QBManager

Integral Calculus

86774 The value of integral $\int_{0}^{π / 2} \frac{ϕ (x)}{ϕ (x) + ϕ (π / 2 - x)} d x$ is

1

π / 4

2

π / 2

3

π

4 None of these

Explanation:

(A) : Let,
$I = \int_{0}^{π / 2} \frac{ϕ (x)}{ϕ (x) + ϕ (\frac{π}{2} - x)} dx$
$\int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x$
$I = \int_{0}^{π / 2} \frac{ϕ (\frac{π}{2} - x)}{ϕ (\frac{π}{2} - x) + ϕ (x)} d x$ On adding equation (i) and (ii), we get -
$2 I = \int_{0}^{π / 2} 1 . dx = [x]_{0}^{π / 2} = π / 2$
$I = π / 4$

BITSAT-2006

Integral Calculus

86775 Evaluate : $\int_{0}^{1} \frac{d x}{\sqrt{2 - x^{2}}}$ .

1

π / 4

2

π

3

π / 2

4

π / 3

Explanation:

(A) : Let,
$I = \int_{0}^{1} \frac{d x}{\sqrt{2 - x^{2}}} \Rightarrow I = \int_{0}^{1} \frac{d x}{\sqrt{(\sqrt{2})^{2} - x^{2}}}$
We know that,
$[∵ \int \frac{1}{\sqrt{a^{2} - x^{2}}} dx = \sin^{- 1} (\frac{x}{a}) + C]$
$= {[\sin^{- 1} \frac{x}{\sqrt{2}}]}_{0}^{1} = \sin^{- 1} (\frac{1}{\sqrt{2}})$
$= \sin^{- 1} (\sin \frac{π}{4}) = \frac{π}{4}$

BITSAT-2009

Integral Calculus

86776 Evaluate $\int_{0}^{π / 2} \frac{\sin x}{1 + \cos^{2} x} d x$

1

π / 2

2

π / 4

3

π / 3

4

π

Explanation:

(B) : Let,
$I = \int_{0}^{π / 2} \frac{\sin x}{1 + \cos^{2} x} d x$
Put, $\cos x = t$
$- \sin xdx = dt$
Now, $x = 0, t = \cos 0 = 1 ∖$
And,
$x = \frac{π}{2}, t = \cos \frac{π}{2} = 0$
$I = - \int_{1}^{0} \frac{dt}{1 + t^{2}} = - {[\tan^{- 1} t]}_{1}^{0} - [\tan^{- 1} 0 - \tan^{- 1} 1]$
$= - [0 - \frac{π}{4}] = \frac{π}{4}$

BITSAT-2013

Integral Calculus

86777 $\int_{0}^{π} [cotx] d x,$ $[.]$ . denotes the greatest integer function, is equal to

1

\frac{π}{2}

2 1

3 -1

4

- \frac{π}{2}

Explanation:

(D):
Let, $I = \int_{0}^{π} [\cot x] d x$
Using $\int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x$
$I = \int_{0}^{π} [\cot (π - x)] dx$
$I = \int_{0}^{π} - \cot x d x$
$On adding equation (i) and (ii), we get -$
$2 I = \int_{0}^{π} [\cot x] dx + \int_{0}^{π} [- \cot x] dx$
$2 I = \int_{0}^{π} (- 1) dx$
$2 I = [- x]_{0}^{π}$
$2 I = - π$
$I = - \frac{π}{2}$

BCECE-2016

Integral Calculus

86774 The value of integral $\int_{0}^{π / 2} \frac{ϕ (x)}{ϕ (x) + ϕ (π / 2 - x)} d x$ is

1

π / 4

2

π / 2

3

π

4 None of these

Explanation:

(A) : Let,
$I = \int_{0}^{π / 2} \frac{ϕ (x)}{ϕ (x) + ϕ (\frac{π}{2} - x)} dx$
$\int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x$
$I = \int_{0}^{π / 2} \frac{ϕ (\frac{π}{2} - x)}{ϕ (\frac{π}{2} - x) + ϕ (x)} d x$ On adding equation (i) and (ii), we get -
$2 I = \int_{0}^{π / 2} 1 . dx = [x]_{0}^{π / 2} = π / 2$
$I = π / 4$

BITSAT-2006

Integral Calculus

86775 Evaluate : $\int_{0}^{1} \frac{d x}{\sqrt{2 - x^{2}}}$ .

1

π / 4

2

π

3

π / 2

4

π / 3

Explanation:

(A) : Let,
$I = \int_{0}^{1} \frac{d x}{\sqrt{2 - x^{2}}} \Rightarrow I = \int_{0}^{1} \frac{d x}{\sqrt{(\sqrt{2})^{2} - x^{2}}}$
We know that,
$[∵ \int \frac{1}{\sqrt{a^{2} - x^{2}}} dx = \sin^{- 1} (\frac{x}{a}) + C]$
$= {[\sin^{- 1} \frac{x}{\sqrt{2}}]}_{0}^{1} = \sin^{- 1} (\frac{1}{\sqrt{2}})$
$= \sin^{- 1} (\sin \frac{π}{4}) = \frac{π}{4}$

BITSAT-2009

Integral Calculus

86776 Evaluate $\int_{0}^{π / 2} \frac{\sin x}{1 + \cos^{2} x} d x$

1

π / 2

2

π / 4

3

π / 3

4

π

Explanation:

(B) : Let,
$I = \int_{0}^{π / 2} \frac{\sin x}{1 + \cos^{2} x} d x$
Put, $\cos x = t$
$- \sin xdx = dt$
Now, $x = 0, t = \cos 0 = 1 ∖$
And,
$x = \frac{π}{2}, t = \cos \frac{π}{2} = 0$
$I = - \int_{1}^{0} \frac{dt}{1 + t^{2}} = - {[\tan^{- 1} t]}_{1}^{0} - [\tan^{- 1} 0 - \tan^{- 1} 1]$
$= - [0 - \frac{π}{4}] = \frac{π}{4}$

BITSAT-2013

Integral Calculus

86777 $\int_{0}^{π} [cotx] d x,$ $[.]$ . denotes the greatest integer function, is equal to

1

\frac{π}{2}

2 1

3 -1

4

- \frac{π}{2}

Explanation:

(D):
Let, $I = \int_{0}^{π} [\cot x] d x$
Using $\int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x$
$I = \int_{0}^{π} [\cot (π - x)] dx$
$I = \int_{0}^{π} - \cot x d x$
$On adding equation (i) and (ii), we get -$
$2 I = \int_{0}^{π} [\cot x] dx + \int_{0}^{π} [- \cot x] dx$
$2 I = \int_{0}^{π} (- 1) dx$
$2 I = [- x]_{0}^{π}$
$2 I = - π$
$I = - \frac{π}{2}$

BCECE-2016

Integral Calculus

86774 The value of integral $\int_{0}^{π / 2} \frac{ϕ (x)}{ϕ (x) + ϕ (π / 2 - x)} d x$ is

1

π / 4

2

π / 2

3

π

4 None of these

Explanation:

(A) : Let,
$I = \int_{0}^{π / 2} \frac{ϕ (x)}{ϕ (x) + ϕ (\frac{π}{2} - x)} dx$
$\int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x$
$I = \int_{0}^{π / 2} \frac{ϕ (\frac{π}{2} - x)}{ϕ (\frac{π}{2} - x) + ϕ (x)} d x$ On adding equation (i) and (ii), we get -
$2 I = \int_{0}^{π / 2} 1 . dx = [x]_{0}^{π / 2} = π / 2$
$I = π / 4$

BITSAT-2006

Integral Calculus

86775 Evaluate : $\int_{0}^{1} \frac{d x}{\sqrt{2 - x^{2}}}$ .

1

π / 4

2

π

3

π / 2

4

π / 3

Explanation:

(A) : Let,
$I = \int_{0}^{1} \frac{d x}{\sqrt{2 - x^{2}}} \Rightarrow I = \int_{0}^{1} \frac{d x}{\sqrt{(\sqrt{2})^{2} - x^{2}}}$
We know that,
$[∵ \int \frac{1}{\sqrt{a^{2} - x^{2}}} dx = \sin^{- 1} (\frac{x}{a}) + C]$
$= {[\sin^{- 1} \frac{x}{\sqrt{2}}]}_{0}^{1} = \sin^{- 1} (\frac{1}{\sqrt{2}})$
$= \sin^{- 1} (\sin \frac{π}{4}) = \frac{π}{4}$

BITSAT-2009

Integral Calculus

86776 Evaluate $\int_{0}^{π / 2} \frac{\sin x}{1 + \cos^{2} x} d x$

1

π / 2

2

π / 4

3

π / 3

4

π

Explanation:

(B) : Let,
$I = \int_{0}^{π / 2} \frac{\sin x}{1 + \cos^{2} x} d x$
Put, $\cos x = t$
$- \sin xdx = dt$
Now, $x = 0, t = \cos 0 = 1 ∖$
And,
$x = \frac{π}{2}, t = \cos \frac{π}{2} = 0$
$I = - \int_{1}^{0} \frac{dt}{1 + t^{2}} = - {[\tan^{- 1} t]}_{1}^{0} - [\tan^{- 1} 0 - \tan^{- 1} 1]$
$= - [0 - \frac{π}{4}] = \frac{π}{4}$

BITSAT-2013

Integral Calculus

86777 $\int_{0}^{π} [cotx] d x,$ $[.]$ . denotes the greatest integer function, is equal to

1

\frac{π}{2}

2 1

3 -1

4

- \frac{π}{2}

Explanation:

(D):
Let, $I = \int_{0}^{π} [\cot x] d x$
Using $\int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x$
$I = \int_{0}^{π} [\cot (π - x)] dx$
$I = \int_{0}^{π} - \cot x d x$
$On adding equation (i) and (ii), we get -$
$2 I = \int_{0}^{π} [\cot x] dx + \int_{0}^{π} [- \cot x] dx$
$2 I = \int_{0}^{π} (- 1) dx$
$2 I = [- x]_{0}^{π}$
$2 I = - π$
$I = - \frac{π}{2}$

BCECE-2016

Integral Calculus

86774 The value of integral $\int_{0}^{π / 2} \frac{ϕ (x)}{ϕ (x) + ϕ (π / 2 - x)} d x$ is

1

π / 4

2

π / 2

3

π

4 None of these

Explanation:

(A) : Let,
$I = \int_{0}^{π / 2} \frac{ϕ (x)}{ϕ (x) + ϕ (\frac{π}{2} - x)} dx$
$\int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x$
$I = \int_{0}^{π / 2} \frac{ϕ (\frac{π}{2} - x)}{ϕ (\frac{π}{2} - x) + ϕ (x)} d x$ On adding equation (i) and (ii), we get -
$2 I = \int_{0}^{π / 2} 1 . dx = [x]_{0}^{π / 2} = π / 2$
$I = π / 4$

BITSAT-2006

Integral Calculus

86775 Evaluate : $\int_{0}^{1} \frac{d x}{\sqrt{2 - x^{2}}}$ .

1

π / 4

2

π

3

π / 2

4

π / 3

Explanation:

(A) : Let,
$I = \int_{0}^{1} \frac{d x}{\sqrt{2 - x^{2}}} \Rightarrow I = \int_{0}^{1} \frac{d x}{\sqrt{(\sqrt{2})^{2} - x^{2}}}$
We know that,
$[∵ \int \frac{1}{\sqrt{a^{2} - x^{2}}} dx = \sin^{- 1} (\frac{x}{a}) + C]$
$= {[\sin^{- 1} \frac{x}{\sqrt{2}}]}_{0}^{1} = \sin^{- 1} (\frac{1}{\sqrt{2}})$
$= \sin^{- 1} (\sin \frac{π}{4}) = \frac{π}{4}$

BITSAT-2009

Integral Calculus

86776 Evaluate $\int_{0}^{π / 2} \frac{\sin x}{1 + \cos^{2} x} d x$

1

π / 2

2

π / 4

3

π / 3

4

π

Explanation:

(B) : Let,
$I = \int_{0}^{π / 2} \frac{\sin x}{1 + \cos^{2} x} d x$
Put, $\cos x = t$
$- \sin xdx = dt$
Now, $x = 0, t = \cos 0 = 1 ∖$
And,
$x = \frac{π}{2}, t = \cos \frac{π}{2} = 0$
$I = - \int_{1}^{0} \frac{dt}{1 + t^{2}} = - {[\tan^{- 1} t]}_{1}^{0} - [\tan^{- 1} 0 - \tan^{- 1} 1]$
$= - [0 - \frac{π}{4}] = \frac{π}{4}$

BITSAT-2013

Integral Calculus

86777 $\int_{0}^{π} [cotx] d x,$ $[.]$ . denotes the greatest integer function, is equal to

1

\frac{π}{2}

2 1

3 -1

4

- \frac{π}{2}

Explanation:

(D):
Let, $I = \int_{0}^{π} [\cot x] d x$
Using $\int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x$
$I = \int_{0}^{π} [\cot (π - x)] dx$
$I = \int_{0}^{π} - \cot x d x$
$On adding equation (i) and (ii), we get -$
$2 I = \int_{0}^{π} [\cot x] dx + \int_{0}^{π} [- \cot x] dx$
$2 I = \int_{0}^{π} (- 1) dx$
$2 I = [- x]_{0}^{π}$
$2 I = - π$
$I = - \frac{π}{2}$

BCECE-2016