Integrals Test

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Integrals Test - 14

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Weekly Quiz Competition

Question 1
1 / -0

$$\int \dfrac {dx}{\sqrt {x^{10} - x^{2}}}; x > 1=$$ ______ $$+ C$$.

A
$$\dfrac {1}{4}\log |\sqrt {x^{10} - x^{2}} + x^{2}|$$

B
$$\dfrac {1}{2}\log |x^{10} - x^{2}|$$

C
$$-\dfrac {1}{4}\sec^{-1} (x^{4})$$

D
$$\dfrac {1}{4}\sec^{-1} (x^{4})$$

Solution

Let $$I=\displaystyle \int \dfrac {dx}{\sqrt {x^{10} - x^{2}}}$$

$$=\displaystyle \int \dfrac {4x^{3}dx}{4x^{3} \times x \sqrt {x^{8} - 1}}$$

$$=\displaystyle \int \dfrac {4x^{3} dx}{4x^{4}\sqrt {x^{8} - 1}}$$
Put $$x^{4} = t$$
$$\Rightarrow 4x^{3}dx = dt$$
$$\displaystyle \int \dfrac {dt}{4t \sqrt {t^{2} - 1}} = \dfrac {1}{4}\sec^{-1} (x^{4}) + C$$.
Question 2
1 / -0

Evaluate the integral
$$\displaystyle \int_{0}^{1}\frac{1}{1+x^{2}}dx$$

A
$$\pi/4$$

B
$$\pi$$

C
$$\pi/3$$

D
$$0$$

Solution

$$\displaystyle \int_{0}^{1}\frac{1}{1+x^2}dx$$

$$\displaystyle \int \frac{1}{1+x^2}dx=\tan^{-1}x +c$$

$$\displaystyle \int_{0}^{1}\frac{1}{1+x^2}dx=\tan^{-1}x+c\displaystyle \vert_{0}^{1}$$

$$=(\tan^{-1}(1)+c)-(\tan^{-1}(0)+c)$$

$$=\dfrac{\pi}{4}$$
Question 3
1 / -0

If $$A= \int_{0}^{1} x^{50}(2-x)^{50} dx, B= \int_{0}^{1}x^{50}(1-x)^{50}dx$$, which of the following is true?

A
$$A=2^{50}B$$

B
$$A=2^{-50}B$$

C
$$A=2^{100}B$$

D
$$A=2^{-100}B$$

Solution

Solution : - given,
$$ \displaystyle \Rightarrow A = \int_{0}^{1}x^{50}(2-2x)^{50}dx$$ $$B = \int_{0}^{1}x^{50}(1-x)^{50}dx$$
$$ \displaystyle \Rightarrow A = \int_{0}^{1} x^{50} \times 2^{50} (1-x)^{30}dx$$
$$ \displaystyle \Rightarrow A = 2^{50}\int_{0}^{1} x^{50} (1-x)^{50}dx $$
$$\displaystyle \Rightarrow A = 2^{50}B $$ proved
Question 4
1 / -0

The value of $$\int _{ }^{ }{ \cfrac { \log { x } }{ { \left( x+1 \right) }^{ 2 } } } dx$$ is

A
$$\cfrac { -\log { x } }{ x+1 } +\log { x } -\log { \left( x+1 \right) } +C $$

B
$$\cfrac { \log { x } }{ x+1 } +\log { x } -\log { \left( x+1 \right) } +C $$

C
$$\cfrac { \log { x } }{ x+1 } -\log { x } -\log { \left( x+1 \right) } + C $$

D
$$\cfrac { -\log { x } }{ x+1 } -\log { x } -\log { \left( x+1 \right) } +C $$

Solution

$$\int \dfrac{lnx}{(x+1)^{2}}$$
$$lnx\int \dfrac{dx}{(x+1)^{2}}-\int \dfrac{1}{x}\int \dfrac{dx}{(x+1)^{2}}$$
$$lnx\dfrac{(x+1)^{-2+1}}{-2+1}-\int \dfrac{1}{x}\dfrac{(x+1)^{-2+1}}{-2+1}dx$$
$$-\dfrac{lnx}{(x+1)}+\int \dfrac{1}{x}\dfrac{1}{(x+1)}dx$$
$$-\dfrac{lnx}{(x+1)}+\int \dfrac{(x+1)-(x)}{x(x+1)}$$
$$-\dfrac{lnx}{(x+1)}+\int \dfrac{1}{x}-\dfrac{1}{1+x}dx$$
$$-\dfrac{lnx}{(1+x)}+lnx-ln(1+x)+C$$
Question 5
1 / -0

$$\displaystyle\int_0^{\pi /2} {\dfrac{{\sin x}}{{\sqrt {1 + {\mathop{\rm cosx}\nolimits} } }}} dx = $$

A
$$\sqrt 2 - 1$$

B
$$2\sqrt 2 $$

C
$$2(\sqrt 2 - 1)$$

D
$$\dfrac{{\sqrt 2 + 1}}{2}$$

Solution

$$\displaystyle I= \int_{0}^{\pi /2}\frac{sin x}{\sqrt{1+cos x}}dx$$

Let $$t= 1+ cos x \Rightarrow dt= -sin xdx$$

limit of t : when $$x=0, t=2$$

when $$x=\dfrac{\pi }{2}, t=1$$

$$\displaystyle \therefore I=\int_{2}^{1}\frac{-dt}{\sqrt{t}}=\int_{1}^{2}\frac{dt}{\sqrt{t}}$$

$$=2[\sqrt{t}]_{1}^{2}$$

$$=2[\sqrt{2}-1]$$
Question 6
1 / -0

What is $$\displaystyle \int \dfrac{dx}{x(1 + ln x)^n}$$ equal to $$(n \neq 1)$$ ?

A
$$\dfrac{1}{(n - 1)(1 + ln x)^{n - 1}} + c$$

B
$$\dfrac{1 - n}{(1 + ln x)^{1- n}} + c$$

C
$$\dfrac{n + 1}{(1 + ln x)^{n+1}} + c$$

D
$$-\dfrac{1}{(n - 1)(1 + ln x)^{n-1}} + c$$

Solution

Given,

$$\int \dfrac{1}{x\left(1+\ln \left(x\right)\right)^n}dx$$

apply $$u=1+\ln \left(x\right)$$

$$=\int \dfrac{1}{u^n}du$$

$$=\int \:u^{-n}du$$

$$=\dfrac{u^{-n+1}}{-n+1}$$

$$=\dfrac{\left(1+\ln \left(x\right)\right)^{-n+1}}{-n+1}$$

$$=\dfrac{\left(1+\ln \left(x\right)\right)^{-n+1}}{-n+1}+C$$
$$-\dfrac{1}{(n - 1)(1 + ln x)^{n-1}} + c$$
Question 7
1 / -0

Evaluate $$\displaystyle\int^1_0\dfrac{1}{(1+x^2)}dx$$

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

B
$$\dfrac{\pi}{3}$$

C
$$\dfrac{\pi}{4}$$

D
None of these

Solution

$$I=\displaystyle\int^1_0\dfrac{dx}{(1+x^2)}\\$$
$$=[\tan^{-1}x]^1_0\\$$ ....... $$\because\displaystyle \int \dfrac {dx}{ {x^2+a^2} }=\tan^{-1}\dfrac xa+C$$

$$=(\tan^{-1}1-\tan^{-1}0)\\$$
$$=\dfrac{\pi}{4}$$.
Question 8
1 / -0

The integral $$\displaystyle \int_{0}^{\pi_/{3}}\frac{\cos x}{3+4\sin x}d_{X=}$$

A
$$\displaystyle \log\left(\frac{3+2\sqrt{3}}{3}\right)$$

B
$$\dfrac{1}{4} \log\left(\displaystyle \frac{3+2\sqrt{3}}{3}\right)$$

C
$$2\displaystyle \log\left(\frac{3+2\sqrt{3}}{3}\right)$$

D
$$\dfrac{1}{2}\log\left(\displaystyle \frac{3+2\sqrt{3}}{2}\right)$$

Solution

$$\displaystyle \int_{0}^{\dfrac{\pi}{3}}\dfrac{\cos x }{3+4 \sin x}dx=\int_{0}^{\dfrac{\pi}{3}}\dfrac{d(\sin x)}{3+4 \sin x}$$
$$\displaystyle =\int_{0}^{\dfrac{\pi}{3}}\dfrac{d (\sin x) }{3+4 \sin x}=\dfrac{1}{4}\left[\log (3+4 \sin x)\right]_{0}^{\dfrac{\pi}{3}}$$
$$\displaystyle =\dfrac{1}{4}\left[\log \left[3+4\left(\sin \dfrac{\pi}{3}\right)\right]-\log (3+4 \sin (0))\right]$$
$$=\dfrac{1}{4}\left (\log\left( 3+\dfrac{4\sqrt{3}}{2}\right)-\log (3)\right)$$
$$=\dfrac{1}{4}(\log (3+2\sqrt{3})-\log 3)$$
$$=\dfrac{1}{4}\log \left ( \dfrac{3+2\sqrt{3}}{3} \right )$$
$$\therefore \displaystyle \int_{0}^{\dfrac{\pi}{3}}\dfrac{\cos x}{3+4 \sin x}dx=\dfrac{1}{4} \log \left(\dfrac{3+2\sqrt{3}}{3}\right)$$
Question 9
1 / -0

Evaluate the integral
$$\displaystyle \int_{1}^{e^{3}}\frac{dx}{x\sqrt{1+\ln x}}$$

A
$$2$$

B
$$2\sqrt{2}$$

C
$$\sqrt{2}$$

D
$$-2$$

Solution

$$\displaystyle \int_{1}^{e^3}\dfrac{dx}{x\sqrt{1+ln x}}=\displaystyle \int_{1}^{e^3}\dfrac{dlnx}{\sqrt{1+ln x}}$$
$$\displaystyle \int_{1}^{e^3}\dfrac{dlnx}{\sqrt{1+ln x}}=2\sqrt{1+lnx}\displaystyle \displaystyle \vert_{1}^{e^3}$$
$$=2\sqrt{1+3}-2\sqrt{1}$$
$$=2\sqrt{4}-2=2(1)$$
$$=2$$
So, $$\displaystyle \int_{1}^{e^3}\dfrac{dx}{x\sqrt{1+ln x}}=2$$
Question 10
1 / -0

Evaluate the integral
$$\displaystyle \int_{0}^{\pi_/{2}}\frac{cosx}{1+sin^{2}x}dx $$

A
$$\pi$$

B
$$\pi/3$$

C
$$\pi/2$$

D
$$\pi/4$$

Solution

$$\displaystyle \int_{0}^{\dfrac{\pi}{2}}\dfrac{cos x}{1+sin^2 x}dx=\displaystyle \int_{0}^{\dfrac{\pi}{2}}\dfrac{d sin x}{1+sin^2 x}$$
$$\displaystyle \int_{0}^{\dfrac{\pi}{2}}\dfrac{d sin x}{1+sin^2 x}=\left [( tan^{-1}(sin x)+ c\right ) \displaystyle ]_{0}^{\dfrac{\pi}{2}}\ (\int\dfrac{1}{1+x^2}=\tan^{-1}x)$$
$$=\left ( tan^{-1}(sin ( \dfrac{\pi}{2})+ c \right )-\left ( tan^{-1}(sin 0)+ c\right )$$
$$=tan^{-1}(1)$$
$$=\dfrac{\pi}{4}$$

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Integrals Test - 14

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Integrals Test - 14

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Question 1

$$\dfrac {1}{4}\log |\sqrt {x^{10} - x^{2}} + x^{2}|$$

$$\dfrac {1}{2}\log |x^{10} - x^{2}|$$

$$-\dfrac {1}{4}\sec^{-1} (x^{4})$$

$$\dfrac {1}{4}\sec^{-1} (x^{4})$$

Solution

Question 2

$$\pi/4$$

$$\pi$$

$$\pi/3$$

$$0$$

Solution

Question 3

$$A=2^{50}B$$

$$A=2^{-50}B$$

$$A=2^{100}B$$

$$A=2^{-100}B$$

Solution

Question 4

$$\cfrac { -\log { x } }{ x+1 } +\log { x } -\log { \left( x+1 \right) } +C $$

$$\cfrac { \log { x } }{ x+1 } +\log { x } -\log { \left( x+1 \right) } +C $$

$$\cfrac { \log { x } }{ x+1 } -\log { x } -\log { \left( x+1 \right) } + C $$

$$\cfrac { -\log { x } }{ x+1 } -\log { x } -\log { \left( x+1 \right) } +C $$

Solution

Question 5

$$\sqrt 2 - 1$$

$$2\sqrt 2 $$

$$2(\sqrt 2 - 1)$$

$$\dfrac{{\sqrt 2 + 1}}{2}$$

Solution

Question 6

$$\dfrac{1}{(n - 1)(1 + ln x)^{n - 1}} + c$$

$$\dfrac{1 - n}{(1 + ln x)^{1- n}} + c$$

$$\dfrac{n + 1}{(1 + ln x)^{n+1}} + c$$

$$-\dfrac{1}{(n - 1)(1 + ln x)^{n-1}} + c$$

Solution

Question 7

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

$$\dfrac{\pi}{3}$$

$$\dfrac{\pi}{4}$$

None of these

Solution

Question 8

$$\displaystyle \log\left(\frac{3+2\sqrt{3}}{3}\right)$$

$$\dfrac{1}{4} \log\left(\displaystyle \frac{3+2\sqrt{3}}{3}\right)$$

$$2\displaystyle \log\left(\frac{3+2\sqrt{3}}{3}\right)$$

$$\dfrac{1}{2}\log\left(\displaystyle \frac{3+2\sqrt{3}}{2}\right)$$

Solution

Question 9

$$2$$

$$2\sqrt{2}$$

$$\sqrt{2}$$

$$-2$$

Solution

Question 10

$$\pi$$

$$\pi/3$$

$$\pi/2$$

$$\pi/4$$

Solution

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