 } $$
$$\therefore I=-\cfrac { 1 }{ 156 } $$
Question 6
1 / -0

$$\displaystyle \int_{0}^{1}{\dfrac{ln(1+x)}{1+{x}^{2}}}dx=$$

A
$$\dfrac{\pi}{4}ln{2}$$

B
$$\dfrac{\pi}{2}ln{2}$$

C
$$\dfrac{\pi}{8}ln{2}$$

D
$$\pi\ ln{2}$$

Solution

$$\displaystyle \int_{0}^{1}{\dfrac{\ln(1+x)}{1+{x}^{2}}}dx=$$
$$\Rightarrow \dfrac{-i\left(-\ln(1-i))\ln(|x+i|)+\ln(i+1)\ln(|x-i|)+li_2\left(\dfrac{x+i}{x-1}\right)-li_2\left(\dfrac{-x+i}{1+i}\right)\right)+c}{2}$$
$$\therefore arc\tan(x)\ln(x+1)-arc\tan|(x)\ln\left(\dfrac{x^2+2x+1}{2}\right)-\tan 2\left(\dfrac{x+1}{2},\dfrac{x+1}{2}\right)$$
$$\dfrac{\ln(x^2+1)-ili_2\left(l\dfrac{i(i+1)x-i+1}{2}\right)+ili_2\left(\dfrac{i(i+1)x-i-1}{2}\right)^{+c}}{2}$$
$$\dfrac{\therefore -ili_2\left(\dfrac{i+1}{2}\right)-ili_2\left(\dfrac{-i-1}{2}\right)+\dfrac{ili2(-1)}{2}-\dfrac{-i(li2-1)}{2}+\dfrac{\pi \ln(2)}{4}}{2}$$
$$=\dfrac{\pi \ln(2)}{4}$$
Question 7
1 / -0

What is $$\displaystyle \int_{0}^{2\pi}\sqrt {1 + \sin \dfrac {x}{2}}dx$$ equal to?

A
$$8$$

B
$$4$$

C
$$2$$

D
$$0$$

Solution

$$1+\sin\frac { x }{ 2 } ={ \sin }^{ 2 }\frac { x }{ 4 } +{ \cos }^{ 2 }\frac { x }{ 4 } +2\sin\frac { x }{ 4 } \cos\frac { x }{ 4 } \\ { (\sin\frac { x }{ 4 } +\cos\frac { x }{ 4 } ) }^{ 2 }\\ \sqrt { 1+\sin\frac { x }{ 2 } } =\sqrt { { (\sin\frac { x }{ 4 } +\cos\frac { x }{ 4 } ) }^{ 2 } } =\sin\frac { x }{ 4 } +\cos\frac { x }{ 4 } $$
$$\displaystyle \int _{ 0 }^{ 2\pi }{ \left(\sin\frac { x }{ 4 } +\cos\frac { x }{ 4 }\right )dx } =4\left(-\cos\frac { x }{ 4 } +\sin\frac { x }{ 4 }\right )+c\\ After\quad putting\quad the\quad limit\quad we\quad get\\\Rightarrow 4(1-(-1))=8\\ $$
So correct answer will be option A
Question 8
1 / -0

$$\int _{ 0 }^{ 5 }{ \sqrt { 25-{ x }^{ 2 } } } dx=.........$$

A
$$25\pi $$

B
$$\cfrac { 25\pi }{ 4 } $$

C
$$\cfrac { \pi }{ 4 } $$

D
$$\cfrac { 25 }{ 4 } $$

Solution

$$\rightarrow I=\int _{ 0 }^{ 5 }{ \sqrt { 25-{ x }^{ 2 } } } dx$$
$$=\int _{ 0 }^{ 5 }{ \sqrt { { (5) }^{ 2 }-{ x }^{ 2 } } } dx\quad $$
Now take $$a=5$$ in the $$\int { \sqrt { { a }^{ 2 }-{ x }^{ 2 } } } dx$$
$$\therefore I={ \left\{ \cfrac { x\sqrt { { 5 }^{ 2 }-{ x }^{ 2 } } }{ 2 } +\cfrac { { 5 }^{ 2 } }{ 2 } \sin ^{ -1 }{ \left( \cfrac { x }{ 5 } \right) } \right\} }_{ 0 }^{ 5 }$$
$$=\left\{ \cfrac { 5\sqrt { { 5 }^{ 2 }-{ 5 }^{ 2 } } }{ 2 } +\cfrac { 25 }{ 2 } \sin ^{ -1 }{ \left( \cfrac { 5 }{ 5 } \right) } \right\} -\left\{ \cfrac { 0\sqrt { { 5 }^{ 2 }-0 } }{ 2 } +\cfrac { 25 }{ 2 } \sin ^{ -1 }{ \left( \cfrac { 0 }{ 5 } \right) } \right\} $$
$$=0+\cfrac { 25 }{ 2 } \left( \cfrac { \pi }{ 2 } \right) -\left\{ 0+\cfrac { 25 }{ 2 } (0) \right\} =\cfrac { 25\pi }{ 4 } -0$$
$$\quad I=\cfrac { 25\pi }{ 4 } $$
Question 9
1 / -0

The value of $$\displaystyle \int_{0}^{\dfrac {\pi}{4}} (\sqrt {\tan x} + \sqrt {\cot x})\,\, dx$$ is equal to

A
$$\dfrac {\pi}{2}$$

B
$$-\dfrac {\pi}{2}$$

C
$$\dfrac {\pi}{\sqrt {2}}$$

D
$$-\dfrac {\pi}{\sqrt {2}}$$

Solution

$$\begin{align} y & =\int_{0}^{\dfrac{\pi}{4}} \left( \sqrt { \tan x } +\sqrt { \cot x } \right) \, dx \\ & =\sqrt { 2 } \int_{0}^{\dfrac{\pi}{4}} \frac { \sin x+\cos x }{ \sqrt { \sin 2x } } \, dx \\ & =\sqrt { 2 } \int_{0}^{\dfrac{\pi}{4}} \frac { (\sin x+\cos x)' }{ \sqrt { 1-(\sin x-\cos x)^{ 2 } } } \, dx \\ & =\sqrt { 2 } \int_{0}^{\dfrac{\pi}{4}} \frac { 1 }{ \sqrt { 1-u^{ 2 } } } \, du \\ & =\sqrt { 2 } \sin ^{ -1 } u \\ & =\sqrt { 2 } \sin ^{ -1 } (\sin x-\cos x) \end{align}$$
Now applying limits, we get
$$y=\sqrt { 2 } \sin ^{ -1 }{ \left(\sin { \dfrac { \pi }{ 4 } } -\cos { \dfrac { \pi }{ 4 } } \right) } -\sqrt{2}\sin ^{ -1 }{ (\sin { 0 } -\cos { 0 } ) } \\ =\sqrt { 2 } \times \dfrac { \pi }{ 2 } =\dfrac { \pi }{ \sqrt { 2 } } $$
Hence, the answer is $$\dfrac{\pi}{\sqrt 2}$$
Question 10
1 / -0

$$\int _{ a }^{ b }{ \cfrac { \log { x } }{ x } } dx=.......\quad $$ (where $$a,b\in { R }^{ + }$$)

A
$$\cfrac { 1 }{ 2 } \log { \left( ab \right) } \log { \left( \cfrac { b }{ a } \right) } $$

B
$$\cfrac { 1 }{ 2 } \log { \left( ab \right) } $$

C
$$\log { \left( \cfrac { b }{ a } \right) } $$

D
$$2\log { \left( \cfrac { b }{ a } \right) } $$

Solution

$$I=\int _{ a }^{ b }{ \cfrac { \log { x } }{ x } } dx\quad (a,b+{ R }^{ + })=\int _{ a }^{ b }{ \log { x } \left( \cfrac { d }{ dx } \log { x } \right) } dx$$
$$=\cfrac { 1 }{ 2}{ { \left[ { \left( \log { x } \right) }^{ 2 } \right] }_{ a }^{ b } } \left( \therefore \int { { \left( f(x) \right) }^{ n }f'(x) } dx=\cfrac { { \left( f(x) \right) }^{ n+1 } }{ n+1 } \right) $$
$$=\cfrac { 1 }{ 2 } { \left[ { \left( \log { a } \right) }^{ 2 }-{ \left( \log { b } \right) }^{ 2 } \right] } =\cfrac { 1 }{ 2 } { \left[ \left( \log { b } +\log { a } \right) \left( \log { b } -\log { a } \right) \right] } $$
$$\therefore I=\cfrac { 1 }{ 2 } \log { \left( ab \right) } \log { \left( \cfrac { b }{ a } \right) } $$

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Re-Attempt Weekly Quiz Competition

Integrals Test - 35

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Integrals Test - 35

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SHARING IS CARING

Question 1

$$(-2, -1)$$

$$(2, -1)$$

$$(-2, 1)$$

$$(2, 1)$$

Solution

Question 2

$$\displaystyle\frac{e^{\cot^{-1}x}}{1+x^2}$$

$$x\cdot e^{\cot^{-1}x}$$

$$e^{\cot^{-1}x}$$

$$-e^{\cot^{-1}x}$$

Solution

Question 3

$$-ln|tan x+\cot \alpha|$$

$$ln|\cot x+\tan\alpha|$$

$$ln|tan x+\cot \alpha|$$

$$-ln|\cot x+\tan\alpha|$$

Solution

Question 4

$$\cfrac { 1 }{ 2 } \le I\le 1\quad $$

$$4\le I\le 2\sqrt { 30 } $$

$$\cfrac { \sqrt { 3 } }{ 8 } \le I\le \cfrac { \sqrt { 2 } }{ 6 } $$

$$1\le I\le \cfrac { 2\sqrt { 3 } }{ \sqrt { 2 } } $$

Solution

Question 5

$$-\cfrac { 1 }{ 132 } $$

$$-\cfrac { 1 }{ 156 } $$

$$-\cfrac { 1 }{ 121 } $$

$$-\cfrac { 1 }{ 12 } $$

Solution

Question 6

$$\dfrac{\pi}{4}ln{2}$$

$$\dfrac{\pi}{2}ln{2}$$

$$\dfrac{\pi}{8}ln{2}$$

$$\pi\ ln{2}$$

Solution

Question 7

$$8$$

$$4$$

$$2$$

$$0$$

Solution

Question 8

$$25\pi $$

$$\cfrac { 25\pi }{ 4 } $$

$$\cfrac { \pi }{ 4 } $$

$$\cfrac { 25 }{ 4 } $$

Solution

Question 9

$$\dfrac {\pi}{2}$$

$$-\dfrac {\pi}{2}$$

$$\dfrac {\pi}{\sqrt {2}}$$

$$-\dfrac {\pi}{\sqrt {2}}$$

Solution

Question 10

$$\cfrac { 1 }{ 2 } \log { \left( ab \right) } \log { \left( \cfrac { b }{ a } \right) } $$

$$\cfrac { 1 }{ 2 } \log { \left( ab \right) } $$

$$\log { \left( \cfrac { b }{ a } \right) } $$

$$2\log { \left( \cfrac { b }{ a } \right) } $$

Solution

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