QBManager

Integral Calculus

86405 $\int_{- π}^{π} \sin x [f (\cos x)] d x =$

1

2 f (π)

2

2 f (2)

3

2 f (1)

4 None of these

Explanation:

(D) : Given,
$\int_{- π}^{π} \sin x [f (\cos x)] d x$
$∵ \int_{- a}^{a} g (x) d x = 0$ , for odd function.
$g (x) = \sin x [f (\cos x)]$
$g (- x) = \sin (- x) [f (\cos (- x))]$
$= - \sin x [f (\cos x)]$
$= - g (x)$
Hence, it is odd function
Then, $\int_{- π}^{π} g (x) d x = 0$
So, $\int_{- π}^{π} \sin x [f (\cos)] = 0$
Hence, option (d) is correct.

AMU-2013

Integral Calculus

86415 $\int_{0}^{π} \frac{x}{\sin x} (3 \cos^{2} x + 2 \sin x + \sin^{3} x - 3) d x =$

1

\frac{π (5 π - 12)}{4}

2

\frac{π}{2}

3

\frac{π}{2} (5 π - 6)

4

\frac{π (5 π - 12)}{6}

Explanation:

(A) : Let,
$I = \int_{0}^{π} \frac{x}{\sin x} (3 \cos^{2} x + 2 \sin x + \sin^{3} x - 3) \cdot d x$
$= \int_{0}^{π} \frac{x}{\sin x} (3 - 3 \sin^{2} x + 2 \sin x + \sin^{3} x - 3) \cdot d x$
$I = \int_{0}^{π} x (\sin^{2} x - 3 \sin x + 2) \cdot d x$
$I = \int_{0}^{π} (π - x) (\sin^{2} (π - x) - 3 \sin (π - x) + 2) \cdot d x$
$I = \int_{0}^{π} (π - x) (\sin^{2} x - 3 \sin x + 2) \cdot d x$
Adding equations (i) and (ii)
$2 I = π \int_{0}^{π} (\sin^{2} x - 3 \sin x + 2) d x$
$I = \frac{π}{2} [\int_{0}^{π} \frac{(1 - \cos 2 x)}{2} \cdot d x + [3 \cos x + 2 x]_{0}^{π}]$
$= \frac{π}{2} {[\frac{1}{2} x - \frac{\sin 2 x}{4}]}_{0}^{π} + (- 3 - 3) + (2 π)$
$= \frac{π}{2} [\frac{π}{2} - 6 + 2 π] = \frac{π}{4} [π - 12 + 4 π] = \frac{π}{4} (5 π - 12)$

AP EAMCET-2022-06.07.2022

Integral Calculus

86416 $\int_{- 1}^{1} \frac{| x |}{x} d x =$

1 1

2 -1

3 0

4

\frac{1}{2}

Explanation:

(C) : Given, $\int_{- 1}^{1} \frac{| x |}{x} d x$
$f (x) = \frac{| x |}{x} {\begin{cases} - \frac{x}{x}, x < 0 \Rightarrow - 1, x < 0 \\ \frac{x}{x}, x > 0 \Rightarrow 1, x > 0 \end{cases}}$
$∴ I = \int_{- 1}^{0} (- 1) dx + \int_{0}^{1} 1 dx$
$= [- x]_{- 1}^{0} + [x]_{0}^{1} = - 1 + 1 = 0$

AP EAMCET-2021-23.08.2021

Integral Calculus

86417 $\int_{2}^{5} \sqrt{x + 2 \sqrt{x - 1}} + \sqrt{x - 2 \sqrt{x - 1}} d x =$

1

16 / 3

2

32 / 3

3

28 / 3

4

4 / 3

Explanation:

(C) : : Let, $I = \int_{2}^{5} [\sqrt{x + 2 \sqrt{x - 1}} + \sqrt{x - 2 \sqrt{x - 1}}] dx$
$= \int^{5} [\sqrt{x - 1 + 1 + 2 \sqrt{x - 1}} + \sqrt{x - 1 + 1 - 2 \sqrt{x - 1}} \cdot dx$
$= \int_{2}^{5} [\sqrt{(\sqrt{x - 1} + 1)^{2}} + \sqrt{(\sqrt{x - 1} - 1)^{2}}] \cdot d x$
$= 2 \int_{2}^{5} \sqrt{x - 1} \cdot d x = 2 \int_{2}^{5} (x - 1)^{1 / 2} \cdot d x$
$= 2 \times \frac{2}{3} {[(x - 1)^{3 / 2}]}_{2}^{5} = \frac{4}{3} [4^{3 / 2} - 1^{3 / 2}]$
$= \frac{4}{3} [8 - 1] = \frac{4 \times 7}{3} = \frac{28}{3}$

AP EAMCET-2022-08.07.2022

Integral Calculus

86405 $\int_{- π}^{π} \sin x [f (\cos x)] d x =$

1

2 f (π)

2

2 f (2)

3

2 f (1)

4 None of these

Explanation:

(D) : Given,
$\int_{- π}^{π} \sin x [f (\cos x)] d x$
$∵ \int_{- a}^{a} g (x) d x = 0$ , for odd function.
$g (x) = \sin x [f (\cos x)]$
$g (- x) = \sin (- x) [f (\cos (- x))]$
$= - \sin x [f (\cos x)]$
$= - g (x)$
Hence, it is odd function
Then, $\int_{- π}^{π} g (x) d x = 0$
So, $\int_{- π}^{π} \sin x [f (\cos)] = 0$
Hence, option (d) is correct.

AMU-2013

Integral Calculus

86415 $\int_{0}^{π} \frac{x}{\sin x} (3 \cos^{2} x + 2 \sin x + \sin^{3} x - 3) d x =$

1

\frac{π (5 π - 12)}{4}

2

\frac{π}{2}

3

\frac{π}{2} (5 π - 6)

4

\frac{π (5 π - 12)}{6}

Explanation:

(A) : Let,
$I = \int_{0}^{π} \frac{x}{\sin x} (3 \cos^{2} x + 2 \sin x + \sin^{3} x - 3) \cdot d x$
$= \int_{0}^{π} \frac{x}{\sin x} (3 - 3 \sin^{2} x + 2 \sin x + \sin^{3} x - 3) \cdot d x$
$I = \int_{0}^{π} x (\sin^{2} x - 3 \sin x + 2) \cdot d x$
$I = \int_{0}^{π} (π - x) (\sin^{2} (π - x) - 3 \sin (π - x) + 2) \cdot d x$
$I = \int_{0}^{π} (π - x) (\sin^{2} x - 3 \sin x + 2) \cdot d x$
Adding equations (i) and (ii)
$2 I = π \int_{0}^{π} (\sin^{2} x - 3 \sin x + 2) d x$
$I = \frac{π}{2} [\int_{0}^{π} \frac{(1 - \cos 2 x)}{2} \cdot d x + [3 \cos x + 2 x]_{0}^{π}]$
$= \frac{π}{2} {[\frac{1}{2} x - \frac{\sin 2 x}{4}]}_{0}^{π} + (- 3 - 3) + (2 π)$
$= \frac{π}{2} [\frac{π}{2} - 6 + 2 π] = \frac{π}{4} [π - 12 + 4 π] = \frac{π}{4} (5 π - 12)$

AP EAMCET-2022-06.07.2022

Integral Calculus

86416 $\int_{- 1}^{1} \frac{| x |}{x} d x =$

1 1

2 -1

3 0

4

\frac{1}{2}

Explanation:

(C) : Given, $\int_{- 1}^{1} \frac{| x |}{x} d x$
$f (x) = \frac{| x |}{x} {\begin{cases} - \frac{x}{x}, x < 0 \Rightarrow - 1, x < 0 \\ \frac{x}{x}, x > 0 \Rightarrow 1, x > 0 \end{cases}}$
$∴ I = \int_{- 1}^{0} (- 1) dx + \int_{0}^{1} 1 dx$
$= [- x]_{- 1}^{0} + [x]_{0}^{1} = - 1 + 1 = 0$

AP EAMCET-2021-23.08.2021

Integral Calculus

86417 $\int_{2}^{5} \sqrt{x + 2 \sqrt{x - 1}} + \sqrt{x - 2 \sqrt{x - 1}} d x =$

1

16 / 3

2

32 / 3

3

28 / 3

4

4 / 3

Explanation:

(C) : : Let, $I = \int_{2}^{5} [\sqrt{x + 2 \sqrt{x - 1}} + \sqrt{x - 2 \sqrt{x - 1}}] dx$
$= \int^{5} [\sqrt{x - 1 + 1 + 2 \sqrt{x - 1}} + \sqrt{x - 1 + 1 - 2 \sqrt{x - 1}} \cdot dx$
$= \int_{2}^{5} [\sqrt{(\sqrt{x - 1} + 1)^{2}} + \sqrt{(\sqrt{x - 1} - 1)^{2}}] \cdot d x$
$= 2 \int_{2}^{5} \sqrt{x - 1} \cdot d x = 2 \int_{2}^{5} (x - 1)^{1 / 2} \cdot d x$
$= 2 \times \frac{2}{3} {[(x - 1)^{3 / 2}]}_{2}^{5} = \frac{4}{3} [4^{3 / 2} - 1^{3 / 2}]$
$= \frac{4}{3} [8 - 1] = \frac{4 \times 7}{3} = \frac{28}{3}$

AP EAMCET-2022-08.07.2022

Integral Calculus

86405 $\int_{- π}^{π} \sin x [f (\cos x)] d x =$

1

2 f (π)

2

2 f (2)

3

2 f (1)

4 None of these

Explanation:

(D) : Given,
$\int_{- π}^{π} \sin x [f (\cos x)] d x$
$∵ \int_{- a}^{a} g (x) d x = 0$ , for odd function.
$g (x) = \sin x [f (\cos x)]$
$g (- x) = \sin (- x) [f (\cos (- x))]$
$= - \sin x [f (\cos x)]$
$= - g (x)$
Hence, it is odd function
Then, $\int_{- π}^{π} g (x) d x = 0$
So, $\int_{- π}^{π} \sin x [f (\cos)] = 0$
Hence, option (d) is correct.

AMU-2013

Integral Calculus

86415 $\int_{0}^{π} \frac{x}{\sin x} (3 \cos^{2} x + 2 \sin x + \sin^{3} x - 3) d x =$

1

\frac{π (5 π - 12)}{4}

2

\frac{π}{2}

3

\frac{π}{2} (5 π - 6)

4

\frac{π (5 π - 12)}{6}

Explanation:

(A) : Let,
$I = \int_{0}^{π} \frac{x}{\sin x} (3 \cos^{2} x + 2 \sin x + \sin^{3} x - 3) \cdot d x$
$= \int_{0}^{π} \frac{x}{\sin x} (3 - 3 \sin^{2} x + 2 \sin x + \sin^{3} x - 3) \cdot d x$
$I = \int_{0}^{π} x (\sin^{2} x - 3 \sin x + 2) \cdot d x$
$I = \int_{0}^{π} (π - x) (\sin^{2} (π - x) - 3 \sin (π - x) + 2) \cdot d x$
$I = \int_{0}^{π} (π - x) (\sin^{2} x - 3 \sin x + 2) \cdot d x$
Adding equations (i) and (ii)
$2 I = π \int_{0}^{π} (\sin^{2} x - 3 \sin x + 2) d x$
$I = \frac{π}{2} [\int_{0}^{π} \frac{(1 - \cos 2 x)}{2} \cdot d x + [3 \cos x + 2 x]_{0}^{π}]$
$= \frac{π}{2} {[\frac{1}{2} x - \frac{\sin 2 x}{4}]}_{0}^{π} + (- 3 - 3) + (2 π)$
$= \frac{π}{2} [\frac{π}{2} - 6 + 2 π] = \frac{π}{4} [π - 12 + 4 π] = \frac{π}{4} (5 π - 12)$

AP EAMCET-2022-06.07.2022

Integral Calculus

86416 $\int_{- 1}^{1} \frac{| x |}{x} d x =$

1 1

2 -1

3 0

4

\frac{1}{2}

Explanation:

(C) : Given, $\int_{- 1}^{1} \frac{| x |}{x} d x$
$f (x) = \frac{| x |}{x} {\begin{cases} - \frac{x}{x}, x < 0 \Rightarrow - 1, x < 0 \\ \frac{x}{x}, x > 0 \Rightarrow 1, x > 0 \end{cases}}$
$∴ I = \int_{- 1}^{0} (- 1) dx + \int_{0}^{1} 1 dx$
$= [- x]_{- 1}^{0} + [x]_{0}^{1} = - 1 + 1 = 0$

AP EAMCET-2021-23.08.2021

Integral Calculus

86417 $\int_{2}^{5} \sqrt{x + 2 \sqrt{x - 1}} + \sqrt{x - 2 \sqrt{x - 1}} d x =$

1

16 / 3

2

32 / 3

3

28 / 3

4

4 / 3

Explanation:

(C) : : Let, $I = \int_{2}^{5} [\sqrt{x + 2 \sqrt{x - 1}} + \sqrt{x - 2 \sqrt{x - 1}}] dx$
$= \int^{5} [\sqrt{x - 1 + 1 + 2 \sqrt{x - 1}} + \sqrt{x - 1 + 1 - 2 \sqrt{x - 1}} \cdot dx$
$= \int_{2}^{5} [\sqrt{(\sqrt{x - 1} + 1)^{2}} + \sqrt{(\sqrt{x - 1} - 1)^{2}}] \cdot d x$
$= 2 \int_{2}^{5} \sqrt{x - 1} \cdot d x = 2 \int_{2}^{5} (x - 1)^{1 / 2} \cdot d x$
$= 2 \times \frac{2}{3} {[(x - 1)^{3 / 2}]}_{2}^{5} = \frac{4}{3} [4^{3 / 2} - 1^{3 / 2}]$
$= \frac{4}{3} [8 - 1] = \frac{4 \times 7}{3} = \frac{28}{3}$

AP EAMCET-2022-08.07.2022

Integral Calculus

86405 $\int_{- π}^{π} \sin x [f (\cos x)] d x =$

1

2 f (π)

2

2 f (2)

3

2 f (1)

4 None of these

Explanation:

(D) : Given,
$\int_{- π}^{π} \sin x [f (\cos x)] d x$
$∵ \int_{- a}^{a} g (x) d x = 0$ , for odd function.
$g (x) = \sin x [f (\cos x)]$
$g (- x) = \sin (- x) [f (\cos (- x))]$
$= - \sin x [f (\cos x)]$
$= - g (x)$
Hence, it is odd function
Then, $\int_{- π}^{π} g (x) d x = 0$
So, $\int_{- π}^{π} \sin x [f (\cos)] = 0$
Hence, option (d) is correct.

AMU-2013

Integral Calculus

86415 $\int_{0}^{π} \frac{x}{\sin x} (3 \cos^{2} x + 2 \sin x + \sin^{3} x - 3) d x =$

1

\frac{π (5 π - 12)}{4}

2

\frac{π}{2}

3

\frac{π}{2} (5 π - 6)

4

\frac{π (5 π - 12)}{6}

Explanation:

(A) : Let,
$I = \int_{0}^{π} \frac{x}{\sin x} (3 \cos^{2} x + 2 \sin x + \sin^{3} x - 3) \cdot d x$
$= \int_{0}^{π} \frac{x}{\sin x} (3 - 3 \sin^{2} x + 2 \sin x + \sin^{3} x - 3) \cdot d x$
$I = \int_{0}^{π} x (\sin^{2} x - 3 \sin x + 2) \cdot d x$
$I = \int_{0}^{π} (π - x) (\sin^{2} (π - x) - 3 \sin (π - x) + 2) \cdot d x$
$I = \int_{0}^{π} (π - x) (\sin^{2} x - 3 \sin x + 2) \cdot d x$
Adding equations (i) and (ii)
$2 I = π \int_{0}^{π} (\sin^{2} x - 3 \sin x + 2) d x$
$I = \frac{π}{2} [\int_{0}^{π} \frac{(1 - \cos 2 x)}{2} \cdot d x + [3 \cos x + 2 x]_{0}^{π}]$
$= \frac{π}{2} {[\frac{1}{2} x - \frac{\sin 2 x}{4}]}_{0}^{π} + (- 3 - 3) + (2 π)$
$= \frac{π}{2} [\frac{π}{2} - 6 + 2 π] = \frac{π}{4} [π - 12 + 4 π] = \frac{π}{4} (5 π - 12)$

AP EAMCET-2022-06.07.2022

Integral Calculus

86416 $\int_{- 1}^{1} \frac{| x |}{x} d x =$

1 1

2 -1

3 0

4

\frac{1}{2}

Explanation:

(C) : Given, $\int_{- 1}^{1} \frac{| x |}{x} d x$
$f (x) = \frac{| x |}{x} {\begin{cases} - \frac{x}{x}, x < 0 \Rightarrow - 1, x < 0 \\ \frac{x}{x}, x > 0 \Rightarrow 1, x > 0 \end{cases}}$
$∴ I = \int_{- 1}^{0} (- 1) dx + \int_{0}^{1} 1 dx$
$= [- x]_{- 1}^{0} + [x]_{0}^{1} = - 1 + 1 = 0$

AP EAMCET-2021-23.08.2021

Integral Calculus

86417 $\int_{2}^{5} \sqrt{x + 2 \sqrt{x - 1}} + \sqrt{x - 2 \sqrt{x - 1}} d x =$

1

16 / 3

2

32 / 3

3

28 / 3

4

4 / 3

Explanation:

(C) : : Let, $I = \int_{2}^{5} [\sqrt{x + 2 \sqrt{x - 1}} + \sqrt{x - 2 \sqrt{x - 1}}] dx$
$= \int^{5} [\sqrt{x - 1 + 1 + 2 \sqrt{x - 1}} + \sqrt{x - 1 + 1 - 2 \sqrt{x - 1}} \cdot dx$
$= \int_{2}^{5} [\sqrt{(\sqrt{x - 1} + 1)^{2}} + \sqrt{(\sqrt{x - 1} - 1)^{2}}] \cdot d x$
$= 2 \int_{2}^{5} \sqrt{x - 1} \cdot d x = 2 \int_{2}^{5} (x - 1)^{1 / 2} \cdot d x$
$= 2 \times \frac{2}{3} {[(x - 1)^{3 / 2}]}_{2}^{5} = \frac{4}{3} [4^{3 / 2} - 1^{3 / 2}]$
$= \frac{4}{3} [8 - 1] = \frac{4 \times 7}{3} = \frac{28}{3}$

AP EAMCET-2022-08.07.2022

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