QBManager

Integral Calculus

86515 If $\int_{0}^{1} \tan^{- 1} x d x = p$ then $\int_{0}^{1} \tan^{- 1} (\frac{1 - x}{1 + x}) d x =$

1

\frac{1 - p}{1 + p}

2

1 - p

3

\frac{π}{4} - p

4

\frac{π}{4} + p

Explanation:

(C) : Given, $\int_{0}^{1} \tan^{- 1} xdx = p$
Then, $\int_{0}^{1} \tan^{- 1} (\frac{1 - x}{1 + x}) d x$
$= \int_{0}^{1} [\tan^{- 1} (1) - \tan^{- 1} x] d x$
$= \int_{0}^{1} \tan^{- 1} (1) dx - \int_{0}^{1} \tan^{- 1} xdx$
$= \frac{π}{4} [x]_{0}^{1} - p = \frac{π}{4} [1 - 0] - p = \frac{π}{4} - p$

MHT CET-2010

Integral Calculus

86516 $\int_{0}^{π / 2} \frac{\sin x + \cos x}{\sqrt{1 + \sin 2 x}} d x =$

1

- \frac{π}{2}

2

\frac{π}{4}

3

\frac{π}{2}

4

π

Explanation:

(C) : $I = \int_{0}^{π / 2} \frac{\sin x + \cos x}{\sqrt{1 + \sin 2 x}} d x$
$I = \int_{0}^{π / 2} \frac{\sin x + \cos x}{\sqrt{\sin^{2} x + \cos^{2} x + 2 \sin x \cos x}} d x$
$I = \int_{0}^{π / 2} \frac{\sin x + \cos x}{\sqrt{(\sin x + \cos x)^{2}}} d x$
$I = \int_{0}^{π / 2} d x = [x]_{0}^{π / 2} = \frac{π}{2} - 0$
So, $I = \frac{π}{2}$

MHT CET-2005

Integral Calculus

86517 $\int_{0}^{π / 2} \frac{1}{1 + \sqrt{\tan x}} d x =$

1

\frac{π}{4}

2

\frac{π}{2}

3

π

4

2 π

Explanation:

(A) : Given,
$I = \int_{0}^{π / 2} \frac{1}{1 + \sqrt{\tan x}} dx$
$I = \int_{0}^{π / 2} \frac{1}{1 + \sqrt{\tan (\frac{π}{2} - x)}} d x$
$I = \int_{0}^{π / 2} \frac{dx}{1 + \sqrt{\cot x}} = \int_{0}^{π / 2} \frac{1}{1 + \sqrt{\frac{1}{\tan x}}} dx$
$I = \int_{0}^{π / 2} \frac{1}{1 + \frac{1}{\sqrt{\tan x}}} dx$
$I = \int_{0}^{π / 2} \frac{\sqrt{\tan x}}{1 + \sqrt{\tan x}} d x$
On adding equation (i) and (ii), we get-
$2 I = \int_{0}^{π / 2} \frac{1 + \sqrt{\tan x}}{1 + \sqrt{\tan x}} d x = \int_{0}^{π / 2} 1 d x$
$2 I = [x]_{0}^{π / 2} \Rightarrow I = \frac{π}{4}$

MHT CET-2006

Integral Calculus

86518 The value of $\int_{0}^{π / 2} \log (\tan x) d x$ is

1 2

2 1

3 0

4 3

Explanation:

(C) : Let,
$I = \int_{0}^{π / 2} \log (\tan x) d x$
Apply rule-
$I = \int_{0}^{π / 2} \log \tan (\frac{π}{2} - x) d x = \int_{0}^{π / 2} \log \cot x d x$
On adding equation (i) and (ii), we get -
$2 I = \int_{0}^{π / 2} \log (\tan x \times \cot x) d x = \int_{0}^{π / 2} \log 1 d x = 0$

Assam CEE-2021

Integral Calculus

86515 If $\int_{0}^{1} \tan^{- 1} x d x = p$ then $\int_{0}^{1} \tan^{- 1} (\frac{1 - x}{1 + x}) d x =$

1

\frac{1 - p}{1 + p}

2

1 - p

3

\frac{π}{4} - p

4

\frac{π}{4} + p

Explanation:

(C) : Given, $\int_{0}^{1} \tan^{- 1} xdx = p$
Then, $\int_{0}^{1} \tan^{- 1} (\frac{1 - x}{1 + x}) d x$
$= \int_{0}^{1} [\tan^{- 1} (1) - \tan^{- 1} x] d x$
$= \int_{0}^{1} \tan^{- 1} (1) dx - \int_{0}^{1} \tan^{- 1} xdx$
$= \frac{π}{4} [x]_{0}^{1} - p = \frac{π}{4} [1 - 0] - p = \frac{π}{4} - p$

MHT CET-2010

Integral Calculus

86516 $\int_{0}^{π / 2} \frac{\sin x + \cos x}{\sqrt{1 + \sin 2 x}} d x =$

1

- \frac{π}{2}

2

\frac{π}{4}

3

\frac{π}{2}

4

π

Explanation:

(C) : $I = \int_{0}^{π / 2} \frac{\sin x + \cos x}{\sqrt{1 + \sin 2 x}} d x$
$I = \int_{0}^{π / 2} \frac{\sin x + \cos x}{\sqrt{\sin^{2} x + \cos^{2} x + 2 \sin x \cos x}} d x$
$I = \int_{0}^{π / 2} \frac{\sin x + \cos x}{\sqrt{(\sin x + \cos x)^{2}}} d x$
$I = \int_{0}^{π / 2} d x = [x]_{0}^{π / 2} = \frac{π}{2} - 0$
So, $I = \frac{π}{2}$

MHT CET-2005

Integral Calculus

86517 $\int_{0}^{π / 2} \frac{1}{1 + \sqrt{\tan x}} d x =$

1

\frac{π}{4}

2

\frac{π}{2}

3

π

4

2 π

Explanation:

(A) : Given,
$I = \int_{0}^{π / 2} \frac{1}{1 + \sqrt{\tan x}} dx$
$I = \int_{0}^{π / 2} \frac{1}{1 + \sqrt{\tan (\frac{π}{2} - x)}} d x$
$I = \int_{0}^{π / 2} \frac{dx}{1 + \sqrt{\cot x}} = \int_{0}^{π / 2} \frac{1}{1 + \sqrt{\frac{1}{\tan x}}} dx$
$I = \int_{0}^{π / 2} \frac{1}{1 + \frac{1}{\sqrt{\tan x}}} dx$
$I = \int_{0}^{π / 2} \frac{\sqrt{\tan x}}{1 + \sqrt{\tan x}} d x$
On adding equation (i) and (ii), we get-
$2 I = \int_{0}^{π / 2} \frac{1 + \sqrt{\tan x}}{1 + \sqrt{\tan x}} d x = \int_{0}^{π / 2} 1 d x$
$2 I = [x]_{0}^{π / 2} \Rightarrow I = \frac{π}{4}$

MHT CET-2006

Integral Calculus

86518 The value of $\int_{0}^{π / 2} \log (\tan x) d x$ is

1 2

2 1

3 0

4 3

Explanation:

(C) : Let,
$I = \int_{0}^{π / 2} \log (\tan x) d x$
Apply rule-
$I = \int_{0}^{π / 2} \log \tan (\frac{π}{2} - x) d x = \int_{0}^{π / 2} \log \cot x d x$
On adding equation (i) and (ii), we get -
$2 I = \int_{0}^{π / 2} \log (\tan x \times \cot x) d x = \int_{0}^{π / 2} \log 1 d x = 0$

Assam CEE-2021

Integral Calculus

86515 If $\int_{0}^{1} \tan^{- 1} x d x = p$ then $\int_{0}^{1} \tan^{- 1} (\frac{1 - x}{1 + x}) d x =$

1

\frac{1 - p}{1 + p}

2

1 - p

3

\frac{π}{4} - p

4

\frac{π}{4} + p

Explanation:

(C) : Given, $\int_{0}^{1} \tan^{- 1} xdx = p$
Then, $\int_{0}^{1} \tan^{- 1} (\frac{1 - x}{1 + x}) d x$
$= \int_{0}^{1} [\tan^{- 1} (1) - \tan^{- 1} x] d x$
$= \int_{0}^{1} \tan^{- 1} (1) dx - \int_{0}^{1} \tan^{- 1} xdx$
$= \frac{π}{4} [x]_{0}^{1} - p = \frac{π}{4} [1 - 0] - p = \frac{π}{4} - p$

MHT CET-2010

Integral Calculus

86516 $\int_{0}^{π / 2} \frac{\sin x + \cos x}{\sqrt{1 + \sin 2 x}} d x =$

1

- \frac{π}{2}

2

\frac{π}{4}

3

\frac{π}{2}

4

π

Explanation:

(C) : $I = \int_{0}^{π / 2} \frac{\sin x + \cos x}{\sqrt{1 + \sin 2 x}} d x$
$I = \int_{0}^{π / 2} \frac{\sin x + \cos x}{\sqrt{\sin^{2} x + \cos^{2} x + 2 \sin x \cos x}} d x$
$I = \int_{0}^{π / 2} \frac{\sin x + \cos x}{\sqrt{(\sin x + \cos x)^{2}}} d x$
$I = \int_{0}^{π / 2} d x = [x]_{0}^{π / 2} = \frac{π}{2} - 0$
So, $I = \frac{π}{2}$

MHT CET-2005

Integral Calculus

86517 $\int_{0}^{π / 2} \frac{1}{1 + \sqrt{\tan x}} d x =$

1

\frac{π}{4}

2

\frac{π}{2}

3

π

4

2 π

Explanation:

(A) : Given,
$I = \int_{0}^{π / 2} \frac{1}{1 + \sqrt{\tan x}} dx$
$I = \int_{0}^{π / 2} \frac{1}{1 + \sqrt{\tan (\frac{π}{2} - x)}} d x$
$I = \int_{0}^{π / 2} \frac{dx}{1 + \sqrt{\cot x}} = \int_{0}^{π / 2} \frac{1}{1 + \sqrt{\frac{1}{\tan x}}} dx$
$I = \int_{0}^{π / 2} \frac{1}{1 + \frac{1}{\sqrt{\tan x}}} dx$
$I = \int_{0}^{π / 2} \frac{\sqrt{\tan x}}{1 + \sqrt{\tan x}} d x$
On adding equation (i) and (ii), we get-
$2 I = \int_{0}^{π / 2} \frac{1 + \sqrt{\tan x}}{1 + \sqrt{\tan x}} d x = \int_{0}^{π / 2} 1 d x$
$2 I = [x]_{0}^{π / 2} \Rightarrow I = \frac{π}{4}$

MHT CET-2006

Integral Calculus

86518 The value of $\int_{0}^{π / 2} \log (\tan x) d x$ is

1 2

2 1

3 0

4 3

Explanation:

(C) : Let,
$I = \int_{0}^{π / 2} \log (\tan x) d x$
Apply rule-
$I = \int_{0}^{π / 2} \log \tan (\frac{π}{2} - x) d x = \int_{0}^{π / 2} \log \cot x d x$
On adding equation (i) and (ii), we get -
$2 I = \int_{0}^{π / 2} \log (\tan x \times \cot x) d x = \int_{0}^{π / 2} \log 1 d x = 0$

Assam CEE-2021

Integral Calculus

86515 If $\int_{0}^{1} \tan^{- 1} x d x = p$ then $\int_{0}^{1} \tan^{- 1} (\frac{1 - x}{1 + x}) d x =$

1

\frac{1 - p}{1 + p}

2

1 - p

3

\frac{π}{4} - p

4

\frac{π}{4} + p

Explanation:

(C) : Given, $\int_{0}^{1} \tan^{- 1} xdx = p$
Then, $\int_{0}^{1} \tan^{- 1} (\frac{1 - x}{1 + x}) d x$
$= \int_{0}^{1} [\tan^{- 1} (1) - \tan^{- 1} x] d x$
$= \int_{0}^{1} \tan^{- 1} (1) dx - \int_{0}^{1} \tan^{- 1} xdx$
$= \frac{π}{4} [x]_{0}^{1} - p = \frac{π}{4} [1 - 0] - p = \frac{π}{4} - p$

MHT CET-2010

Integral Calculus

86516 $\int_{0}^{π / 2} \frac{\sin x + \cos x}{\sqrt{1 + \sin 2 x}} d x =$

1

- \frac{π}{2}

2

\frac{π}{4}

3

\frac{π}{2}

4

π

Explanation:

(C) : $I = \int_{0}^{π / 2} \frac{\sin x + \cos x}{\sqrt{1 + \sin 2 x}} d x$
$I = \int_{0}^{π / 2} \frac{\sin x + \cos x}{\sqrt{\sin^{2} x + \cos^{2} x + 2 \sin x \cos x}} d x$
$I = \int_{0}^{π / 2} \frac{\sin x + \cos x}{\sqrt{(\sin x + \cos x)^{2}}} d x$
$I = \int_{0}^{π / 2} d x = [x]_{0}^{π / 2} = \frac{π}{2} - 0$
So, $I = \frac{π}{2}$

MHT CET-2005

Integral Calculus

86517 $\int_{0}^{π / 2} \frac{1}{1 + \sqrt{\tan x}} d x =$

1

\frac{π}{4}

2

\frac{π}{2}

3

π

4

2 π

Explanation:

(A) : Given,
$I = \int_{0}^{π / 2} \frac{1}{1 + \sqrt{\tan x}} dx$
$I = \int_{0}^{π / 2} \frac{1}{1 + \sqrt{\tan (\frac{π}{2} - x)}} d x$
$I = \int_{0}^{π / 2} \frac{dx}{1 + \sqrt{\cot x}} = \int_{0}^{π / 2} \frac{1}{1 + \sqrt{\frac{1}{\tan x}}} dx$
$I = \int_{0}^{π / 2} \frac{1}{1 + \frac{1}{\sqrt{\tan x}}} dx$
$I = \int_{0}^{π / 2} \frac{\sqrt{\tan x}}{1 + \sqrt{\tan x}} d x$
On adding equation (i) and (ii), we get-
$2 I = \int_{0}^{π / 2} \frac{1 + \sqrt{\tan x}}{1 + \sqrt{\tan x}} d x = \int_{0}^{π / 2} 1 d x$
$2 I = [x]_{0}^{π / 2} \Rightarrow I = \frac{π}{4}$

MHT CET-2006

Integral Calculus

86518 The value of $\int_{0}^{π / 2} \log (\tan x) d x$ is

1 2

2 1

3 0

4 3

Explanation:

(C) : Let,
$I = \int_{0}^{π / 2} \log (\tan x) d x$
Apply rule-
$I = \int_{0}^{π / 2} \log \tan (\frac{π}{2} - x) d x = \int_{0}^{π / 2} \log \cot x d x$
On adding equation (i) and (ii), we get -
$2 I = \int_{0}^{π / 2} \log (\tan x \times \cot x) d x = \int_{0}^{π / 2} \log 1 d x = 0$

Assam CEE-2021

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