Integrals Test

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Integrals Test - 37

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Question 1
1 / -0

$$\displaystyle \int \dfrac{\{f(x) \cdot \phi' (x) - f'(x) \cdot \phi(x) \}}{f(x) \cdot \phi(x)} \{log \,\phi(x) - log \,f(x) \}dx$$ is equal to

A
$$log \dfrac{\phi(x)}{f(x)} + k$$

B
$$\dfrac{1}{2} \left \{log \dfrac{\phi(x)}{f(x)} \right \}^2 + k$$

C
$$\dfrac{\phi(x)}{f(x)} log \dfrac{\phi(x)}{f(x)} + k$$

D
None of these

Solution

$$I=\displaystyle\int\left\{\dfrac{f(x)\phi'(x)-f'(x).\phi(x)}{f(x).\phi(x)}\right\}\left\{\log\phi(x)-\log f(x)\right\}dx$$
$$=\displaystyle\int \left\{\dfrac{f(x)\phi'(x)-f'(x)\phi(x)}{f(x)\phi(x)}\right\}f(x)\log\dfrac{\phi(x)}{f(x)}dx$$
Let $$\log\dfrac{\phi (x)}{f(x)}=t$$
$$\dfrac{f(x)}{\phi(x)}\left(\dfrac{f(x)\phi'(x)-f'(x)\phi(x)}{f^2(x)}\right)dx=dt$$
$$=\displaystyle\int t dt$$
$$=\dfrac{t^2}{2}+k$$
$$=\dfrac{1}{2}\left[\log \dfrac{(\phi(x)}{f(x)}\right]^2+k$$
$$\therefore I=\dfrac{1}{2}\left[\log \dfrac{\phi(x)}{f(x)}\right]^2+k$$.
Question 2
1 / -0

$$\int \log _ { 10 } x d x =$$

A
$$x \log _ { 10 } x + c$$

B
$$x \left( \log _ { 10 } x + \log _ { 10 } e \right) + c$$

C
$$\log _ { 10 } x + c$$

D
$$x \left( \log _ { 10 } x - \log _ { 10 } e \right) + c$$

Solution

$$\int log_{10}xdx\\=\int(\dfrac{log_ex}{log_e10})dx\\=(\dfrac{1}{log_e10})\int logx\cdot1\cdot dx\\(\dfrac{1}{log_e10})[(logx)\cdot x-\int(\dfrac{1}{x}\cdot x dx)]\\=(\dfrac{d1}{log_e10})(xlogx-x)+c\\=x((\dfrac{log_ex}{log_e10})-(\dfrac{1}{log_e10}))+c\\=x(log_{10}x-log_{10}e)+C$$
Question 3
1 / -0

Evaluate: $$\displaystyle\int \frac { 1 } { ( x + 2 ) \sqrt { x + 1 } } d x $$

A
$$2 \tan ^ { - 1 } ( \sqrt { x + 1 } ) + c$$

B
$$\tan ^ { - 1 } ( \sqrt { x - 1 } ) + c$$

C
$$2 \tanh ^ { - 1 } ( \sqrt { x + 1 } ) - c$$

D
$$\tan ^ { - 1 } ( \sqrt { x + 1 } ) + c$$

Solution

$$I=\displaystyle\int _{ }^{ }{ \frac { 1 }{ { (x+2)\sqrt { x+1 } } } dx } $$

$$ Substitute\, u=\sqrt { x+1 } $$

$$du=\dfrac 1{2\sqrt{x+1}}dx$$

$$=2\displaystyle\int _{ }^{ }{ \dfrac { 1 }{ { 1+{ u^{ 2 } } } } du } $$

$$ =2{ \tan ^{ -1 } }u+C $$

$$ =2{ \tan ^{ -1 } }\left( { \sqrt { x+1 } } \right) +C$$

Hence the correct option is $$A,\,2{ \tan ^{ -1 } }\left( { \sqrt { x+1 } } \right) +C$$
Question 4
1 / -0

Evaluate: $$\int _{ 1/3 }^{ 1 }{ \cfrac { { \left( x-{ x }^{ 3 } \right) }^{ 1/3 } }{ { x }^{ 4 } } } dx=$$

A
$$3$$

B
$$0$$

C
$$6$$

D
$$4$$

Solution

$$I=\displaystyle\int^1_{1/3}\dfrac{x^{1/3}(1-x^2)^{1/3}}{x^4}dx$$
Let $$x=\sin\theta, dx=\cos\theta d\theta$$
$$I=\displaystyle\int^{\pi/2}_{\sin^{-1}1/3}\dfrac{(\sin\theta)^{1/3}(\cos^{2/3}\theta)}{\sin^4\theta}\times \cos\theta d\theta$$
$$=\displaystyle\int \dfrac{\cos^{5/3}\theta}{\sin^{11/3}\theta}d\theta \times \dfrac{\cos^2\theta}{\cos^2\theta}$$
$$=\displaystyle\int \dfrac{\cos^{11/3}\theta}{\sin^{11/3}\theta\cos^2\theta}d\theta$$
$$=\displaystyle\cot^{11/3}\theta .\sec^2\theta d\theta$$
Let $$\tan\theta =t$$
$$\Rightarrow \sec^2\theta d\theta =dt$$
$$=\displaystyle\int t^{-11/3}dt$$
$$=\dfrac{t^{-8/3}}{(-8/3)}=\dfrac{-3}{8}t^{-8/3}$$
$$=\left.\dfrac{-3}{8}(\tan\theta)^{-8/3}\right|^{\pi/2}_{\sin^{-11}1/3}$$
As $$\sin^{-11/3}=\tan^{-1}\dfrac{1}{2\sqrt{2}}$$
$$\Rightarrow I=\dfrac{-3}{8}\left(\dfrac{1}{(\infty)^{8/3}}-\left(\dfrac{1}{2\sqrt{2}}\right)^{-8/3}\right)$$
$$=\dfrac{3}{8}\times (2\sqrt{2})^{8/3}=\dfrac{3}{8}(2^{3/2})^{8/3}$$
$$=\dfrac{3}{8}\times 16=6\Rightarrow (C)$$.
Question 5
1 / -0

Evaluate:
$$\int x ^ { x } \ln ( e ^x ) d x$$

A
$$x ^ { x } + C$$

B
$$x \cdot \ln x + C$$

C
$$( \ln x ) ^ { x } + C$$

D
$$x ^ { \ln x } + C$$

Solution

Let $$x^x=z$$

So, $$x\log x=\log z$$

Now on differentiating

$$(1+\log x)=\cfrac{1}{z}\cfrac{dz}{dx}$$

$$\implies dz=x^x(1+\log x)dx$$

Let $$I=\int x^x(\log e^x)dx$$

$$\implies I=\int x^x(\log e+\log e^x)dx$$

$$\implies I=\int x^x(1+\log e^x)dx$$

Let $$x^x=z$$

We get $$x^x(1+\log x)dx=dz$$

So, $$I=\int dz=z+c=x^x+C$$
Question 6
1 / -0

Evaluate: $$\displaystyle \int _ { 0 } ^ { \pi / 4 } \sec ^ { 7 } {\theta} \sin ^ { 3 } {\theta} {d \theta} =$$

A
$$\dfrac { 1 } { 12 }$$

B
$$\dfrac { 3 } { 12 }$$

C
$$\dfrac { 5 } { 12 }$$

D
$$\dfrac { 7 } { 12 }$$

Solution

$$\displaystyle \int_{0}^{\pi /4} sec^{7}\theta \sin^{3}\theta d\theta $$

$$=\displaystyle \int_{0}^{\pi /4} \frac{\sec^{4}\theta }{\cos^{3}\theta }sin^{3}\theta d\theta $$

$$=\displaystyle\int_{0}^{\pi /4} \tan^{3}\theta \sec^{4}\theta d\theta $$

$$= \displaystyle\int_{0}^{\pi /4} \tan^{3}\theta \sec^{2}\theta .\sec^{2}\theta d\theta $$

$$ =\displaystyle\int_{0}^{\pi /4} tan^{3}\theta (1+tan^{2}\theta )\sec^{2}\theta d\theta $$

$$ u = tan\theta \Rightarrow du = \sec^{2}\theta d\theta $$

$$ =\displaystyle\int_{0}^{1} u^{3}(1+u^{2})du $$

$$ =\displaystyle\int_{0}^{1} (u^{5}+u^{3})du $$

$$ = \dfrac{u^{6}}{6}+\dfrac{u^{4}}{4}|_{0}^{1} $$

$$ = \dfrac{1}{6}+\dfrac{1}{4} = \dfrac{5}{12} $$
Question 7
1 / -0

The value of the defined integral $$\displaystyle \int^{\pi/2}_{0}(\sin x+\cos x)\sqrt {\dfrac {e^{x}}{\sin x}}dx$$ equals

A
$$2\sqrt {e^{\pi/2}}$$

B
$$\sqrt {e^{\pi/2}}$$

C
$$2\sqrt {e^{\pi/2}}.\cos 1$$

D
$$\dfrac {1}{2}e^{\pi/4}$$
Question 8
1 / -0

$$\displaystyle \int \frac { 1 - x ^ { 2 } } { \left( 1 + x ^ { 2 } \right) \sqrt { 1 + x ^ { 4 } } } d x$$ is equal to

A
$$\sqrt { 2 } \sin ^ { - 1 } \left\{ \frac { \sqrt { 2 } x } { x ^ { 2 } + 1 } \right\} + c$$

B
$$\frac { 1 } { \sqrt { 2 } } \sin ^ { - 1 } \left\{ \frac { \sqrt { 2 } x } { x ^ { 2 } + 1 } \right\} + c$$

C
$$\frac { 1 } { 2 } \sin ^ { - 1 } \left\{ \frac { \sqrt { 2 } x } { x ^ { 2 } + 1 } \right\} + c$$

D
$$\frac { 1 } { \sqrt { 2 } } \sin ^ { - 1 } \left\{ \frac { x ^ { 2 } + 1 } { \sqrt { 2 } x } \right\} + c$$
Question 9
1 / -0

$$\displaystyle \int \dfrac { d x } { \sin ^ { 2 } x \cos ^ { 2 } x }$$ equals-

A
$$\tan x - \cot x + C$$

B
$$\tan x + \cot x + c$$

C
$$\tan x \cot x + c$$

D
$$\tan x - \cot 2 x + c$$

Solution

$$I=\int \dfrac{dx}{\sin^{2}x\,\cos^{2}x} $$

$$I= \int \dfrac{\sin^{2}x+cos^{2}x}{\sin^{2}x\,cos^{2}x}dx $$

$$I= \int (\dfrac{1}{cos^{2}x}+\dfrac{1}{sin^{2}x})dx = \int (\sec^{2}x+cosec^{2}x)dx $$

$$I=\ tanx-\cot x+c $$
Question 10
1 / -0

$$\int { { e }^{ x^{ 3 } }+{ x }^{ 2-1 }(3{ x }^{ 4 }+{ 2x }^{ 3 }+{ 2x }^{ 2 }\quad x=h(x)+c } $$ then the value of $$h(1)h(-1)$$.

A
1

B
-1

C
2

D
-2

Solution

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Re-Attempt Weekly Quiz Competition

Integrals Test - 37

Result

Integrals Test - 37

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SHARING IS CARING

Question 1

$$log \dfrac{\phi(x)}{f(x)} + k$$

$$\dfrac{1}{2} \left \{log \dfrac{\phi(x)}{f(x)} \right \}^2 + k$$

$$\dfrac{\phi(x)}{f(x)} log \dfrac{\phi(x)}{f(x)} + k$$

None of these

Solution

Question 2

$$x \log _ { 10 } x + c$$

$$x \left( \log _ { 10 } x + \log _ { 10 } e \right) + c$$

$$\log _ { 10 } x + c$$

$$x \left( \log _ { 10 } x - \log _ { 10 } e \right) + c$$

Solution

Question 3

$$2 \tan ^ { - 1 } ( \sqrt { x + 1 } ) + c$$

$$\tan ^ { - 1 } ( \sqrt { x - 1 } ) + c$$

$$2 \tanh ^ { - 1 } ( \sqrt { x + 1 } ) - c$$

$$\tan ^ { - 1 } ( \sqrt { x + 1 } ) + c$$

Solution

Question 4

$$3$$

$$0$$

$$6$$

$$4$$

Solution

Question 5

$$x ^ { x } + C$$

$$x \cdot \ln x + C$$

$$( \ln x ) ^ { x } + C$$

$$x ^ { \ln x } + C$$

Solution

Question 6

$$\dfrac { 1 } { 12 }$$

$$\dfrac { 3 } { 12 }$$

$$\dfrac { 5 } { 12 }$$

$$\dfrac { 7 } { 12 }$$

Solution

Question 7

$$2\sqrt {e^{\pi/2}}$$

$$\sqrt {e^{\pi/2}}$$

$$2\sqrt {e^{\pi/2}}.\cos 1$$

$$\dfrac {1}{2}e^{\pi/4}$$

Question 8

$$\sqrt { 2 } \sin ^ { - 1 } \left\{ \frac { \sqrt { 2 } x } { x ^ { 2 } + 1 } \right\} + c$$

$$\frac { 1 } { \sqrt { 2 } } \sin ^ { - 1 } \left\{ \frac { \sqrt { 2 } x } { x ^ { 2 } + 1 } \right\} + c$$

$$\frac { 1 } { 2 } \sin ^ { - 1 } \left\{ \frac { \sqrt { 2 } x } { x ^ { 2 } + 1 } \right\} + c$$

$$\frac { 1 } { \sqrt { 2 } } \sin ^ { - 1 } \left\{ \frac { x ^ { 2 } + 1 } { \sqrt { 2 } x } \right\} + c$$

Question 9

$$\tan x - \cot x + C$$

$$\tan x + \cot x + c$$

$$\tan x \cot x + c$$

$$\tan x - \cot 2 x + c$$

Solution

Question 10

1

-1

2

-2

Solution

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