QBManager

Integral Calculus

86462 $\int_{0}^{π / 2} \log (\tan x) d x$

1 zero

2 2

3

π / 3

4

π / 4

Explanation:

(A) : Let,
$I = \int_{0}^{π / 2} \log (\tan x) d x$
$I = \int_{0}^{π / 2} \log [\tan (\frac{π}{2} - x)] d x = \int_{0}^{π / 2} \log (\cot x) d x$
$I = - \int_{0}^{π / 2} \log (\tan x) d x$
Adding equation (i) and (ii), we get -
$2 I = \int_{0}^{π / 2} \log (\tan x) d x - \int_{0}^{π / 2} \log (\tan x) d x$
$2 I = 0$
$I = 0 (zero)$

Karnataka CET-2000

Integral Calculus

86463 The value of $\int_{- 1}^{3} [\tan^{- 1} (\frac{x}{x^{2} + 1}) + \tan^{- 1} (\frac{x^{2} + 1}{x})] d x$ is :

1

2 π

2

π

3

π / 2

4

π / 4

Explanation:

(A) : Let,
$I = \int_{- 1}^{3} [\tan^{- 1} (\frac{x}{x^{2} + 1}) + \tan^{- 1} (\frac{x^{2} + 1}{x})] d x$
We know that,
$\tan^{- 1} (\frac{x}{x^{2} + 1}) + \tan^{- 1} (\frac{x^{2} + 1}{x}) = \frac{π}{2}$
$∵ \tan^{- 1} p + \tan^{- 1} (\frac{1}{p}) = \frac{π}{2}$
$∴ I = \int_{- 1}^{3} \frac{π}{2} dx = \frac{π}{2} \int_{- 1}^{3} 1 . dx$
$= \frac{π}{2} [x]_{- 1}^{3} = \frac{π}{2} [3 + 1] = \frac{π}{2} \times 4 = 2 π$

Karnataka CET-2000

Integral Calculus

86464 The value of $\int_{0}^{π} \frac{d x}{5 + 3 \cos x}$ is :

1

π / 4

2

π / 8

3

π / 2

4 zero

Explanation:

(A) : Let, $I = \int_{0}^{π} \frac{d x}{5 + 3 \cos x}$
Put, $\cos x = \frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}$
$I = \int_{0}^{π} \frac{1}{5 + \frac{3 \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}} d x$
$I = \int_{0}^{π} \frac{1}{\frac{5 + 5 \tan^{2} \frac{x}{2} + 3 - 3 \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}} d x$
$I = \int_{0}^{π} \frac{1 + \tan^{2} \frac{x}{2}}{8 + 2 \tan^{2} \frac{x}{2}} d x$
$∴ I = \int_{0}^{π} \frac{\sec^{2} \frac{x}{2}}{8 + 2 \tan^{2} \frac{x}{2}} d x$
Let, $\tan \frac{x}{2} = t$
$\frac{1}{2} \sec^{2} \frac{x}{2} dx = dt$
$I = \int_{0}^{\infty} \frac{dt}{4 + t^{2}} I = \int_{0}^{\infty} \frac{1}{(t)^{2} + (2)^{2}} dt$
$= {[\frac{1}{2} \tan^{- 1} (\frac{t}{2})]}_{0}^{\infty} = \frac{1}{2} [\tan^{- 1} (\infty) - \tan^{- 1} (0)]$
$= \frac{1}{2} [\frac{π}{2} - 0] = \frac{π}{4}$

Karnataka CET-2002

Integral Calculus

86466 If $I_{n} = \int_{0}^{\frac{π}{4}} \tan^{n} x d x$ , where $n$ is positive integer, than $I_{10} + I_{8}$ is equal to

1 9

2

\frac{1}{7}

3

\frac{1}{8}

4

\frac{1}{9}

Explanation:

(D) : Given, $I_{n} = \int_{0}^{π / 4} \tan^{n} x d x$
$I_{n + 2} = \int_{0}^{π / 4} \tan^{n + 2} x d x$
Adding equation (i) and (ii), we get -
$I_{n} + I_{n + 2} = \int_{0}^{π / 4} (\tan^{n} x + \tan^{n + 2} x) d x$
$= \int_{0}^{π / 4} \tan^{n} x (1 + \tan^{2} x) d x$
$= \int_{0}^{π / 4} \tan^{n} x \times \sec^{2} x d x$
Put, $\tan x = t$
$\sec^{2} x d x = d t$
$If, t = 0, \tan x = 0$
And, $t = \frac{π}{4}, \tan x = 1$
$∵ I_{n} + I_{n + 2} = \int_{0}^{1} (t)^{n} dt$
$= {[\frac{t^{n + 1}}{n + 1}]}_{0}^{1} = [\frac{1}{n + 1} - 0]$
$I_{n} + I_{n + 2} = (\frac{1}{n + 1})$
If, $n = 8$
So, $I_{8} + I_{10} = \frac{1}{8 + 1} = \frac{1}{9}$

Karnataka CET-2021

Integral Calculus

86467 $\int \frac{\cos 2 x - \cos 2 θ}{\cos x - \cos θ} d x$ is equal to

1

2 (\sin x + x \cos θ) + C

2

2 (\sin x - x \cos θ) + C

3

2 (\sin x + 2 x \cos θ) + C

4

2 (\sin x - 2 x \cos θ) + C

Explanation:

(A) : $I = \int \frac{\cos 2 x - \cos 2 θ}{\cos x - \cos θ} d x$
$= \int \frac{(2 \cos^{2} x - 1) - (2 \cos^{2} θ - 1)}{\cos x - \cos θ} dx$
$= 2 \int \frac{\cos^{2} x - \cos^{2} θ}{\cos x - \cos θ} d x$
$= 2 \int \frac{(\cos x - \cos θ) (\cos x + \cos θ)}{\cos x - \cos θ} d x$
$= 2 \int (\cos x + \cos θ) d x$
$= 2 [\sin x + x \cos θ] + C$

Karnataka CET-2017

Integral Calculus

86462 $\int_{0}^{π / 2} \log (\tan x) d x$

1 zero

2 2

3

π / 3

4

π / 4

Explanation:

(A) : Let,
$I = \int_{0}^{π / 2} \log (\tan x) d x$
$I = \int_{0}^{π / 2} \log [\tan (\frac{π}{2} - x)] d x = \int_{0}^{π / 2} \log (\cot x) d x$
$I = - \int_{0}^{π / 2} \log (\tan x) d x$
Adding equation (i) and (ii), we get -
$2 I = \int_{0}^{π / 2} \log (\tan x) d x - \int_{0}^{π / 2} \log (\tan x) d x$
$2 I = 0$
$I = 0 (zero)$

Karnataka CET-2000

Integral Calculus

86463 The value of $\int_{- 1}^{3} [\tan^{- 1} (\frac{x}{x^{2} + 1}) + \tan^{- 1} (\frac{x^{2} + 1}{x})] d x$ is :

1

2 π

2

π

3

π / 2

4

π / 4

Explanation:

(A) : Let,
$I = \int_{- 1}^{3} [\tan^{- 1} (\frac{x}{x^{2} + 1}) + \tan^{- 1} (\frac{x^{2} + 1}{x})] d x$
We know that,
$\tan^{- 1} (\frac{x}{x^{2} + 1}) + \tan^{- 1} (\frac{x^{2} + 1}{x}) = \frac{π}{2}$
$∵ \tan^{- 1} p + \tan^{- 1} (\frac{1}{p}) = \frac{π}{2}$
$∴ I = \int_{- 1}^{3} \frac{π}{2} dx = \frac{π}{2} \int_{- 1}^{3} 1 . dx$
$= \frac{π}{2} [x]_{- 1}^{3} = \frac{π}{2} [3 + 1] = \frac{π}{2} \times 4 = 2 π$

Karnataka CET-2000

Integral Calculus

86464 The value of $\int_{0}^{π} \frac{d x}{5 + 3 \cos x}$ is :

1

π / 4

2

π / 8

3

π / 2

4 zero

Explanation:

(A) : Let, $I = \int_{0}^{π} \frac{d x}{5 + 3 \cos x}$
Put, $\cos x = \frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}$
$I = \int_{0}^{π} \frac{1}{5 + \frac{3 \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}} d x$
$I = \int_{0}^{π} \frac{1}{\frac{5 + 5 \tan^{2} \frac{x}{2} + 3 - 3 \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}} d x$
$I = \int_{0}^{π} \frac{1 + \tan^{2} \frac{x}{2}}{8 + 2 \tan^{2} \frac{x}{2}} d x$
$∴ I = \int_{0}^{π} \frac{\sec^{2} \frac{x}{2}}{8 + 2 \tan^{2} \frac{x}{2}} d x$
Let, $\tan \frac{x}{2} = t$
$\frac{1}{2} \sec^{2} \frac{x}{2} dx = dt$
$I = \int_{0}^{\infty} \frac{dt}{4 + t^{2}} I = \int_{0}^{\infty} \frac{1}{(t)^{2} + (2)^{2}} dt$
$= {[\frac{1}{2} \tan^{- 1} (\frac{t}{2})]}_{0}^{\infty} = \frac{1}{2} [\tan^{- 1} (\infty) - \tan^{- 1} (0)]$
$= \frac{1}{2} [\frac{π}{2} - 0] = \frac{π}{4}$

Karnataka CET-2002

Integral Calculus

86466 If $I_{n} = \int_{0}^{\frac{π}{4}} \tan^{n} x d x$ , where $n$ is positive integer, than $I_{10} + I_{8}$ is equal to

1 9

2

\frac{1}{7}

3

\frac{1}{8}

4

\frac{1}{9}

Explanation:

(D) : Given, $I_{n} = \int_{0}^{π / 4} \tan^{n} x d x$
$I_{n + 2} = \int_{0}^{π / 4} \tan^{n + 2} x d x$
Adding equation (i) and (ii), we get -
$I_{n} + I_{n + 2} = \int_{0}^{π / 4} (\tan^{n} x + \tan^{n + 2} x) d x$
$= \int_{0}^{π / 4} \tan^{n} x (1 + \tan^{2} x) d x$
$= \int_{0}^{π / 4} \tan^{n} x \times \sec^{2} x d x$
Put, $\tan x = t$
$\sec^{2} x d x = d t$
$If, t = 0, \tan x = 0$
And, $t = \frac{π}{4}, \tan x = 1$
$∵ I_{n} + I_{n + 2} = \int_{0}^{1} (t)^{n} dt$
$= {[\frac{t^{n + 1}}{n + 1}]}_{0}^{1} = [\frac{1}{n + 1} - 0]$
$I_{n} + I_{n + 2} = (\frac{1}{n + 1})$
If, $n = 8$
So, $I_{8} + I_{10} = \frac{1}{8 + 1} = \frac{1}{9}$

Karnataka CET-2021

Integral Calculus

86467 $\int \frac{\cos 2 x - \cos 2 θ}{\cos x - \cos θ} d x$ is equal to

1

2 (\sin x + x \cos θ) + C

2

2 (\sin x - x \cos θ) + C

3

2 (\sin x + 2 x \cos θ) + C

4

2 (\sin x - 2 x \cos θ) + C

Explanation:

(A) : $I = \int \frac{\cos 2 x - \cos 2 θ}{\cos x - \cos θ} d x$
$= \int \frac{(2 \cos^{2} x - 1) - (2 \cos^{2} θ - 1)}{\cos x - \cos θ} dx$
$= 2 \int \frac{\cos^{2} x - \cos^{2} θ}{\cos x - \cos θ} d x$
$= 2 \int \frac{(\cos x - \cos θ) (\cos x + \cos θ)}{\cos x - \cos θ} d x$
$= 2 \int (\cos x + \cos θ) d x$
$= 2 [\sin x + x \cos θ] + C$

Karnataka CET-2017

Integral Calculus

86462 $\int_{0}^{π / 2} \log (\tan x) d x$

1 zero

2 2

3

π / 3

4

π / 4

Explanation:

(A) : Let,
$I = \int_{0}^{π / 2} \log (\tan x) d x$
$I = \int_{0}^{π / 2} \log [\tan (\frac{π}{2} - x)] d x = \int_{0}^{π / 2} \log (\cot x) d x$
$I = - \int_{0}^{π / 2} \log (\tan x) d x$
Adding equation (i) and (ii), we get -
$2 I = \int_{0}^{π / 2} \log (\tan x) d x - \int_{0}^{π / 2} \log (\tan x) d x$
$2 I = 0$
$I = 0 (zero)$

Karnataka CET-2000

Integral Calculus

86463 The value of $\int_{- 1}^{3} [\tan^{- 1} (\frac{x}{x^{2} + 1}) + \tan^{- 1} (\frac{x^{2} + 1}{x})] d x$ is :

1

2 π

2

π

3

π / 2

4

π / 4

Explanation:

(A) : Let,
$I = \int_{- 1}^{3} [\tan^{- 1} (\frac{x}{x^{2} + 1}) + \tan^{- 1} (\frac{x^{2} + 1}{x})] d x$
We know that,
$\tan^{- 1} (\frac{x}{x^{2} + 1}) + \tan^{- 1} (\frac{x^{2} + 1}{x}) = \frac{π}{2}$
$∵ \tan^{- 1} p + \tan^{- 1} (\frac{1}{p}) = \frac{π}{2}$
$∴ I = \int_{- 1}^{3} \frac{π}{2} dx = \frac{π}{2} \int_{- 1}^{3} 1 . dx$
$= \frac{π}{2} [x]_{- 1}^{3} = \frac{π}{2} [3 + 1] = \frac{π}{2} \times 4 = 2 π$

Karnataka CET-2000

Integral Calculus

86464 The value of $\int_{0}^{π} \frac{d x}{5 + 3 \cos x}$ is :

1

π / 4

2

π / 8

3

π / 2

4 zero

Explanation:

(A) : Let, $I = \int_{0}^{π} \frac{d x}{5 + 3 \cos x}$
Put, $\cos x = \frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}$
$I = \int_{0}^{π} \frac{1}{5 + \frac{3 \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}} d x$
$I = \int_{0}^{π} \frac{1}{\frac{5 + 5 \tan^{2} \frac{x}{2} + 3 - 3 \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}} d x$
$I = \int_{0}^{π} \frac{1 + \tan^{2} \frac{x}{2}}{8 + 2 \tan^{2} \frac{x}{2}} d x$
$∴ I = \int_{0}^{π} \frac{\sec^{2} \frac{x}{2}}{8 + 2 \tan^{2} \frac{x}{2}} d x$
Let, $\tan \frac{x}{2} = t$
$\frac{1}{2} \sec^{2} \frac{x}{2} dx = dt$
$I = \int_{0}^{\infty} \frac{dt}{4 + t^{2}} I = \int_{0}^{\infty} \frac{1}{(t)^{2} + (2)^{2}} dt$
$= {[\frac{1}{2} \tan^{- 1} (\frac{t}{2})]}_{0}^{\infty} = \frac{1}{2} [\tan^{- 1} (\infty) - \tan^{- 1} (0)]$
$= \frac{1}{2} [\frac{π}{2} - 0] = \frac{π}{4}$

Karnataka CET-2002

Integral Calculus

86466 If $I_{n} = \int_{0}^{\frac{π}{4}} \tan^{n} x d x$ , where $n$ is positive integer, than $I_{10} + I_{8}$ is equal to

1 9

2

\frac{1}{7}

3

\frac{1}{8}

4

\frac{1}{9}

Explanation:

(D) : Given, $I_{n} = \int_{0}^{π / 4} \tan^{n} x d x$
$I_{n + 2} = \int_{0}^{π / 4} \tan^{n + 2} x d x$
Adding equation (i) and (ii), we get -
$I_{n} + I_{n + 2} = \int_{0}^{π / 4} (\tan^{n} x + \tan^{n + 2} x) d x$
$= \int_{0}^{π / 4} \tan^{n} x (1 + \tan^{2} x) d x$
$= \int_{0}^{π / 4} \tan^{n} x \times \sec^{2} x d x$
Put, $\tan x = t$
$\sec^{2} x d x = d t$
$If, t = 0, \tan x = 0$
And, $t = \frac{π}{4}, \tan x = 1$
$∵ I_{n} + I_{n + 2} = \int_{0}^{1} (t)^{n} dt$
$= {[\frac{t^{n + 1}}{n + 1}]}_{0}^{1} = [\frac{1}{n + 1} - 0]$
$I_{n} + I_{n + 2} = (\frac{1}{n + 1})$
If, $n = 8$
So, $I_{8} + I_{10} = \frac{1}{8 + 1} = \frac{1}{9}$

Karnataka CET-2021

Integral Calculus

86467 $\int \frac{\cos 2 x - \cos 2 θ}{\cos x - \cos θ} d x$ is equal to

1

2 (\sin x + x \cos θ) + C

2

2 (\sin x - x \cos θ) + C

3

2 (\sin x + 2 x \cos θ) + C

4

2 (\sin x - 2 x \cos θ) + C

Explanation:

(A) : $I = \int \frac{\cos 2 x - \cos 2 θ}{\cos x - \cos θ} d x$
$= \int \frac{(2 \cos^{2} x - 1) - (2 \cos^{2} θ - 1)}{\cos x - \cos θ} dx$
$= 2 \int \frac{\cos^{2} x - \cos^{2} θ}{\cos x - \cos θ} d x$
$= 2 \int \frac{(\cos x - \cos θ) (\cos x + \cos θ)}{\cos x - \cos θ} d x$
$= 2 \int (\cos x + \cos θ) d x$
$= 2 [\sin x + x \cos θ] + C$

Karnataka CET-2017

Integral Calculus

86462 $\int_{0}^{π / 2} \log (\tan x) d x$

1 zero

2 2

3

π / 3

4

π / 4

Explanation:

(A) : Let,
$I = \int_{0}^{π / 2} \log (\tan x) d x$
$I = \int_{0}^{π / 2} \log [\tan (\frac{π}{2} - x)] d x = \int_{0}^{π / 2} \log (\cot x) d x$
$I = - \int_{0}^{π / 2} \log (\tan x) d x$
Adding equation (i) and (ii), we get -
$2 I = \int_{0}^{π / 2} \log (\tan x) d x - \int_{0}^{π / 2} \log (\tan x) d x$
$2 I = 0$
$I = 0 (zero)$

Karnataka CET-2000

Integral Calculus

86463 The value of $\int_{- 1}^{3} [\tan^{- 1} (\frac{x}{x^{2} + 1}) + \tan^{- 1} (\frac{x^{2} + 1}{x})] d x$ is :

1

2 π

2

π

3

π / 2

4

π / 4

Explanation:

(A) : Let,
$I = \int_{- 1}^{3} [\tan^{- 1} (\frac{x}{x^{2} + 1}) + \tan^{- 1} (\frac{x^{2} + 1}{x})] d x$
We know that,
$\tan^{- 1} (\frac{x}{x^{2} + 1}) + \tan^{- 1} (\frac{x^{2} + 1}{x}) = \frac{π}{2}$
$∵ \tan^{- 1} p + \tan^{- 1} (\frac{1}{p}) = \frac{π}{2}$
$∴ I = \int_{- 1}^{3} \frac{π}{2} dx = \frac{π}{2} \int_{- 1}^{3} 1 . dx$
$= \frac{π}{2} [x]_{- 1}^{3} = \frac{π}{2} [3 + 1] = \frac{π}{2} \times 4 = 2 π$

Karnataka CET-2000

Integral Calculus

86464 The value of $\int_{0}^{π} \frac{d x}{5 + 3 \cos x}$ is :

1

π / 4

2

π / 8

3

π / 2

4 zero

Explanation:

(A) : Let, $I = \int_{0}^{π} \frac{d x}{5 + 3 \cos x}$
Put, $\cos x = \frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}$
$I = \int_{0}^{π} \frac{1}{5 + \frac{3 \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}} d x$
$I = \int_{0}^{π} \frac{1}{\frac{5 + 5 \tan^{2} \frac{x}{2} + 3 - 3 \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}} d x$
$I = \int_{0}^{π} \frac{1 + \tan^{2} \frac{x}{2}}{8 + 2 \tan^{2} \frac{x}{2}} d x$
$∴ I = \int_{0}^{π} \frac{\sec^{2} \frac{x}{2}}{8 + 2 \tan^{2} \frac{x}{2}} d x$
Let, $\tan \frac{x}{2} = t$
$\frac{1}{2} \sec^{2} \frac{x}{2} dx = dt$
$I = \int_{0}^{\infty} \frac{dt}{4 + t^{2}} I = \int_{0}^{\infty} \frac{1}{(t)^{2} + (2)^{2}} dt$
$= {[\frac{1}{2} \tan^{- 1} (\frac{t}{2})]}_{0}^{\infty} = \frac{1}{2} [\tan^{- 1} (\infty) - \tan^{- 1} (0)]$
$= \frac{1}{2} [\frac{π}{2} - 0] = \frac{π}{4}$

Karnataka CET-2002

Integral Calculus

86466 If $I_{n} = \int_{0}^{\frac{π}{4}} \tan^{n} x d x$ , where $n$ is positive integer, than $I_{10} + I_{8}$ is equal to

1 9

2

\frac{1}{7}

3

\frac{1}{8}

4

\frac{1}{9}

Explanation:

(D) : Given, $I_{n} = \int_{0}^{π / 4} \tan^{n} x d x$
$I_{n + 2} = \int_{0}^{π / 4} \tan^{n + 2} x d x$
Adding equation (i) and (ii), we get -
$I_{n} + I_{n + 2} = \int_{0}^{π / 4} (\tan^{n} x + \tan^{n + 2} x) d x$
$= \int_{0}^{π / 4} \tan^{n} x (1 + \tan^{2} x) d x$
$= \int_{0}^{π / 4} \tan^{n} x \times \sec^{2} x d x$
Put, $\tan x = t$
$\sec^{2} x d x = d t$
$If, t = 0, \tan x = 0$
And, $t = \frac{π}{4}, \tan x = 1$
$∵ I_{n} + I_{n + 2} = \int_{0}^{1} (t)^{n} dt$
$= {[\frac{t^{n + 1}}{n + 1}]}_{0}^{1} = [\frac{1}{n + 1} - 0]$
$I_{n} + I_{n + 2} = (\frac{1}{n + 1})$
If, $n = 8$
So, $I_{8} + I_{10} = \frac{1}{8 + 1} = \frac{1}{9}$

Karnataka CET-2021

Integral Calculus

86467 $\int \frac{\cos 2 x - \cos 2 θ}{\cos x - \cos θ} d x$ is equal to

1

2 (\sin x + x \cos θ) + C

2

2 (\sin x - x \cos θ) + C

3

2 (\sin x + 2 x \cos θ) + C

4

2 (\sin x - 2 x \cos θ) + C

Explanation:

(A) : $I = \int \frac{\cos 2 x - \cos 2 θ}{\cos x - \cos θ} d x$
$= \int \frac{(2 \cos^{2} x - 1) - (2 \cos^{2} θ - 1)}{\cos x - \cos θ} dx$
$= 2 \int \frac{\cos^{2} x - \cos^{2} θ}{\cos x - \cos θ} d x$
$= 2 \int \frac{(\cos x - \cos θ) (\cos x + \cos θ)}{\cos x - \cos θ} d x$
$= 2 \int (\cos x + \cos θ) d x$
$= 2 [\sin x + x \cos θ] + C$

Karnataka CET-2017

Integral Calculus

86462 $\int_{0}^{π / 2} \log (\tan x) d x$

1 zero

2 2

3

π / 3

4

π / 4

Explanation:

(A) : Let,
$I = \int_{0}^{π / 2} \log (\tan x) d x$
$I = \int_{0}^{π / 2} \log [\tan (\frac{π}{2} - x)] d x = \int_{0}^{π / 2} \log (\cot x) d x$
$I = - \int_{0}^{π / 2} \log (\tan x) d x$
Adding equation (i) and (ii), we get -
$2 I = \int_{0}^{π / 2} \log (\tan x) d x - \int_{0}^{π / 2} \log (\tan x) d x$
$2 I = 0$
$I = 0 (zero)$

Karnataka CET-2000

Integral Calculus

86463 The value of $\int_{- 1}^{3} [\tan^{- 1} (\frac{x}{x^{2} + 1}) + \tan^{- 1} (\frac{x^{2} + 1}{x})] d x$ is :

1

2 π

2

π

3

π / 2

4

π / 4

Explanation:

(A) : Let,
$I = \int_{- 1}^{3} [\tan^{- 1} (\frac{x}{x^{2} + 1}) + \tan^{- 1} (\frac{x^{2} + 1}{x})] d x$
We know that,
$\tan^{- 1} (\frac{x}{x^{2} + 1}) + \tan^{- 1} (\frac{x^{2} + 1}{x}) = \frac{π}{2}$
$∵ \tan^{- 1} p + \tan^{- 1} (\frac{1}{p}) = \frac{π}{2}$
$∴ I = \int_{- 1}^{3} \frac{π}{2} dx = \frac{π}{2} \int_{- 1}^{3} 1 . dx$
$= \frac{π}{2} [x]_{- 1}^{3} = \frac{π}{2} [3 + 1] = \frac{π}{2} \times 4 = 2 π$

Karnataka CET-2000

Integral Calculus

86464 The value of $\int_{0}^{π} \frac{d x}{5 + 3 \cos x}$ is :

1

π / 4

2

π / 8

3

π / 2

4 zero

Explanation:

(A) : Let, $I = \int_{0}^{π} \frac{d x}{5 + 3 \cos x}$
Put, $\cos x = \frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}$
$I = \int_{0}^{π} \frac{1}{5 + \frac{3 \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}} d x$
$I = \int_{0}^{π} \frac{1}{\frac{5 + 5 \tan^{2} \frac{x}{2} + 3 - 3 \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}} d x$
$I = \int_{0}^{π} \frac{1 + \tan^{2} \frac{x}{2}}{8 + 2 \tan^{2} \frac{x}{2}} d x$
$∴ I = \int_{0}^{π} \frac{\sec^{2} \frac{x}{2}}{8 + 2 \tan^{2} \frac{x}{2}} d x$
Let, $\tan \frac{x}{2} = t$
$\frac{1}{2} \sec^{2} \frac{x}{2} dx = dt$
$I = \int_{0}^{\infty} \frac{dt}{4 + t^{2}} I = \int_{0}^{\infty} \frac{1}{(t)^{2} + (2)^{2}} dt$
$= {[\frac{1}{2} \tan^{- 1} (\frac{t}{2})]}_{0}^{\infty} = \frac{1}{2} [\tan^{- 1} (\infty) - \tan^{- 1} (0)]$
$= \frac{1}{2} [\frac{π}{2} - 0] = \frac{π}{4}$

Karnataka CET-2002

Integral Calculus

86466 If $I_{n} = \int_{0}^{\frac{π}{4}} \tan^{n} x d x$ , where $n$ is positive integer, than $I_{10} + I_{8}$ is equal to

1 9

2

\frac{1}{7}

3

\frac{1}{8}

4

\frac{1}{9}

Explanation:

(D) : Given, $I_{n} = \int_{0}^{π / 4} \tan^{n} x d x$
$I_{n + 2} = \int_{0}^{π / 4} \tan^{n + 2} x d x$
Adding equation (i) and (ii), we get -
$I_{n} + I_{n + 2} = \int_{0}^{π / 4} (\tan^{n} x + \tan^{n + 2} x) d x$
$= \int_{0}^{π / 4} \tan^{n} x (1 + \tan^{2} x) d x$
$= \int_{0}^{π / 4} \tan^{n} x \times \sec^{2} x d x$
Put, $\tan x = t$
$\sec^{2} x d x = d t$
$If, t = 0, \tan x = 0$
And, $t = \frac{π}{4}, \tan x = 1$
$∵ I_{n} + I_{n + 2} = \int_{0}^{1} (t)^{n} dt$
$= {[\frac{t^{n + 1}}{n + 1}]}_{0}^{1} = [\frac{1}{n + 1} - 0]$
$I_{n} + I_{n + 2} = (\frac{1}{n + 1})$
If, $n = 8$
So, $I_{8} + I_{10} = \frac{1}{8 + 1} = \frac{1}{9}$

Karnataka CET-2021

Integral Calculus

86467 $\int \frac{\cos 2 x - \cos 2 θ}{\cos x - \cos θ} d x$ is equal to

1

2 (\sin x + x \cos θ) + C

2

2 (\sin x - x \cos θ) + C

3

2 (\sin x + 2 x \cos θ) + C

4

2 (\sin x - 2 x \cos θ) + C

Explanation:

(A) : $I = \int \frac{\cos 2 x - \cos 2 θ}{\cos x - \cos θ} d x$
$= \int \frac{(2 \cos^{2} x - 1) - (2 \cos^{2} θ - 1)}{\cos x - \cos θ} dx$
$= 2 \int \frac{\cos^{2} x - \cos^{2} θ}{\cos x - \cos θ} d x$
$= 2 \int \frac{(\cos x - \cos θ) (\cos x + \cos θ)}{\cos x - \cos θ} d x$
$= 2 \int (\cos x + \cos θ) d x$
$= 2 [\sin x + x \cos θ] + C$

Karnataka CET-2017