QBManager

Integral Calculus

86526 $\int_{0}^{\frac{π}{2}} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} d x$ is equal to

1 0

2 1

3

\frac{π}{4}

4

\frac{π}{2}

Explanation:

(C) : $I = \int_{0}^{\frac{π}{2}} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} d x$
Apply property,
$I = \int_{0}^{\frac{π}{2}} \frac{\sqrt{\sin (\frac{π}{2} - x)}}{\sqrt{\sin (\frac{π}{2} - x)} + \sqrt{\cos (\frac{π}{2} - x)}} d x$
$I = \int_{0}^{\frac{π}{2}} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} d x$
On adding equation (i) \& (ii), we get -
$2 I = \int_{0}^{\frac{π}{2}} (\frac{\sqrt{\sin x} + \sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}}) d x$
$= \int_{0}^{\frac{π}{2}} 1 dx$
$I = \frac{1}{2} [x]_{0}^{π / 2} = [\frac{π}{2}] = \frac{π}{4}$

SRM JEE-2016

Integral Calculus

86527 $\int_{0}^{\infty} \frac{d x}{(x^{2} + a^{2}) (x^{2} + b^{2})}$ is

1

\frac{π a b}{a + b}

2

\frac{π}{2 (a + b)}

3

\frac{π}{2 a b (a + b)}

4

\frac{π (a + b)}{2 a b}

Explanation:

(C) : Given,
$\int_{0}^{\infty} \frac{d x}{(x^{2} + a^{2}) (x^{2} + b^{2})} = \frac{1}{b^{2} - a^{2}} \int_{0}^{\infty} \frac{(x^{2} + b^{2}) - (x^{2} + a^{2})}{(x^{2} + a^{2}) (x^{2} + b^{2})} d x$
$= \frac{1}{b^{2} - a^{2}} \int_{0}^{\infty} [\frac{1}{x^{2} + a^{2}} - \frac{1}{x^{2} + b^{2}}] dx$
$= \frac{1}{b^{2} - a^{2}} {[\frac{1}{a} \tan^{- 1} \frac{x}{a} - \frac{1}{b} \tan^{- 1} \frac{x}{b}]}_{0}^{\infty}$
$= \frac{1}{b^{2} - a^{2}} [\frac{π}{2 a} - \frac{π}{2 b}] = \frac{π}{2 a b (a + b)}$

BITSAT-2020

Integral Calculus

86528 The value of definite integral $\int_{0}^{\frac{π}{2}} \log (\tan x) d x$ is

1 0

2

\frac{π}{4}

3

\frac{π}{2}

4

π

Explanation:

(A) : Let,
$= \int_{0}^{\frac{π}{2}} \log (\tan x) d x$
$= \int_{0}^{\frac{π}{2}} \log {\tan (\frac{π}{2} - x)} d x$
$= \int_{0}^{\frac{π}{2}} \log (\cot x) d x$
On adding equation (i) and (ii), we get -
$∴ 2 I = \int_{0}^{\frac{π}{2}} \log (\tan x) d x + \int_{0}^{\frac{π}{2}} \log (\cot x) d x$
$2 I = \int_{0}^{\frac{π}{2}} [\log \tan x + \log \cot x] d x$
$2 I = \int_{0}^{\frac{π}{2}} \log (\tan x \cdot \cot x) d x \Rightarrow 2 I = \int_{0}^{\frac{π}{2}} \log (1) d x$
$I = 0$

BITSAT-2019

Integral Calculus

86529 The value of $I = \int_{0}^{1} x | x - \frac{1}{2} | d x$ is

1

\frac{1}{3}

2

\frac{1}{4}

3

\frac{1}{8}

4 None of these

Explanation:

(C) : Given,
$I = \int_{0}^{1} x | x - \frac{1}{2} | d x$
$= - \int_{0}^{1 / 2} x (x - \frac{1}{2}) d x + \int_{1 / 2}^{1} x (x - \frac{1}{2}) d x$
$= \int_{0}^{1 / 2} (\frac{x}{2} - x^{2}) d x + \int_{1 / 2}^{1} (x^{2} - \frac{x}{2}) d x$
$= {[\frac{x^{2}}{4} - \frac{x^{3}}{3}]}_{0}^{1 / 2} + {[\frac{x^{3}}{3} - \frac{x^{2}}{4}]}_{1 / 2}^{1}$
$= (\frac{1}{16} - \frac{1}{24}) + [(\frac{1}{3} - \frac{1}{4}) - \frac{1}{24} + \frac{1}{16})]$
$= (\frac{6 - 4}{96}) + (\frac{32 - 24 - 4 + 6}{96}) = \frac{12}{96} = \frac{1}{8}$

VITEEE-2012

Integral Calculus

86526 $\int_{0}^{\frac{π}{2}} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} d x$ is equal to

1 0

2 1

3

\frac{π}{4}

4

\frac{π}{2}

Explanation:

(C) : $I = \int_{0}^{\frac{π}{2}} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} d x$
Apply property,
$I = \int_{0}^{\frac{π}{2}} \frac{\sqrt{\sin (\frac{π}{2} - x)}}{\sqrt{\sin (\frac{π}{2} - x)} + \sqrt{\cos (\frac{π}{2} - x)}} d x$
$I = \int_{0}^{\frac{π}{2}} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} d x$
On adding equation (i) \& (ii), we get -
$2 I = \int_{0}^{\frac{π}{2}} (\frac{\sqrt{\sin x} + \sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}}) d x$
$= \int_{0}^{\frac{π}{2}} 1 dx$
$I = \frac{1}{2} [x]_{0}^{π / 2} = [\frac{π}{2}] = \frac{π}{4}$

SRM JEE-2016

Integral Calculus

86527 $\int_{0}^{\infty} \frac{d x}{(x^{2} + a^{2}) (x^{2} + b^{2})}$ is

1

\frac{π a b}{a + b}

2

\frac{π}{2 (a + b)}

3

\frac{π}{2 a b (a + b)}

4

\frac{π (a + b)}{2 a b}

Explanation:

(C) : Given,
$\int_{0}^{\infty} \frac{d x}{(x^{2} + a^{2}) (x^{2} + b^{2})} = \frac{1}{b^{2} - a^{2}} \int_{0}^{\infty} \frac{(x^{2} + b^{2}) - (x^{2} + a^{2})}{(x^{2} + a^{2}) (x^{2} + b^{2})} d x$
$= \frac{1}{b^{2} - a^{2}} \int_{0}^{\infty} [\frac{1}{x^{2} + a^{2}} - \frac{1}{x^{2} + b^{2}}] dx$
$= \frac{1}{b^{2} - a^{2}} {[\frac{1}{a} \tan^{- 1} \frac{x}{a} - \frac{1}{b} \tan^{- 1} \frac{x}{b}]}_{0}^{\infty}$
$= \frac{1}{b^{2} - a^{2}} [\frac{π}{2 a} - \frac{π}{2 b}] = \frac{π}{2 a b (a + b)}$

BITSAT-2020

Integral Calculus

86528 The value of definite integral $\int_{0}^{\frac{π}{2}} \log (\tan x) d x$ is

1 0

2

\frac{π}{4}

3

\frac{π}{2}

4

π

Explanation:

(A) : Let,
$= \int_{0}^{\frac{π}{2}} \log (\tan x) d x$
$= \int_{0}^{\frac{π}{2}} \log {\tan (\frac{π}{2} - x)} d x$
$= \int_{0}^{\frac{π}{2}} \log (\cot x) d x$
On adding equation (i) and (ii), we get -
$∴ 2 I = \int_{0}^{\frac{π}{2}} \log (\tan x) d x + \int_{0}^{\frac{π}{2}} \log (\cot x) d x$
$2 I = \int_{0}^{\frac{π}{2}} [\log \tan x + \log \cot x] d x$
$2 I = \int_{0}^{\frac{π}{2}} \log (\tan x \cdot \cot x) d x \Rightarrow 2 I = \int_{0}^{\frac{π}{2}} \log (1) d x$
$I = 0$

BITSAT-2019

Integral Calculus

86529 The value of $I = \int_{0}^{1} x | x - \frac{1}{2} | d x$ is

1

\frac{1}{3}

2

\frac{1}{4}

3

\frac{1}{8}

4 None of these

Explanation:

(C) : Given,
$I = \int_{0}^{1} x | x - \frac{1}{2} | d x$
$= - \int_{0}^{1 / 2} x (x - \frac{1}{2}) d x + \int_{1 / 2}^{1} x (x - \frac{1}{2}) d x$
$= \int_{0}^{1 / 2} (\frac{x}{2} - x^{2}) d x + \int_{1 / 2}^{1} (x^{2} - \frac{x}{2}) d x$
$= {[\frac{x^{2}}{4} - \frac{x^{3}}{3}]}_{0}^{1 / 2} + {[\frac{x^{3}}{3} - \frac{x^{2}}{4}]}_{1 / 2}^{1}$
$= (\frac{1}{16} - \frac{1}{24}) + [(\frac{1}{3} - \frac{1}{4}) - \frac{1}{24} + \frac{1}{16})]$
$= (\frac{6 - 4}{96}) + (\frac{32 - 24 - 4 + 6}{96}) = \frac{12}{96} = \frac{1}{8}$

VITEEE-2012

Integral Calculus

86526 $\int_{0}^{\frac{π}{2}} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} d x$ is equal to

1 0

2 1

3

\frac{π}{4}

4

\frac{π}{2}

Explanation:

(C) : $I = \int_{0}^{\frac{π}{2}} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} d x$
Apply property,
$I = \int_{0}^{\frac{π}{2}} \frac{\sqrt{\sin (\frac{π}{2} - x)}}{\sqrt{\sin (\frac{π}{2} - x)} + \sqrt{\cos (\frac{π}{2} - x)}} d x$
$I = \int_{0}^{\frac{π}{2}} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} d x$
On adding equation (i) \& (ii), we get -
$2 I = \int_{0}^{\frac{π}{2}} (\frac{\sqrt{\sin x} + \sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}}) d x$
$= \int_{0}^{\frac{π}{2}} 1 dx$
$I = \frac{1}{2} [x]_{0}^{π / 2} = [\frac{π}{2}] = \frac{π}{4}$

SRM JEE-2016

Integral Calculus

86527 $\int_{0}^{\infty} \frac{d x}{(x^{2} + a^{2}) (x^{2} + b^{2})}$ is

1

\frac{π a b}{a + b}

2

\frac{π}{2 (a + b)}

3

\frac{π}{2 a b (a + b)}

4

\frac{π (a + b)}{2 a b}

Explanation:

(C) : Given,
$\int_{0}^{\infty} \frac{d x}{(x^{2} + a^{2}) (x^{2} + b^{2})} = \frac{1}{b^{2} - a^{2}} \int_{0}^{\infty} \frac{(x^{2} + b^{2}) - (x^{2} + a^{2})}{(x^{2} + a^{2}) (x^{2} + b^{2})} d x$
$= \frac{1}{b^{2} - a^{2}} \int_{0}^{\infty} [\frac{1}{x^{2} + a^{2}} - \frac{1}{x^{2} + b^{2}}] dx$
$= \frac{1}{b^{2} - a^{2}} {[\frac{1}{a} \tan^{- 1} \frac{x}{a} - \frac{1}{b} \tan^{- 1} \frac{x}{b}]}_{0}^{\infty}$
$= \frac{1}{b^{2} - a^{2}} [\frac{π}{2 a} - \frac{π}{2 b}] = \frac{π}{2 a b (a + b)}$

BITSAT-2020

Integral Calculus

86528 The value of definite integral $\int_{0}^{\frac{π}{2}} \log (\tan x) d x$ is

1 0

2

\frac{π}{4}

3

\frac{π}{2}

4

π

Explanation:

(A) : Let,
$= \int_{0}^{\frac{π}{2}} \log (\tan x) d x$
$= \int_{0}^{\frac{π}{2}} \log {\tan (\frac{π}{2} - x)} d x$
$= \int_{0}^{\frac{π}{2}} \log (\cot x) d x$
On adding equation (i) and (ii), we get -
$∴ 2 I = \int_{0}^{\frac{π}{2}} \log (\tan x) d x + \int_{0}^{\frac{π}{2}} \log (\cot x) d x$
$2 I = \int_{0}^{\frac{π}{2}} [\log \tan x + \log \cot x] d x$
$2 I = \int_{0}^{\frac{π}{2}} \log (\tan x \cdot \cot x) d x \Rightarrow 2 I = \int_{0}^{\frac{π}{2}} \log (1) d x$
$I = 0$

BITSAT-2019

Integral Calculus

86529 The value of $I = \int_{0}^{1} x | x - \frac{1}{2} | d x$ is

1

\frac{1}{3}

2

\frac{1}{4}

3

\frac{1}{8}

4 None of these

Explanation:

(C) : Given,
$I = \int_{0}^{1} x | x - \frac{1}{2} | d x$
$= - \int_{0}^{1 / 2} x (x - \frac{1}{2}) d x + \int_{1 / 2}^{1} x (x - \frac{1}{2}) d x$
$= \int_{0}^{1 / 2} (\frac{x}{2} - x^{2}) d x + \int_{1 / 2}^{1} (x^{2} - \frac{x}{2}) d x$
$= {[\frac{x^{2}}{4} - \frac{x^{3}}{3}]}_{0}^{1 / 2} + {[\frac{x^{3}}{3} - \frac{x^{2}}{4}]}_{1 / 2}^{1}$
$= (\frac{1}{16} - \frac{1}{24}) + [(\frac{1}{3} - \frac{1}{4}) - \frac{1}{24} + \frac{1}{16})]$
$= (\frac{6 - 4}{96}) + (\frac{32 - 24 - 4 + 6}{96}) = \frac{12}{96} = \frac{1}{8}$

VITEEE-2012

Integral Calculus

86526 $\int_{0}^{\frac{π}{2}} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} d x$ is equal to

1 0

2 1

3

\frac{π}{4}

4

\frac{π}{2}

Explanation:

(C) : $I = \int_{0}^{\frac{π}{2}} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} d x$
Apply property,
$I = \int_{0}^{\frac{π}{2}} \frac{\sqrt{\sin (\frac{π}{2} - x)}}{\sqrt{\sin (\frac{π}{2} - x)} + \sqrt{\cos (\frac{π}{2} - x)}} d x$
$I = \int_{0}^{\frac{π}{2}} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} d x$
On adding equation (i) \& (ii), we get -
$2 I = \int_{0}^{\frac{π}{2}} (\frac{\sqrt{\sin x} + \sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}}) d x$
$= \int_{0}^{\frac{π}{2}} 1 dx$
$I = \frac{1}{2} [x]_{0}^{π / 2} = [\frac{π}{2}] = \frac{π}{4}$

SRM JEE-2016

Integral Calculus

86527 $\int_{0}^{\infty} \frac{d x}{(x^{2} + a^{2}) (x^{2} + b^{2})}$ is

1

\frac{π a b}{a + b}

2

\frac{π}{2 (a + b)}

3

\frac{π}{2 a b (a + b)}

4

\frac{π (a + b)}{2 a b}

Explanation:

(C) : Given,
$\int_{0}^{\infty} \frac{d x}{(x^{2} + a^{2}) (x^{2} + b^{2})} = \frac{1}{b^{2} - a^{2}} \int_{0}^{\infty} \frac{(x^{2} + b^{2}) - (x^{2} + a^{2})}{(x^{2} + a^{2}) (x^{2} + b^{2})} d x$
$= \frac{1}{b^{2} - a^{2}} \int_{0}^{\infty} [\frac{1}{x^{2} + a^{2}} - \frac{1}{x^{2} + b^{2}}] dx$
$= \frac{1}{b^{2} - a^{2}} {[\frac{1}{a} \tan^{- 1} \frac{x}{a} - \frac{1}{b} \tan^{- 1} \frac{x}{b}]}_{0}^{\infty}$
$= \frac{1}{b^{2} - a^{2}} [\frac{π}{2 a} - \frac{π}{2 b}] = \frac{π}{2 a b (a + b)}$

BITSAT-2020

Integral Calculus

86528 The value of definite integral $\int_{0}^{\frac{π}{2}} \log (\tan x) d x$ is

1 0

2

\frac{π}{4}

3

\frac{π}{2}

4

π

Explanation:

(A) : Let,
$= \int_{0}^{\frac{π}{2}} \log (\tan x) d x$
$= \int_{0}^{\frac{π}{2}} \log {\tan (\frac{π}{2} - x)} d x$
$= \int_{0}^{\frac{π}{2}} \log (\cot x) d x$
On adding equation (i) and (ii), we get -
$∴ 2 I = \int_{0}^{\frac{π}{2}} \log (\tan x) d x + \int_{0}^{\frac{π}{2}} \log (\cot x) d x$
$2 I = \int_{0}^{\frac{π}{2}} [\log \tan x + \log \cot x] d x$
$2 I = \int_{0}^{\frac{π}{2}} \log (\tan x \cdot \cot x) d x \Rightarrow 2 I = \int_{0}^{\frac{π}{2}} \log (1) d x$
$I = 0$

BITSAT-2019

Integral Calculus

86529 The value of $I = \int_{0}^{1} x | x - \frac{1}{2} | d x$ is

1

\frac{1}{3}

2

\frac{1}{4}

3

\frac{1}{8}

4 None of these

Explanation:

(C) : Given,
$I = \int_{0}^{1} x | x - \frac{1}{2} | d x$
$= - \int_{0}^{1 / 2} x (x - \frac{1}{2}) d x + \int_{1 / 2}^{1} x (x - \frac{1}{2}) d x$
$= \int_{0}^{1 / 2} (\frac{x}{2} - x^{2}) d x + \int_{1 / 2}^{1} (x^{2} - \frac{x}{2}) d x$
$= {[\frac{x^{2}}{4} - \frac{x^{3}}{3}]}_{0}^{1 / 2} + {[\frac{x^{3}}{3} - \frac{x^{2}}{4}]}_{1 / 2}^{1}$
$= (\frac{1}{16} - \frac{1}{24}) + [(\frac{1}{3} - \frac{1}{4}) - \frac{1}{24} + \frac{1}{16})]$
$= (\frac{6 - 4}{96}) + (\frac{32 - 24 - 4 + 6}{96}) = \frac{12}{96} = \frac{1}{8}$

VITEEE-2012

Question Bank Series

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