QBManager

Integral Calculus

86761 The value of $\int_{- π}^{π} \frac{\sin^{2} x}{1 + 7^{x}} d x$ is equal to

1

7^{π}

2

π

3

\frac{π}{2}

4

2^{π}

5

7 π

Explanation:

(C) : Given,
$I = \int_{- π}^{π} \frac{\sin^{2} x}{1 + 7^{x}} d x$
$I = \int_{- π}^{π} \frac{\sin^{2} (π - π - x)}{1 + 7^{(π - π - x)}} d x$
$I = \int_{- π}^{π} \frac{\sin^{2} (- x)}{1 + 7^{- x}} d x$
$I = \int_{- π}^{π} \frac{7^{x} \sin^{2} x}{7^{x} + 1} d x$
From equation (i) and (ii), we get-
$2 I = \int_{- π}^{π} \frac{(1 + 7)^{x} \sin^{2} x}{(1 + 7^{x})} d x$
$2 I = \int_{0}^{π / 2} \sin^{2} xdx$
$2 I = \int_{- π}^{π} (\frac{1}{2} - \frac{1}{2} \cos 2 x) d x$
$= \frac{1}{2} \int_{- π}^{π} dx - \frac{1}{2} \int_{- π}^{π} \cos 2 x dx$
$= \frac{1}{2} [x]_{- π}^{π} - \frac{1}{4} [\sin 2 x]_{- π}^{π}$
$2 I = \frac{1}{2} [π + π] - \frac{1}{4} [\sin 2 π - \sin (- 2 π)]$
$2 I = π \Rightarrow I = \frac{π}{2}$

Kerala CEE-2015

Integral Calculus

86762 $\int_{0}^{π / 2} \frac{d x}{1 + \tan^{3} x}$ is equal to

1 1

2

π

3

\frac{π}{2}

4

\frac{π}{4}

5 0

Explanation:

(D) : Given,
$I = \int_{0}^{π / 2} \frac{d x}{1 + \tan^{3} x}$
$I = \int_{0}^{π / 2} \frac{d x}{1 + \tan^{3} (\frac{π}{2} - x)}$
$I = \int_{0}^{π / 2} \frac{d x}{1 + \cot^{3} x}$
$I = \int_{0}^{π / 2} \frac{d x}{1 + \frac{1}{\tan^{3} x}} d x$
$I = \int_{0}^{π / 2} \frac{\tan^{3} x}{1 + \tan^{3} x} d x$
From equation (i) and (ii) we get-
$2 I = \int_{0}^{π / 2} \frac{1 + \tan^{3} x}{1 + \tan^{3} x} dx$
$2 I = \int_{0}^{π / 2} dx$
$2 I = [x]_{0}^{\frac{π}{2}}$
$I = \frac{1}{2} [\frac{π}{2} - 0] = \frac{π}{4}$

MHT CET-2022

Integral Calculus

86763 $\int_{0}^{π / 2} \frac{\sin 2 x}{1 + 2 \cos^{2} x} d x$ is equal to

1

\frac{1}{2} \log 2

2

\log 2

3

\frac{1}{2} \log 3

4

\log 3

5

\frac{1}{3} \log 3

Explanation:

(C) : Given,
$I = \int_{0}^{π / 2} \frac{\sin 2 x}{1 + 2 \cos^{2} x} d x$
$= \int_{0}^{π / 2} \frac{\sin 2 x}{1 + 2 \cos^{2} x - 1 + 1} d x$
$= \int_{0}^{π / 2} \frac{\sin 2 x}{2 + \cos 2 x} d x$
$(2 \cos^{2} x - 1 = \cos 2 x)$
$2 + \cos 2 x = t$
$- 2 \sin 2 x d x = d t$
$\sin 2 x d x = \frac{- 1}{2} d t$
When, $x = 0, t = 3$
$x = \frac{π}{2}, t = 1$
Now, $I = \frac{- 1}{2} \int_{3}^{1} \frac{1}{t} dt$
$= \frac{- 1}{2} [\log t]_{3}^{1}$
$= \frac{- 1}{2} [\log 1 - \log 3] = \frac{1}{2} \log 3$

Kerala CEE-2013

Integral Calculus

86705 The value of $\int_{0}^{\sqrt{2}} [x^{2}] dx$ , where $[\cdot]$ is the greatest integer function, is

1

2 - \sqrt{2}

2

2 + \sqrt{2}

3

\sqrt{2} - 1

4

\sqrt{2} - 2

Explanation:

(C) : Let , $I = \int_{0}^{\sqrt{2}} [x^{2}] d x = \int_{0}^{1} [x^{2}] d x + \int_{1}^{\sqrt{2}} [x^{2}] d x$
$= \int_{0}^{1} 0 d x + \int_{1}^{\sqrt{2}} 1 d x = [x]_{1}^{\sqrt{2}}$
$I = \sqrt{2} - 1$

VITEEE-2014

Integral Calculus

86761 The value of $\int_{- π}^{π} \frac{\sin^{2} x}{1 + 7^{x}} d x$ is equal to

1

7^{π}

2

π

3

\frac{π}{2}

4

2^{π}

5

7 π

Explanation:

(C) : Given,
$I = \int_{- π}^{π} \frac{\sin^{2} x}{1 + 7^{x}} d x$
$I = \int_{- π}^{π} \frac{\sin^{2} (π - π - x)}{1 + 7^{(π - π - x)}} d x$
$I = \int_{- π}^{π} \frac{\sin^{2} (- x)}{1 + 7^{- x}} d x$
$I = \int_{- π}^{π} \frac{7^{x} \sin^{2} x}{7^{x} + 1} d x$
From equation (i) and (ii), we get-
$2 I = \int_{- π}^{π} \frac{(1 + 7)^{x} \sin^{2} x}{(1 + 7^{x})} d x$
$2 I = \int_{0}^{π / 2} \sin^{2} xdx$
$2 I = \int_{- π}^{π} (\frac{1}{2} - \frac{1}{2} \cos 2 x) d x$
$= \frac{1}{2} \int_{- π}^{π} dx - \frac{1}{2} \int_{- π}^{π} \cos 2 x dx$
$= \frac{1}{2} [x]_{- π}^{π} - \frac{1}{4} [\sin 2 x]_{- π}^{π}$
$2 I = \frac{1}{2} [π + π] - \frac{1}{4} [\sin 2 π - \sin (- 2 π)]$
$2 I = π \Rightarrow I = \frac{π}{2}$

Kerala CEE-2015

Integral Calculus

86762 $\int_{0}^{π / 2} \frac{d x}{1 + \tan^{3} x}$ is equal to

1 1

2

π

3

\frac{π}{2}

4

\frac{π}{4}

5 0

Explanation:

(D) : Given,
$I = \int_{0}^{π / 2} \frac{d x}{1 + \tan^{3} x}$
$I = \int_{0}^{π / 2} \frac{d x}{1 + \tan^{3} (\frac{π}{2} - x)}$
$I = \int_{0}^{π / 2} \frac{d x}{1 + \cot^{3} x}$
$I = \int_{0}^{π / 2} \frac{d x}{1 + \frac{1}{\tan^{3} x}} d x$
$I = \int_{0}^{π / 2} \frac{\tan^{3} x}{1 + \tan^{3} x} d x$
From equation (i) and (ii) we get-
$2 I = \int_{0}^{π / 2} \frac{1 + \tan^{3} x}{1 + \tan^{3} x} dx$
$2 I = \int_{0}^{π / 2} dx$
$2 I = [x]_{0}^{\frac{π}{2}}$
$I = \frac{1}{2} [\frac{π}{2} - 0] = \frac{π}{4}$

MHT CET-2022

Integral Calculus

86763 $\int_{0}^{π / 2} \frac{\sin 2 x}{1 + 2 \cos^{2} x} d x$ is equal to

1

\frac{1}{2} \log 2

2

\log 2

3

\frac{1}{2} \log 3

4

\log 3

5

\frac{1}{3} \log 3

Explanation:

(C) : Given,
$I = \int_{0}^{π / 2} \frac{\sin 2 x}{1 + 2 \cos^{2} x} d x$
$= \int_{0}^{π / 2} \frac{\sin 2 x}{1 + 2 \cos^{2} x - 1 + 1} d x$
$= \int_{0}^{π / 2} \frac{\sin 2 x}{2 + \cos 2 x} d x$
$(2 \cos^{2} x - 1 = \cos 2 x)$
$2 + \cos 2 x = t$
$- 2 \sin 2 x d x = d t$
$\sin 2 x d x = \frac{- 1}{2} d t$
When, $x = 0, t = 3$
$x = \frac{π}{2}, t = 1$
Now, $I = \frac{- 1}{2} \int_{3}^{1} \frac{1}{t} dt$
$= \frac{- 1}{2} [\log t]_{3}^{1}$
$= \frac{- 1}{2} [\log 1 - \log 3] = \frac{1}{2} \log 3$

Kerala CEE-2013

Integral Calculus

86705 The value of $\int_{0}^{\sqrt{2}} [x^{2}] dx$ , where $[\cdot]$ is the greatest integer function, is

1

2 - \sqrt{2}

2

2 + \sqrt{2}

3

\sqrt{2} - 1

4

\sqrt{2} - 2

Explanation:

(C) : Let , $I = \int_{0}^{\sqrt{2}} [x^{2}] d x = \int_{0}^{1} [x^{2}] d x + \int_{1}^{\sqrt{2}} [x^{2}] d x$
$= \int_{0}^{1} 0 d x + \int_{1}^{\sqrt{2}} 1 d x = [x]_{1}^{\sqrt{2}}$
$I = \sqrt{2} - 1$

VITEEE-2014

Integral Calculus

86761 The value of $\int_{- π}^{π} \frac{\sin^{2} x}{1 + 7^{x}} d x$ is equal to

1

7^{π}

2

π

3

\frac{π}{2}

4

2^{π}

5

7 π

Explanation:

(C) : Given,
$I = \int_{- π}^{π} \frac{\sin^{2} x}{1 + 7^{x}} d x$
$I = \int_{- π}^{π} \frac{\sin^{2} (π - π - x)}{1 + 7^{(π - π - x)}} d x$
$I = \int_{- π}^{π} \frac{\sin^{2} (- x)}{1 + 7^{- x}} d x$
$I = \int_{- π}^{π} \frac{7^{x} \sin^{2} x}{7^{x} + 1} d x$
From equation (i) and (ii), we get-
$2 I = \int_{- π}^{π} \frac{(1 + 7)^{x} \sin^{2} x}{(1 + 7^{x})} d x$
$2 I = \int_{0}^{π / 2} \sin^{2} xdx$
$2 I = \int_{- π}^{π} (\frac{1}{2} - \frac{1}{2} \cos 2 x) d x$
$= \frac{1}{2} \int_{- π}^{π} dx - \frac{1}{2} \int_{- π}^{π} \cos 2 x dx$
$= \frac{1}{2} [x]_{- π}^{π} - \frac{1}{4} [\sin 2 x]_{- π}^{π}$
$2 I = \frac{1}{2} [π + π] - \frac{1}{4} [\sin 2 π - \sin (- 2 π)]$
$2 I = π \Rightarrow I = \frac{π}{2}$

Kerala CEE-2015

Integral Calculus

86762 $\int_{0}^{π / 2} \frac{d x}{1 + \tan^{3} x}$ is equal to

1 1

2

π

3

\frac{π}{2}

4

\frac{π}{4}

5 0

Explanation:

(D) : Given,
$I = \int_{0}^{π / 2} \frac{d x}{1 + \tan^{3} x}$
$I = \int_{0}^{π / 2} \frac{d x}{1 + \tan^{3} (\frac{π}{2} - x)}$
$I = \int_{0}^{π / 2} \frac{d x}{1 + \cot^{3} x}$
$I = \int_{0}^{π / 2} \frac{d x}{1 + \frac{1}{\tan^{3} x}} d x$
$I = \int_{0}^{π / 2} \frac{\tan^{3} x}{1 + \tan^{3} x} d x$
From equation (i) and (ii) we get-
$2 I = \int_{0}^{π / 2} \frac{1 + \tan^{3} x}{1 + \tan^{3} x} dx$
$2 I = \int_{0}^{π / 2} dx$
$2 I = [x]_{0}^{\frac{π}{2}}$
$I = \frac{1}{2} [\frac{π}{2} - 0] = \frac{π}{4}$

MHT CET-2022

Integral Calculus

86763 $\int_{0}^{π / 2} \frac{\sin 2 x}{1 + 2 \cos^{2} x} d x$ is equal to

1

\frac{1}{2} \log 2

2

\log 2

3

\frac{1}{2} \log 3

4

\log 3

5

\frac{1}{3} \log 3

Explanation:

(C) : Given,
$I = \int_{0}^{π / 2} \frac{\sin 2 x}{1 + 2 \cos^{2} x} d x$
$= \int_{0}^{π / 2} \frac{\sin 2 x}{1 + 2 \cos^{2} x - 1 + 1} d x$
$= \int_{0}^{π / 2} \frac{\sin 2 x}{2 + \cos 2 x} d x$
$(2 \cos^{2} x - 1 = \cos 2 x)$
$2 + \cos 2 x = t$
$- 2 \sin 2 x d x = d t$
$\sin 2 x d x = \frac{- 1}{2} d t$
When, $x = 0, t = 3$
$x = \frac{π}{2}, t = 1$
Now, $I = \frac{- 1}{2} \int_{3}^{1} \frac{1}{t} dt$
$= \frac{- 1}{2} [\log t]_{3}^{1}$
$= \frac{- 1}{2} [\log 1 - \log 3] = \frac{1}{2} \log 3$

Kerala CEE-2013

Integral Calculus

86705 The value of $\int_{0}^{\sqrt{2}} [x^{2}] dx$ , where $[\cdot]$ is the greatest integer function, is

1

2 - \sqrt{2}

2

2 + \sqrt{2}

3

\sqrt{2} - 1

4

\sqrt{2} - 2

Explanation:

(C) : Let , $I = \int_{0}^{\sqrt{2}} [x^{2}] d x = \int_{0}^{1} [x^{2}] d x + \int_{1}^{\sqrt{2}} [x^{2}] d x$
$= \int_{0}^{1} 0 d x + \int_{1}^{\sqrt{2}} 1 d x = [x]_{1}^{\sqrt{2}}$
$I = \sqrt{2} - 1$

VITEEE-2014

Integral Calculus

86761 The value of $\int_{- π}^{π} \frac{\sin^{2} x}{1 + 7^{x}} d x$ is equal to

1

7^{π}

2

π

3

\frac{π}{2}

4

2^{π}

5

7 π

Explanation:

(C) : Given,
$I = \int_{- π}^{π} \frac{\sin^{2} x}{1 + 7^{x}} d x$
$I = \int_{- π}^{π} \frac{\sin^{2} (π - π - x)}{1 + 7^{(π - π - x)}} d x$
$I = \int_{- π}^{π} \frac{\sin^{2} (- x)}{1 + 7^{- x}} d x$
$I = \int_{- π}^{π} \frac{7^{x} \sin^{2} x}{7^{x} + 1} d x$
From equation (i) and (ii), we get-
$2 I = \int_{- π}^{π} \frac{(1 + 7)^{x} \sin^{2} x}{(1 + 7^{x})} d x$
$2 I = \int_{0}^{π / 2} \sin^{2} xdx$
$2 I = \int_{- π}^{π} (\frac{1}{2} - \frac{1}{2} \cos 2 x) d x$
$= \frac{1}{2} \int_{- π}^{π} dx - \frac{1}{2} \int_{- π}^{π} \cos 2 x dx$
$= \frac{1}{2} [x]_{- π}^{π} - \frac{1}{4} [\sin 2 x]_{- π}^{π}$
$2 I = \frac{1}{2} [π + π] - \frac{1}{4} [\sin 2 π - \sin (- 2 π)]$
$2 I = π \Rightarrow I = \frac{π}{2}$

Kerala CEE-2015

Integral Calculus

86762 $\int_{0}^{π / 2} \frac{d x}{1 + \tan^{3} x}$ is equal to

1 1

2

π

3

\frac{π}{2}

4

\frac{π}{4}

5 0

Explanation:

(D) : Given,
$I = \int_{0}^{π / 2} \frac{d x}{1 + \tan^{3} x}$
$I = \int_{0}^{π / 2} \frac{d x}{1 + \tan^{3} (\frac{π}{2} - x)}$
$I = \int_{0}^{π / 2} \frac{d x}{1 + \cot^{3} x}$
$I = \int_{0}^{π / 2} \frac{d x}{1 + \frac{1}{\tan^{3} x}} d x$
$I = \int_{0}^{π / 2} \frac{\tan^{3} x}{1 + \tan^{3} x} d x$
From equation (i) and (ii) we get-
$2 I = \int_{0}^{π / 2} \frac{1 + \tan^{3} x}{1 + \tan^{3} x} dx$
$2 I = \int_{0}^{π / 2} dx$
$2 I = [x]_{0}^{\frac{π}{2}}$
$I = \frac{1}{2} [\frac{π}{2} - 0] = \frac{π}{4}$

MHT CET-2022

Integral Calculus

86763 $\int_{0}^{π / 2} \frac{\sin 2 x}{1 + 2 \cos^{2} x} d x$ is equal to

1

\frac{1}{2} \log 2

2

\log 2

3

\frac{1}{2} \log 3

4

\log 3

5

\frac{1}{3} \log 3

Explanation:

(C) : Given,
$I = \int_{0}^{π / 2} \frac{\sin 2 x}{1 + 2 \cos^{2} x} d x$
$= \int_{0}^{π / 2} \frac{\sin 2 x}{1 + 2 \cos^{2} x - 1 + 1} d x$
$= \int_{0}^{π / 2} \frac{\sin 2 x}{2 + \cos 2 x} d x$
$(2 \cos^{2} x - 1 = \cos 2 x)$
$2 + \cos 2 x = t$
$- 2 \sin 2 x d x = d t$
$\sin 2 x d x = \frac{- 1}{2} d t$
When, $x = 0, t = 3$
$x = \frac{π}{2}, t = 1$
Now, $I = \frac{- 1}{2} \int_{3}^{1} \frac{1}{t} dt$
$= \frac{- 1}{2} [\log t]_{3}^{1}$
$= \frac{- 1}{2} [\log 1 - \log 3] = \frac{1}{2} \log 3$

Kerala CEE-2013

Integral Calculus

86705 The value of $\int_{0}^{\sqrt{2}} [x^{2}] dx$ , where $[\cdot]$ is the greatest integer function, is

1

2 - \sqrt{2}

2

2 + \sqrt{2}

3

\sqrt{2} - 1

4

\sqrt{2} - 2

Explanation:

(C) : Let , $I = \int_{0}^{\sqrt{2}} [x^{2}] d x = \int_{0}^{1} [x^{2}] d x + \int_{1}^{\sqrt{2}} [x^{2}] d x$
$= \int_{0}^{1} 0 d x + \int_{1}^{\sqrt{2}} 1 d x = [x]_{1}^{\sqrt{2}}$
$I = \sqrt{2} - 1$

VITEEE-2014