QBManager

Integral Calculus

86559 $\int_{- 1}^{2} | x^{3} - x | d x$ is equal to

1 11

2 4

3

\frac{11}{4}

4

\frac{4}{11}

Explanation:

(C) : We have,
$\int_{- 1}^{2} | x^{3} - x | d x$
$= \int_{- 1}^{0} (x^{3} - x) d x - \int_{0}^{1} (x^{3} - x) d x + \int_{1}^{2} (x^{3} - x) d x$
$= {[\frac{x^{4}}{4} - \frac{x^{2}}{2}]}_{- 1}^{0} - {[\frac{x^{4}}{4} - \frac{x^{2}}{2}]}_{0}^{1} + {[\frac{x^{4}}{4} - \frac{x^{2}}{2}]}_{1}^{2}$
$= [(0 - 0) - (\frac{1}{4} - \frac{1}{2})] - [(\frac{1}{4} - \frac{1}{2}) - (0 - 0)] + [(4 - 2) - (\frac{1}{4} - \frac{1}{2})]$
$= \frac{1}{4} + \frac{1}{4} + 2 + \frac{1}{4} = \frac{11}{4}$

CG PET-2013

Integral Calculus

86567 The value of $\int_{\frac{π}{3}}^{\frac{π}{2}} \frac{(2 + 3 \sin x)}{\sin x (1 + \cos x)} d x$ is equal to

1

\frac{7}{2} - \sqrt{3} - \log_{e} \sqrt{3}

2

\frac{10}{3} - \sqrt{3} - \log_{e} \sqrt{3}

3

\frac{10}{3} - \sqrt{3} + \log_{e} \sqrt{3}

4

- 2 + 3 \sqrt{3} + \log_{e} \sqrt{3}

Explanation:

(C) : $I = \int_{π / 3}^{π / 2} \frac{2 + 3 \sin x}{\sin x (1 + \cos x)} d x$
$I = \int_{π / 3}^{π / 2} \frac{2}{\sin x (1 + \cos x)} d x + 3 \int_{π / 3}^{π / 2} \frac{\sin x}{\sin x (1 + \cos x)} d x$
$I = 2 \int_{π / 3}^{π / 2} \frac{d x}{\sin x (1 + \cos x)} + 3 \int_{π / 3}^{π / 2} \frac{d x}{(1 + \cos x)}$
$I = 2 I_{1} + 3 I_{2}$
$I_{1} = \int_{π / 3}^{π / 2} \frac{d x}{\sin x (1 + \cos x)} = \int_{π / 3}^{π / 2} \frac{d x}{\frac{2 \tan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}} \times 2 \cos^{2} \frac{x}{2}}$
$I_{1} = \int_{π / 3}^{π / 2} \frac{(1 + \tan^{2} \frac{x}{2}) \cdot \sec^{2} \frac{x}{2}}{4 \tan \frac{x}{2}}$
Let,
$\tan \frac{x}{2} = t \Rightarrow \frac{1}{2} \sec \frac{x}{2} dx = dt$
When, $x = \frac{π}{2}, t = 1$
$x = \frac{π}{3}, t = \frac{1}{\sqrt{3}}$
$I_{1} = \frac{1}{2} \int_{1 / \sqrt{3}}^{1} \frac{1 + t^{2}}{t} d t = \frac{1}{2} {[\ln t + \frac{t^{2}}{2}]}_{\frac{1}{\sqrt{3}}}^{1}$
$I_{1} = \frac{1}{2} [(0 + \frac{1}{2}) - (\ln \frac{1}{\sqrt{3}} + \frac{1}{6})] = \frac{1}{2} (\frac{1}{3} + \ln \sqrt{3})$
$I_{1} = \frac{1}{6} + \frac{1}{2} \ln \sqrt{3}$
$I_{2} = \int_{π / 3}^{π / 2} \frac{d x}{1 + \cos x} = \int_{π / 3}^{π / 2} \frac{1}{1 + \cos x} \times \frac{1 - \cos x}{1 - \cos x} d x$
$I_{2} = \int_{π / 3}^{π / 2} \frac{1 - \cos x}{1 - \cos x} d x$
$I_{2} = \int_{π / 3}^{π / 2} \frac{1 - \cos^{2} x}{\sin^{2} x} d x$
$I_{2} = \int_{π / 3}^{π / 2} ({cosec}^{2} x - \cot x cosec x) d x$
$I_{2} = [cosec x - \cot x]_{π / 3}^{π / 2}$
$I_{2} = 1 - (\frac{2}{\sqrt{3}} - \frac{1}{\sqrt{3}}) = 1 - \frac{1}{\sqrt{3}}$
$I_{2} = 1 - \frac{1}{\sqrt{3}}$
Putting value of $I_{1}$ and $I_{2}$ from equation (ii) and (iii) to equation (i),
$I = 2 I_{1} + 3 I_{2}$
$I = 2 \times (\frac{1}{6} + \frac{1}{2} \ln \sqrt{3}) + 3 (1 - \frac{1}{\sqrt{3}})$
$I = \frac{1}{3} + \ln \sqrt{3} + 3 - \sqrt{3}$
$I = \frac{10}{3} - \sqrt{3} + \log_{e} \sqrt{3}$

JEE Main-2023-31.01.2023

Integral Calculus

86575 $\int_{2}^{4} {| x - 2 | + | x - 3 |} d x =$

1 1

2 2

3 3

4 4

Explanation:

(C): Given,
$I = \int_{2}^{4} {| x - 2 | + | x - 3 |} d x$
$| x - 2 | = {\begin{cases} x - 2, x \geq 2 \\ - (x - 2), x < 2 \end{cases}$ ,
$| x - 3 | = {\begin{cases} x - 3, x \geq 3 \\ - (x - 3), x < 3 \end{cases}$
$= \int_{1}^{3} (| x - 2 | + | x - 3 |) d x + \int_{3}^{4} (| x - 2 | + | x - 3 |) d x$
$= \int_{2}^{3} ((x - 2) + (- (x - 3))) d x + \int_{3}^{4} ((x - 2) + (x - 3)) d x$
$= \int_{2}^{3} [x - 2 - x + 3] d x + \int_{3}^{4} (x - 2 + x - 3) d x$
$= \int_{2}^{3} 1 d x + \int_{3}^{4} (2 x - 5) d x$
$= [x]_{2}^{3} + {[\frac{2 x^{2}}{2} - 5 x]}_{3}^{4}$
$= (3 - 2) + (4^{2} - 5 (4)) - (3^{2} - 5 (3))$
$= 1 + (16 - 20) - (9 - 15)$
$= 1 + (- 4) - (- 6) = 1 - 4 + 6 = 7 - 4 = 3$

AP EAMCET-2021-19.08.2021

Integral Calculus

86576 The value of the integral $\int_{- 2}^{0} \frac{d x}{\sqrt{12 - x^{2} - 4 x}}$ is

1

π / 2

2

π / 6

3

π / 3

4

- π / 6

Explanation:

(B) : Given,
$I = \int_{- 2}^{0} \frac{d x}{\sqrt{12 - x^{2} - 4 x}}$
$= \int_{- 2}^{0} \frac{dx}{\sqrt{12 - x^{2} - 4 x - 4 + 4}}$
$= \int_{- 2}^{0} \frac{dx}{\sqrt{16 - (x + 2)^{2}}}$
$= {[\sin^{- 1} (\frac{x + 2}{4})]}_{- 2}^{0} = \sin^{- 1} (\frac{2}{4}) = \frac{π}{6}$

AMU-2016

Integral Calculus

86559 $\int_{- 1}^{2} | x^{3} - x | d x$ is equal to

1 11

2 4

3

\frac{11}{4}

4

\frac{4}{11}

Explanation:

(C) : We have,
$\int_{- 1}^{2} | x^{3} - x | d x$
$= \int_{- 1}^{0} (x^{3} - x) d x - \int_{0}^{1} (x^{3} - x) d x + \int_{1}^{2} (x^{3} - x) d x$
$= {[\frac{x^{4}}{4} - \frac{x^{2}}{2}]}_{- 1}^{0} - {[\frac{x^{4}}{4} - \frac{x^{2}}{2}]}_{0}^{1} + {[\frac{x^{4}}{4} - \frac{x^{2}}{2}]}_{1}^{2}$
$= [(0 - 0) - (\frac{1}{4} - \frac{1}{2})] - [(\frac{1}{4} - \frac{1}{2}) - (0 - 0)] + [(4 - 2) - (\frac{1}{4} - \frac{1}{2})]$
$= \frac{1}{4} + \frac{1}{4} + 2 + \frac{1}{4} = \frac{11}{4}$

CG PET-2013

Integral Calculus

86567 The value of $\int_{\frac{π}{3}}^{\frac{π}{2}} \frac{(2 + 3 \sin x)}{\sin x (1 + \cos x)} d x$ is equal to

1

\frac{7}{2} - \sqrt{3} - \log_{e} \sqrt{3}

2

\frac{10}{3} - \sqrt{3} - \log_{e} \sqrt{3}

3

\frac{10}{3} - \sqrt{3} + \log_{e} \sqrt{3}

4

- 2 + 3 \sqrt{3} + \log_{e} \sqrt{3}

Explanation:

(C) : $I = \int_{π / 3}^{π / 2} \frac{2 + 3 \sin x}{\sin x (1 + \cos x)} d x$
$I = \int_{π / 3}^{π / 2} \frac{2}{\sin x (1 + \cos x)} d x + 3 \int_{π / 3}^{π / 2} \frac{\sin x}{\sin x (1 + \cos x)} d x$
$I = 2 \int_{π / 3}^{π / 2} \frac{d x}{\sin x (1 + \cos x)} + 3 \int_{π / 3}^{π / 2} \frac{d x}{(1 + \cos x)}$
$I = 2 I_{1} + 3 I_{2}$
$I_{1} = \int_{π / 3}^{π / 2} \frac{d x}{\sin x (1 + \cos x)} = \int_{π / 3}^{π / 2} \frac{d x}{\frac{2 \tan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}} \times 2 \cos^{2} \frac{x}{2}}$
$I_{1} = \int_{π / 3}^{π / 2} \frac{(1 + \tan^{2} \frac{x}{2}) \cdot \sec^{2} \frac{x}{2}}{4 \tan \frac{x}{2}}$
Let,
$\tan \frac{x}{2} = t \Rightarrow \frac{1}{2} \sec \frac{x}{2} dx = dt$
When, $x = \frac{π}{2}, t = 1$
$x = \frac{π}{3}, t = \frac{1}{\sqrt{3}}$
$I_{1} = \frac{1}{2} \int_{1 / \sqrt{3}}^{1} \frac{1 + t^{2}}{t} d t = \frac{1}{2} {[\ln t + \frac{t^{2}}{2}]}_{\frac{1}{\sqrt{3}}}^{1}$
$I_{1} = \frac{1}{2} [(0 + \frac{1}{2}) - (\ln \frac{1}{\sqrt{3}} + \frac{1}{6})] = \frac{1}{2} (\frac{1}{3} + \ln \sqrt{3})$
$I_{1} = \frac{1}{6} + \frac{1}{2} \ln \sqrt{3}$
$I_{2} = \int_{π / 3}^{π / 2} \frac{d x}{1 + \cos x} = \int_{π / 3}^{π / 2} \frac{1}{1 + \cos x} \times \frac{1 - \cos x}{1 - \cos x} d x$
$I_{2} = \int_{π / 3}^{π / 2} \frac{1 - \cos x}{1 - \cos x} d x$
$I_{2} = \int_{π / 3}^{π / 2} \frac{1 - \cos^{2} x}{\sin^{2} x} d x$
$I_{2} = \int_{π / 3}^{π / 2} ({cosec}^{2} x - \cot x cosec x) d x$
$I_{2} = [cosec x - \cot x]_{π / 3}^{π / 2}$
$I_{2} = 1 - (\frac{2}{\sqrt{3}} - \frac{1}{\sqrt{3}}) = 1 - \frac{1}{\sqrt{3}}$
$I_{2} = 1 - \frac{1}{\sqrt{3}}$
Putting value of $I_{1}$ and $I_{2}$ from equation (ii) and (iii) to equation (i),
$I = 2 I_{1} + 3 I_{2}$
$I = 2 \times (\frac{1}{6} + \frac{1}{2} \ln \sqrt{3}) + 3 (1 - \frac{1}{\sqrt{3}})$
$I = \frac{1}{3} + \ln \sqrt{3} + 3 - \sqrt{3}$
$I = \frac{10}{3} - \sqrt{3} + \log_{e} \sqrt{3}$

JEE Main-2023-31.01.2023

Integral Calculus

86575 $\int_{2}^{4} {| x - 2 | + | x - 3 |} d x =$

1 1

2 2

3 3

4 4

Explanation:

(C): Given,
$I = \int_{2}^{4} {| x - 2 | + | x - 3 |} d x$
$| x - 2 | = {\begin{cases} x - 2, x \geq 2 \\ - (x - 2), x < 2 \end{cases}$ ,
$| x - 3 | = {\begin{cases} x - 3, x \geq 3 \\ - (x - 3), x < 3 \end{cases}$
$= \int_{1}^{3} (| x - 2 | + | x - 3 |) d x + \int_{3}^{4} (| x - 2 | + | x - 3 |) d x$
$= \int_{2}^{3} ((x - 2) + (- (x - 3))) d x + \int_{3}^{4} ((x - 2) + (x - 3)) d x$
$= \int_{2}^{3} [x - 2 - x + 3] d x + \int_{3}^{4} (x - 2 + x - 3) d x$
$= \int_{2}^{3} 1 d x + \int_{3}^{4} (2 x - 5) d x$
$= [x]_{2}^{3} + {[\frac{2 x^{2}}{2} - 5 x]}_{3}^{4}$
$= (3 - 2) + (4^{2} - 5 (4)) - (3^{2} - 5 (3))$
$= 1 + (16 - 20) - (9 - 15)$
$= 1 + (- 4) - (- 6) = 1 - 4 + 6 = 7 - 4 = 3$

AP EAMCET-2021-19.08.2021

Integral Calculus

86576 The value of the integral $\int_{- 2}^{0} \frac{d x}{\sqrt{12 - x^{2} - 4 x}}$ is

1

π / 2

2

π / 6

3

π / 3

4

- π / 6

Explanation:

(B) : Given,
$I = \int_{- 2}^{0} \frac{d x}{\sqrt{12 - x^{2} - 4 x}}$
$= \int_{- 2}^{0} \frac{dx}{\sqrt{12 - x^{2} - 4 x - 4 + 4}}$
$= \int_{- 2}^{0} \frac{dx}{\sqrt{16 - (x + 2)^{2}}}$
$= {[\sin^{- 1} (\frac{x + 2}{4})]}_{- 2}^{0} = \sin^{- 1} (\frac{2}{4}) = \frac{π}{6}$

AMU-2016

Integral Calculus

86559 $\int_{- 1}^{2} | x^{3} - x | d x$ is equal to

1 11

2 4

3

\frac{11}{4}

4

\frac{4}{11}

Explanation:

(C) : We have,
$\int_{- 1}^{2} | x^{3} - x | d x$
$= \int_{- 1}^{0} (x^{3} - x) d x - \int_{0}^{1} (x^{3} - x) d x + \int_{1}^{2} (x^{3} - x) d x$
$= {[\frac{x^{4}}{4} - \frac{x^{2}}{2}]}_{- 1}^{0} - {[\frac{x^{4}}{4} - \frac{x^{2}}{2}]}_{0}^{1} + {[\frac{x^{4}}{4} - \frac{x^{2}}{2}]}_{1}^{2}$
$= [(0 - 0) - (\frac{1}{4} - \frac{1}{2})] - [(\frac{1}{4} - \frac{1}{2}) - (0 - 0)] + [(4 - 2) - (\frac{1}{4} - \frac{1}{2})]$
$= \frac{1}{4} + \frac{1}{4} + 2 + \frac{1}{4} = \frac{11}{4}$

CG PET-2013

Integral Calculus

86567 The value of $\int_{\frac{π}{3}}^{\frac{π}{2}} \frac{(2 + 3 \sin x)}{\sin x (1 + \cos x)} d x$ is equal to

1

\frac{7}{2} - \sqrt{3} - \log_{e} \sqrt{3}

2

\frac{10}{3} - \sqrt{3} - \log_{e} \sqrt{3}

3

\frac{10}{3} - \sqrt{3} + \log_{e} \sqrt{3}

4

- 2 + 3 \sqrt{3} + \log_{e} \sqrt{3}

Explanation:

(C) : $I = \int_{π / 3}^{π / 2} \frac{2 + 3 \sin x}{\sin x (1 + \cos x)} d x$
$I = \int_{π / 3}^{π / 2} \frac{2}{\sin x (1 + \cos x)} d x + 3 \int_{π / 3}^{π / 2} \frac{\sin x}{\sin x (1 + \cos x)} d x$
$I = 2 \int_{π / 3}^{π / 2} \frac{d x}{\sin x (1 + \cos x)} + 3 \int_{π / 3}^{π / 2} \frac{d x}{(1 + \cos x)}$
$I = 2 I_{1} + 3 I_{2}$
$I_{1} = \int_{π / 3}^{π / 2} \frac{d x}{\sin x (1 + \cos x)} = \int_{π / 3}^{π / 2} \frac{d x}{\frac{2 \tan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}} \times 2 \cos^{2} \frac{x}{2}}$
$I_{1} = \int_{π / 3}^{π / 2} \frac{(1 + \tan^{2} \frac{x}{2}) \cdot \sec^{2} \frac{x}{2}}{4 \tan \frac{x}{2}}$
Let,
$\tan \frac{x}{2} = t \Rightarrow \frac{1}{2} \sec \frac{x}{2} dx = dt$
When, $x = \frac{π}{2}, t = 1$
$x = \frac{π}{3}, t = \frac{1}{\sqrt{3}}$
$I_{1} = \frac{1}{2} \int_{1 / \sqrt{3}}^{1} \frac{1 + t^{2}}{t} d t = \frac{1}{2} {[\ln t + \frac{t^{2}}{2}]}_{\frac{1}{\sqrt{3}}}^{1}$
$I_{1} = \frac{1}{2} [(0 + \frac{1}{2}) - (\ln \frac{1}{\sqrt{3}} + \frac{1}{6})] = \frac{1}{2} (\frac{1}{3} + \ln \sqrt{3})$
$I_{1} = \frac{1}{6} + \frac{1}{2} \ln \sqrt{3}$
$I_{2} = \int_{π / 3}^{π / 2} \frac{d x}{1 + \cos x} = \int_{π / 3}^{π / 2} \frac{1}{1 + \cos x} \times \frac{1 - \cos x}{1 - \cos x} d x$
$I_{2} = \int_{π / 3}^{π / 2} \frac{1 - \cos x}{1 - \cos x} d x$
$I_{2} = \int_{π / 3}^{π / 2} \frac{1 - \cos^{2} x}{\sin^{2} x} d x$
$I_{2} = \int_{π / 3}^{π / 2} ({cosec}^{2} x - \cot x cosec x) d x$
$I_{2} = [cosec x - \cot x]_{π / 3}^{π / 2}$
$I_{2} = 1 - (\frac{2}{\sqrt{3}} - \frac{1}{\sqrt{3}}) = 1 - \frac{1}{\sqrt{3}}$
$I_{2} = 1 - \frac{1}{\sqrt{3}}$
Putting value of $I_{1}$ and $I_{2}$ from equation (ii) and (iii) to equation (i),
$I = 2 I_{1} + 3 I_{2}$
$I = 2 \times (\frac{1}{6} + \frac{1}{2} \ln \sqrt{3}) + 3 (1 - \frac{1}{\sqrt{3}})$
$I = \frac{1}{3} + \ln \sqrt{3} + 3 - \sqrt{3}$
$I = \frac{10}{3} - \sqrt{3} + \log_{e} \sqrt{3}$

JEE Main-2023-31.01.2023

Integral Calculus

86575 $\int_{2}^{4} {| x - 2 | + | x - 3 |} d x =$

1 1

2 2

3 3

4 4

Explanation:

(C): Given,
$I = \int_{2}^{4} {| x - 2 | + | x - 3 |} d x$
$| x - 2 | = {\begin{cases} x - 2, x \geq 2 \\ - (x - 2), x < 2 \end{cases}$ ,
$| x - 3 | = {\begin{cases} x - 3, x \geq 3 \\ - (x - 3), x < 3 \end{cases}$
$= \int_{1}^{3} (| x - 2 | + | x - 3 |) d x + \int_{3}^{4} (| x - 2 | + | x - 3 |) d x$
$= \int_{2}^{3} ((x - 2) + (- (x - 3))) d x + \int_{3}^{4} ((x - 2) + (x - 3)) d x$
$= \int_{2}^{3} [x - 2 - x + 3] d x + \int_{3}^{4} (x - 2 + x - 3) d x$
$= \int_{2}^{3} 1 d x + \int_{3}^{4} (2 x - 5) d x$
$= [x]_{2}^{3} + {[\frac{2 x^{2}}{2} - 5 x]}_{3}^{4}$
$= (3 - 2) + (4^{2} - 5 (4)) - (3^{2} - 5 (3))$
$= 1 + (16 - 20) - (9 - 15)$
$= 1 + (- 4) - (- 6) = 1 - 4 + 6 = 7 - 4 = 3$

AP EAMCET-2021-19.08.2021

Integral Calculus

86576 The value of the integral $\int_{- 2}^{0} \frac{d x}{\sqrt{12 - x^{2} - 4 x}}$ is

1

π / 2

2

π / 6

3

π / 3

4

- π / 6

Explanation:

(B) : Given,
$I = \int_{- 2}^{0} \frac{d x}{\sqrt{12 - x^{2} - 4 x}}$
$= \int_{- 2}^{0} \frac{dx}{\sqrt{12 - x^{2} - 4 x - 4 + 4}}$
$= \int_{- 2}^{0} \frac{dx}{\sqrt{16 - (x + 2)^{2}}}$
$= {[\sin^{- 1} (\frac{x + 2}{4})]}_{- 2}^{0} = \sin^{- 1} (\frac{2}{4}) = \frac{π}{6}$

AMU-2016

Integral Calculus

86559 $\int_{- 1}^{2} | x^{3} - x | d x$ is equal to

1 11

2 4

3

\frac{11}{4}

4

\frac{4}{11}

Explanation:

(C) : We have,
$\int_{- 1}^{2} | x^{3} - x | d x$
$= \int_{- 1}^{0} (x^{3} - x) d x - \int_{0}^{1} (x^{3} - x) d x + \int_{1}^{2} (x^{3} - x) d x$
$= {[\frac{x^{4}}{4} - \frac{x^{2}}{2}]}_{- 1}^{0} - {[\frac{x^{4}}{4} - \frac{x^{2}}{2}]}_{0}^{1} + {[\frac{x^{4}}{4} - \frac{x^{2}}{2}]}_{1}^{2}$
$= [(0 - 0) - (\frac{1}{4} - \frac{1}{2})] - [(\frac{1}{4} - \frac{1}{2}) - (0 - 0)] + [(4 - 2) - (\frac{1}{4} - \frac{1}{2})]$
$= \frac{1}{4} + \frac{1}{4} + 2 + \frac{1}{4} = \frac{11}{4}$

CG PET-2013

Integral Calculus

86567 The value of $\int_{\frac{π}{3}}^{\frac{π}{2}} \frac{(2 + 3 \sin x)}{\sin x (1 + \cos x)} d x$ is equal to

1

\frac{7}{2} - \sqrt{3} - \log_{e} \sqrt{3}

2

\frac{10}{3} - \sqrt{3} - \log_{e} \sqrt{3}

3

\frac{10}{3} - \sqrt{3} + \log_{e} \sqrt{3}

4

- 2 + 3 \sqrt{3} + \log_{e} \sqrt{3}

Explanation:

(C) : $I = \int_{π / 3}^{π / 2} \frac{2 + 3 \sin x}{\sin x (1 + \cos x)} d x$
$I = \int_{π / 3}^{π / 2} \frac{2}{\sin x (1 + \cos x)} d x + 3 \int_{π / 3}^{π / 2} \frac{\sin x}{\sin x (1 + \cos x)} d x$
$I = 2 \int_{π / 3}^{π / 2} \frac{d x}{\sin x (1 + \cos x)} + 3 \int_{π / 3}^{π / 2} \frac{d x}{(1 + \cos x)}$
$I = 2 I_{1} + 3 I_{2}$
$I_{1} = \int_{π / 3}^{π / 2} \frac{d x}{\sin x (1 + \cos x)} = \int_{π / 3}^{π / 2} \frac{d x}{\frac{2 \tan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}} \times 2 \cos^{2} \frac{x}{2}}$
$I_{1} = \int_{π / 3}^{π / 2} \frac{(1 + \tan^{2} \frac{x}{2}) \cdot \sec^{2} \frac{x}{2}}{4 \tan \frac{x}{2}}$
Let,
$\tan \frac{x}{2} = t \Rightarrow \frac{1}{2} \sec \frac{x}{2} dx = dt$
When, $x = \frac{π}{2}, t = 1$
$x = \frac{π}{3}, t = \frac{1}{\sqrt{3}}$
$I_{1} = \frac{1}{2} \int_{1 / \sqrt{3}}^{1} \frac{1 + t^{2}}{t} d t = \frac{1}{2} {[\ln t + \frac{t^{2}}{2}]}_{\frac{1}{\sqrt{3}}}^{1}$
$I_{1} = \frac{1}{2} [(0 + \frac{1}{2}) - (\ln \frac{1}{\sqrt{3}} + \frac{1}{6})] = \frac{1}{2} (\frac{1}{3} + \ln \sqrt{3})$
$I_{1} = \frac{1}{6} + \frac{1}{2} \ln \sqrt{3}$
$I_{2} = \int_{π / 3}^{π / 2} \frac{d x}{1 + \cos x} = \int_{π / 3}^{π / 2} \frac{1}{1 + \cos x} \times \frac{1 - \cos x}{1 - \cos x} d x$
$I_{2} = \int_{π / 3}^{π / 2} \frac{1 - \cos x}{1 - \cos x} d x$
$I_{2} = \int_{π / 3}^{π / 2} \frac{1 - \cos^{2} x}{\sin^{2} x} d x$
$I_{2} = \int_{π / 3}^{π / 2} ({cosec}^{2} x - \cot x cosec x) d x$
$I_{2} = [cosec x - \cot x]_{π / 3}^{π / 2}$
$I_{2} = 1 - (\frac{2}{\sqrt{3}} - \frac{1}{\sqrt{3}}) = 1 - \frac{1}{\sqrt{3}}$
$I_{2} = 1 - \frac{1}{\sqrt{3}}$
Putting value of $I_{1}$ and $I_{2}$ from equation (ii) and (iii) to equation (i),
$I = 2 I_{1} + 3 I_{2}$
$I = 2 \times (\frac{1}{6} + \frac{1}{2} \ln \sqrt{3}) + 3 (1 - \frac{1}{\sqrt{3}})$
$I = \frac{1}{3} + \ln \sqrt{3} + 3 - \sqrt{3}$
$I = \frac{10}{3} - \sqrt{3} + \log_{e} \sqrt{3}$

JEE Main-2023-31.01.2023

Integral Calculus

86575 $\int_{2}^{4} {| x - 2 | + | x - 3 |} d x =$

1 1

2 2

3 3

4 4

Explanation:

(C): Given,
$I = \int_{2}^{4} {| x - 2 | + | x - 3 |} d x$
$| x - 2 | = {\begin{cases} x - 2, x \geq 2 \\ - (x - 2), x < 2 \end{cases}$ ,
$| x - 3 | = {\begin{cases} x - 3, x \geq 3 \\ - (x - 3), x < 3 \end{cases}$
$= \int_{1}^{3} (| x - 2 | + | x - 3 |) d x + \int_{3}^{4} (| x - 2 | + | x - 3 |) d x$
$= \int_{2}^{3} ((x - 2) + (- (x - 3))) d x + \int_{3}^{4} ((x - 2) + (x - 3)) d x$
$= \int_{2}^{3} [x - 2 - x + 3] d x + \int_{3}^{4} (x - 2 + x - 3) d x$
$= \int_{2}^{3} 1 d x + \int_{3}^{4} (2 x - 5) d x$
$= [x]_{2}^{3} + {[\frac{2 x^{2}}{2} - 5 x]}_{3}^{4}$
$= (3 - 2) + (4^{2} - 5 (4)) - (3^{2} - 5 (3))$
$= 1 + (16 - 20) - (9 - 15)$
$= 1 + (- 4) - (- 6) = 1 - 4 + 6 = 7 - 4 = 3$

AP EAMCET-2021-19.08.2021

Integral Calculus

86576 The value of the integral $\int_{- 2}^{0} \frac{d x}{\sqrt{12 - x^{2} - 4 x}}$ is

1

π / 2

2

π / 6

3

π / 3

4

- π / 6

Explanation:

(B) : Given,
$I = \int_{- 2}^{0} \frac{d x}{\sqrt{12 - x^{2} - 4 x}}$
$= \int_{- 2}^{0} \frac{dx}{\sqrt{12 - x^{2} - 4 x - 4 + 4}}$
$= \int_{- 2}^{0} \frac{dx}{\sqrt{16 - (x + 2)^{2}}}$
$= {[\sin^{- 1} (\frac{x + 2}{4})]}_{- 2}^{0} = \sin^{- 1} (\frac{2}{4}) = \frac{π}{6}$

AMU-2016