QBManager

Integral Calculus

86414 The integral $\int_{0}^{π} \frac{x}{2 cosec x - \sin x} d x$ is equal to

1

\frac{π}{4}

2

\frac{π}{2}

3

\frac{π^{2}}{2}

4

\frac{π^{2}}{4}

Explanation:

(D) : Let, $I = \int_{0}^{π} \frac{x}{2 cosec x - \sin x} d x$
$I = \int_{0}^{π} \frac{π - x}{2 cosec (π - x) - \sin (π - x)} d x$
$I = \int_{0}^{π} \frac{π - x}{2 cosec x - \sin x} \cdot dx$
From equation (i) and equation (ii)
$2 I = π \int_{0}^{π} \frac{\sin x}{2 - \sin^{2} x} \cdot d x$
$2 I = π \int_{0}^{π} \frac{\sin x}{1 + \cos^{2} x} \cdot d x$
Let, $\cos x = t$
$- \sin x . d x = d t$
When, $x = 0, t = 1$
$x = π, t = - 1$
$2 I = π \int_{1}^{- 1} \frac{- dt}{1 + t^{2}} = π \int_{- 1}^{1} \frac{1}{1 + t^{2}} \cdot dt$
$2 I = π {[\tan^{- 1} t]}_{- 1}^{1} = π [\frac{π}{4} - (\frac{- π}{4})]$
$2 I = \frac{π^{2}}{2}, I = \frac{π^{2}}{4}$

J&K CET-2015

Integral Calculus

86425 $\int_{0}^{1} x (1 - x)^{12} d x$ is equal to

1

\frac{1}{132}

2

\frac{1}{156}

3

\frac{1}{182}

4 None of the above

Explanation:

(C) : Given,
$\int_{0}^{1} x (1 - x)^{12} dx$
$= {[\frac{x (1 - x)^{13}}{- 13}]}_{0}^{1} + \int_{0}^{1} \frac{(1 - x)^{13}}{13} dx$
$= {[\frac{x (1 - x)^{13}}{- 13}]}_{0}^{1} + {[\frac{(1 - x)^{14}}{- 14 \times 13}]}_{0}^{1}$
$= [0 - 0] + [0 - \frac{1}{- 14 \times 13}] = \frac{1}{182}$

Manipal-2017

Integral Calculus

86427 If $a < 0 < b$ , then $\int_{a}^{b} \frac{| x |}{x} d x$

1

a - b

2

b - a

3

a + b

4

- a - b

Explanation:

(C) : Given,
If $a < 0 < b$ then,
$\int_{a}^{b} \frac{| x |}{x} d x = \int_{a}^{0} [\frac{| x |}{x}] d x + \int_{0}^{b} [\frac{| x |}{x}] d x = \int_{a}^{0} - 1 d x + \int_{0}^{b} 1 . d x$
$= [- x]_{a}^{0} + [x]_{0}^{b} . = (- 1) (0 - a) + b = a + b$

Rajasthan PET-2011

Integral Calculus

86378 $\int \frac{e^{x^{2}} (2 x + x^{3})}{{(3 + x^{2})}^{2}} dx$ is equal to :

1

\frac{e^{x^{2}}}{(3 + x^{2})} + k

2

\frac{1}{2} \frac{e^{x^{2}}}{{(3 + x^{2})}^{2}} + k

3

\frac{1}{4} \frac{e^{x^{2}}}{{(3 + x^{2})}^{2}} + k

4

\frac{1}{2} \frac{e^{x^{2}}}{(3 + x^{2})} + k

Explanation:

(D) : Given,
$\int \frac{e^{x^{2}} (2 x + x^{3})}{{(3 + x^{2})}^{2}} dx$
Put,
$x^{2} = t \Rightarrow 2 xdx = dt$
$I = \int \frac{e^{x^{2}} (2 + x^{2}) x d x}{{(3 + x^{2})}^{2}} = \frac{1}{2} \int e^{t} \frac{(2 + t)}{(3 + t)^{2}} d t$
$= \frac{1}{2} \int \frac{e^{t} (3 + t - 1)}{(3 + t)^{2}} dt = \frac{1}{2} \int e^{t} [\frac{1}{3 + t} - \frac{1}{(3 + t)^{2}}] dt$
$= \frac{1}{2} e^{t} \cdot \frac{1}{3 + t} + k [∵ \frac{d}{dt} (\frac{1}{3 + t}) = \frac{- 1}{(3 + t)^{2}}]$
$= \frac{1}{2} \frac{e^{x^{2}}}{(3 + x^{2})} + k$

BITSAT-2018

Integral Calculus

86414 The integral $\int_{0}^{π} \frac{x}{2 cosec x - \sin x} d x$ is equal to

1

\frac{π}{4}

2

\frac{π}{2}

3

\frac{π^{2}}{2}

4

\frac{π^{2}}{4}

Explanation:

(D) : Let, $I = \int_{0}^{π} \frac{x}{2 cosec x - \sin x} d x$
$I = \int_{0}^{π} \frac{π - x}{2 cosec (π - x) - \sin (π - x)} d x$
$I = \int_{0}^{π} \frac{π - x}{2 cosec x - \sin x} \cdot dx$
From equation (i) and equation (ii)
$2 I = π \int_{0}^{π} \frac{\sin x}{2 - \sin^{2} x} \cdot d x$
$2 I = π \int_{0}^{π} \frac{\sin x}{1 + \cos^{2} x} \cdot d x$
Let, $\cos x = t$
$- \sin x . d x = d t$
When, $x = 0, t = 1$
$x = π, t = - 1$
$2 I = π \int_{1}^{- 1} \frac{- dt}{1 + t^{2}} = π \int_{- 1}^{1} \frac{1}{1 + t^{2}} \cdot dt$
$2 I = π {[\tan^{- 1} t]}_{- 1}^{1} = π [\frac{π}{4} - (\frac{- π}{4})]$
$2 I = \frac{π^{2}}{2}, I = \frac{π^{2}}{4}$

J&K CET-2015

Integral Calculus

86425 $\int_{0}^{1} x (1 - x)^{12} d x$ is equal to

1

\frac{1}{132}

2

\frac{1}{156}

3

\frac{1}{182}

4 None of the above

Explanation:

(C) : Given,
$\int_{0}^{1} x (1 - x)^{12} dx$
$= {[\frac{x (1 - x)^{13}}{- 13}]}_{0}^{1} + \int_{0}^{1} \frac{(1 - x)^{13}}{13} dx$
$= {[\frac{x (1 - x)^{13}}{- 13}]}_{0}^{1} + {[\frac{(1 - x)^{14}}{- 14 \times 13}]}_{0}^{1}$
$= [0 - 0] + [0 - \frac{1}{- 14 \times 13}] = \frac{1}{182}$

Manipal-2017

Integral Calculus

86427 If $a < 0 < b$ , then $\int_{a}^{b} \frac{| x |}{x} d x$

1

a - b

2

b - a

3

a + b

4

- a - b

Explanation:

(C) : Given,
If $a < 0 < b$ then,
$\int_{a}^{b} \frac{| x |}{x} d x = \int_{a}^{0} [\frac{| x |}{x}] d x + \int_{0}^{b} [\frac{| x |}{x}] d x = \int_{a}^{0} - 1 d x + \int_{0}^{b} 1 . d x$
$= [- x]_{a}^{0} + [x]_{0}^{b} . = (- 1) (0 - a) + b = a + b$

Rajasthan PET-2011

Integral Calculus

86378 $\int \frac{e^{x^{2}} (2 x + x^{3})}{{(3 + x^{2})}^{2}} dx$ is equal to :

1

\frac{e^{x^{2}}}{(3 + x^{2})} + k

2

\frac{1}{2} \frac{e^{x^{2}}}{{(3 + x^{2})}^{2}} + k

3

\frac{1}{4} \frac{e^{x^{2}}}{{(3 + x^{2})}^{2}} + k

4

\frac{1}{2} \frac{e^{x^{2}}}{(3 + x^{2})} + k

Explanation:

(D) : Given,
$\int \frac{e^{x^{2}} (2 x + x^{3})}{{(3 + x^{2})}^{2}} dx$
Put,
$x^{2} = t \Rightarrow 2 xdx = dt$
$I = \int \frac{e^{x^{2}} (2 + x^{2}) x d x}{{(3 + x^{2})}^{2}} = \frac{1}{2} \int e^{t} \frac{(2 + t)}{(3 + t)^{2}} d t$
$= \frac{1}{2} \int \frac{e^{t} (3 + t - 1)}{(3 + t)^{2}} dt = \frac{1}{2} \int e^{t} [\frac{1}{3 + t} - \frac{1}{(3 + t)^{2}}] dt$
$= \frac{1}{2} e^{t} \cdot \frac{1}{3 + t} + k [∵ \frac{d}{dt} (\frac{1}{3 + t}) = \frac{- 1}{(3 + t)^{2}}]$
$= \frac{1}{2} \frac{e^{x^{2}}}{(3 + x^{2})} + k$

BITSAT-2018

Integral Calculus

86414 The integral $\int_{0}^{π} \frac{x}{2 cosec x - \sin x} d x$ is equal to

1

\frac{π}{4}

2

\frac{π}{2}

3

\frac{π^{2}}{2}

4

\frac{π^{2}}{4}

Explanation:

(D) : Let, $I = \int_{0}^{π} \frac{x}{2 cosec x - \sin x} d x$
$I = \int_{0}^{π} \frac{π - x}{2 cosec (π - x) - \sin (π - x)} d x$
$I = \int_{0}^{π} \frac{π - x}{2 cosec x - \sin x} \cdot dx$
From equation (i) and equation (ii)
$2 I = π \int_{0}^{π} \frac{\sin x}{2 - \sin^{2} x} \cdot d x$
$2 I = π \int_{0}^{π} \frac{\sin x}{1 + \cos^{2} x} \cdot d x$
Let, $\cos x = t$
$- \sin x . d x = d t$
When, $x = 0, t = 1$
$x = π, t = - 1$
$2 I = π \int_{1}^{- 1} \frac{- dt}{1 + t^{2}} = π \int_{- 1}^{1} \frac{1}{1 + t^{2}} \cdot dt$
$2 I = π {[\tan^{- 1} t]}_{- 1}^{1} = π [\frac{π}{4} - (\frac{- π}{4})]$
$2 I = \frac{π^{2}}{2}, I = \frac{π^{2}}{4}$

J&K CET-2015

Integral Calculus

86425 $\int_{0}^{1} x (1 - x)^{12} d x$ is equal to

1

\frac{1}{132}

2

\frac{1}{156}

3

\frac{1}{182}

4 None of the above

Explanation:

(C) : Given,
$\int_{0}^{1} x (1 - x)^{12} dx$
$= {[\frac{x (1 - x)^{13}}{- 13}]}_{0}^{1} + \int_{0}^{1} \frac{(1 - x)^{13}}{13} dx$
$= {[\frac{x (1 - x)^{13}}{- 13}]}_{0}^{1} + {[\frac{(1 - x)^{14}}{- 14 \times 13}]}_{0}^{1}$
$= [0 - 0] + [0 - \frac{1}{- 14 \times 13}] = \frac{1}{182}$

Manipal-2017

Integral Calculus

86427 If $a < 0 < b$ , then $\int_{a}^{b} \frac{| x |}{x} d x$

1

a - b

2

b - a

3

a + b

4

- a - b

Explanation:

(C) : Given,
If $a < 0 < b$ then,
$\int_{a}^{b} \frac{| x |}{x} d x = \int_{a}^{0} [\frac{| x |}{x}] d x + \int_{0}^{b} [\frac{| x |}{x}] d x = \int_{a}^{0} - 1 d x + \int_{0}^{b} 1 . d x$
$= [- x]_{a}^{0} + [x]_{0}^{b} . = (- 1) (0 - a) + b = a + b$

Rajasthan PET-2011

Integral Calculus

86378 $\int \frac{e^{x^{2}} (2 x + x^{3})}{{(3 + x^{2})}^{2}} dx$ is equal to :

1

\frac{e^{x^{2}}}{(3 + x^{2})} + k

2

\frac{1}{2} \frac{e^{x^{2}}}{{(3 + x^{2})}^{2}} + k

3

\frac{1}{4} \frac{e^{x^{2}}}{{(3 + x^{2})}^{2}} + k

4

\frac{1}{2} \frac{e^{x^{2}}}{(3 + x^{2})} + k

Explanation:

(D) : Given,
$\int \frac{e^{x^{2}} (2 x + x^{3})}{{(3 + x^{2})}^{2}} dx$
Put,
$x^{2} = t \Rightarrow 2 xdx = dt$
$I = \int \frac{e^{x^{2}} (2 + x^{2}) x d x}{{(3 + x^{2})}^{2}} = \frac{1}{2} \int e^{t} \frac{(2 + t)}{(3 + t)^{2}} d t$
$= \frac{1}{2} \int \frac{e^{t} (3 + t - 1)}{(3 + t)^{2}} dt = \frac{1}{2} \int e^{t} [\frac{1}{3 + t} - \frac{1}{(3 + t)^{2}}] dt$
$= \frac{1}{2} e^{t} \cdot \frac{1}{3 + t} + k [∵ \frac{d}{dt} (\frac{1}{3 + t}) = \frac{- 1}{(3 + t)^{2}}]$
$= \frac{1}{2} \frac{e^{x^{2}}}{(3 + x^{2})} + k$

BITSAT-2018

Integral Calculus

86414 The integral $\int_{0}^{π} \frac{x}{2 cosec x - \sin x} d x$ is equal to

1

\frac{π}{4}

2

\frac{π}{2}

3

\frac{π^{2}}{2}

4

\frac{π^{2}}{4}

Explanation:

(D) : Let, $I = \int_{0}^{π} \frac{x}{2 cosec x - \sin x} d x$
$I = \int_{0}^{π} \frac{π - x}{2 cosec (π - x) - \sin (π - x)} d x$
$I = \int_{0}^{π} \frac{π - x}{2 cosec x - \sin x} \cdot dx$
From equation (i) and equation (ii)
$2 I = π \int_{0}^{π} \frac{\sin x}{2 - \sin^{2} x} \cdot d x$
$2 I = π \int_{0}^{π} \frac{\sin x}{1 + \cos^{2} x} \cdot d x$
Let, $\cos x = t$
$- \sin x . d x = d t$
When, $x = 0, t = 1$
$x = π, t = - 1$
$2 I = π \int_{1}^{- 1} \frac{- dt}{1 + t^{2}} = π \int_{- 1}^{1} \frac{1}{1 + t^{2}} \cdot dt$
$2 I = π {[\tan^{- 1} t]}_{- 1}^{1} = π [\frac{π}{4} - (\frac{- π}{4})]$
$2 I = \frac{π^{2}}{2}, I = \frac{π^{2}}{4}$

J&K CET-2015

Integral Calculus

86425 $\int_{0}^{1} x (1 - x)^{12} d x$ is equal to

1

\frac{1}{132}

2

\frac{1}{156}

3

\frac{1}{182}

4 None of the above

Explanation:

(C) : Given,
$\int_{0}^{1} x (1 - x)^{12} dx$
$= {[\frac{x (1 - x)^{13}}{- 13}]}_{0}^{1} + \int_{0}^{1} \frac{(1 - x)^{13}}{13} dx$
$= {[\frac{x (1 - x)^{13}}{- 13}]}_{0}^{1} + {[\frac{(1 - x)^{14}}{- 14 \times 13}]}_{0}^{1}$
$= [0 - 0] + [0 - \frac{1}{- 14 \times 13}] = \frac{1}{182}$

Manipal-2017

Integral Calculus

86427 If $a < 0 < b$ , then $\int_{a}^{b} \frac{| x |}{x} d x$

1

a - b

2

b - a

3

a + b

4

- a - b

Explanation:

(C) : Given,
If $a < 0 < b$ then,
$\int_{a}^{b} \frac{| x |}{x} d x = \int_{a}^{0} [\frac{| x |}{x}] d x + \int_{0}^{b} [\frac{| x |}{x}] d x = \int_{a}^{0} - 1 d x + \int_{0}^{b} 1 . d x$
$= [- x]_{a}^{0} + [x]_{0}^{b} . = (- 1) (0 - a) + b = a + b$

Rajasthan PET-2011

Integral Calculus

86378 $\int \frac{e^{x^{2}} (2 x + x^{3})}{{(3 + x^{2})}^{2}} dx$ is equal to :

1

\frac{e^{x^{2}}}{(3 + x^{2})} + k

2

\frac{1}{2} \frac{e^{x^{2}}}{{(3 + x^{2})}^{2}} + k

3

\frac{1}{4} \frac{e^{x^{2}}}{{(3 + x^{2})}^{2}} + k

4

\frac{1}{2} \frac{e^{x^{2}}}{(3 + x^{2})} + k

Explanation:

(D) : Given,
$\int \frac{e^{x^{2}} (2 x + x^{3})}{{(3 + x^{2})}^{2}} dx$
Put,
$x^{2} = t \Rightarrow 2 xdx = dt$
$I = \int \frac{e^{x^{2}} (2 + x^{2}) x d x}{{(3 + x^{2})}^{2}} = \frac{1}{2} \int e^{t} \frac{(2 + t)}{(3 + t)^{2}} d t$
$= \frac{1}{2} \int \frac{e^{t} (3 + t - 1)}{(3 + t)^{2}} dt = \frac{1}{2} \int e^{t} [\frac{1}{3 + t} - \frac{1}{(3 + t)^{2}}] dt$
$= \frac{1}{2} e^{t} \cdot \frac{1}{3 + t} + k [∵ \frac{d}{dt} (\frac{1}{3 + t}) = \frac{- 1}{(3 + t)^{2}}]$
$= \frac{1}{2} \frac{e^{x^{2}}}{(3 + x^{2})} + k$

BITSAT-2018

Question Bank Series

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