QBManager

Integral Calculus

86256 If $x > 0, \int \frac{x^{3} - 7 x + 6}{x (x + 3)} d x = a x + b x^{2} + c \ln x + d$ , then $a + b + c =$

1 0

2

\frac{1}{2}

3

- \frac{1}{2}

4 -1

Explanation:

(C) : Given,
$\int \frac{x^{3} - 7 x + 6}{x (x + 3)} d x = a x + b x^{2} + c \ln x + d$
L.H.S., $I = \int \frac{x^{3} - 7 x + 6}{x (x + 3)} d x$
$= \int \frac{x^{3} - 7 x + 6}{x^{2} + 3 x} d x$
$= \int (x - 3 + \frac{2 x + 6}{x^{2} + 3 x}) d x$
$= \int x \cdot d x - \int 3 d x + \int \frac{2 x + 3}{x^{2} + 3 x} d x + 3 \int \frac{1}{x^{2} + 3 x} d x$
$= \frac{x^{2}}{2} - 3 x + \ln (x^{2} + 3 x) + 3 \int \frac{1}{x (x + 3)} d x$
$= \frac{x^{2}}{2} - 3 x + \ln (x^{2} + 3 x) + \frac{3 \times 1}{3} \int (\frac{1}{x} - \frac{1}{x + 3}) d x$
$= \frac{x^{2}}{2} - 3 x + \ln (x^{2} + 3 x) + \ln x - \ln (x + 3) + C$
$= \frac{x^{2}}{2} - 3 x + \ln (x^{2} + 3 x \times \frac{x}{x + 3}) + C$
$= \frac{x^{2}}{2} - 3 x + \ln (x (x + 3) \times \frac{x}{x + 3}) + C$
$= \frac{x^{2}}{2} - 3 x + \ln x^{2} + C$
$= \frac{x^{2}}{2} - 3 x + 2 \ln x + C$
On comparing L.H.S. and R.H.S., we get -
$a = - 3, b = \frac{1}{2}, c = 2$
Then, $a + b + c = - 3 + \frac{1}{2} + 2 = \frac{- 1}{2}$

COMEDK-2016

Integral Calculus

86257 $\int_{0}^{π / 4} \frac{\sin x \cos x}{\cos^{4} x + \sin^{4} x} d x$ is equal to

1

8 π

2

π / 4

3

4 π

4

π / 8

Explanation:

(D) : Let,
$I = \int_{0}^{π / 4} \frac{\sin x \cdot \cos x}{\cos^{4} x + \sin^{4} x} d x$
$= \int_{0}^{π / 4} \frac{\sin x \cdot \cos x}{\cos^{4} x + \sin^{4} x + 2 \sin^{2} \cdot \cos^{2} x - 2 \sin^{2} x \cdot \cos^{2} x} \cdot d x$
$= \int_{0}^{π / 4} \frac{\sin x \cdot \cos x}{{(\cos^{2} x + \sin^{2} x)}^{2} - 2 \sin^{2} x \cdot \cos^{2} x} d x$
$= \int_{0}^{π / 4} \frac{\sin x \cdot \cos x}{1 - \frac{4 \sin^{2} x \cdot \cos^{2} x}{2}} d x$
$= \int_{0}^{π / 4} \frac{2 \sin x \cdot \cos x}{2 - \sin^{2} 2 x} d x = \int_{0}^{π / 4} \frac{\sin 2 x}{1 + \cos^{2} 2 x} d x$
Let, $\cos 2 x = t$
$- 2 \sin 2 x . dx = dt$
$∴ I = \frac{- 1}{2} \int_{1}^{0} \frac{dt}{1 + t^{2}}$
$= \frac{- 1}{2} {[\tan^{- 1} (t)]}_{1}^{0} = \frac{- 1}{2} [\tan^{- 1} (0) - \tan^{- 1} (1)]$
$= \frac{- 1}{2} [0 - \frac{π}{4}] = \frac{π}{8}$

COMEDK-2019

Integral Calculus

86258 $\int \frac{d x}{1 - \cos x - \sin x}$ is equal to

1

\log | 1 + \cot \frac{x}{2} | + C

2

\log | 1 - \tan \frac{x}{2} | + C

3

\log | 1 - \cot \frac{x}{2} | + C

4

\log | 1 + \tan \frac{x}{2} | + C

Explanation:

(C) :Let, $I = \int \frac{d x}{1 - \cos x - \sin x}$
Put, $\cos x = \frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}$ and $\sin x = \frac{2 \tan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}$
$∴ I = \int \frac{d x}{1 - (\frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}) - \frac{2 \tan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}}$
$= \int \frac{\sec^{2} \frac{x}{2} d x}{2 \tan^{2} \frac{x}{2} - 2 \tan \frac{x}{2}} = \int \frac{\frac{1}{2} \sec^{2} \frac{x}{2} d x}{\tan^{2} \frac{x}{2} - \tan \frac{x}{2}}$
Put, $\tan \frac{x}{2} = t \Rightarrow \frac{1}{2} \sec^{2} \frac{x}{2} dx = dt$
$∴ I = \int \frac{dt}{t^{2} - t} = \int [\frac{1}{t - 1} - \frac{1}{t}] dt$
$= \log (t - 1) - \log t + C = \log | \frac{t - 1}{t} | + C$
$= \log | \frac{\tan \frac{x}{2} - 1}{\tan \frac{x}{2}} | + C = \log | 1 - \cot \frac{x}{2} | + C$

COMEDK-2020

Integral Calculus

86259 Evaluate : $\int_{0}^{π / 2} \frac{1}{a^{2} \sin^{2} x + b^{2} \cos^{2} x} d x$

1

\frac{π a}{4 b}

2

\frac{π a}{2 b}

3

\frac{π b}{4 a}

4

\frac{π}{2 ab}

Explanation:

(D) : Let,
$I = \int_{0}^{π / 2} \frac{1}{a^{2} \sin^{2} x + b^{2} \cos^{2} x} d x$
Dividing numerator and denominator by $\cos^{2} x$ we get
$I = \int_{0}^{π / 2} \frac{\sec^{2} x d x}{a^{2} \tan^{2} x + b^{2}}$
Put, $\tan x = t \Rightarrow \sec^{2} x d x = d t$
When, $x = 0 \Rightarrow t = 0$ and when $x = \frac{π}{2} \Rightarrow t = \infty$
$\Rightarrow I = \int_{0}^{\infty} \frac{dt}{a^{2} t^{2} + b^{2}} = \frac{1}{a^{2}} \int_{0}^{\infty} \frac{dt}{t^{2} + {(\frac{b}{a})}^{2}}$
$= \frac{1}{a^{2}} \times \frac{1}{(\frac{b}{a})} {[\tan^{- 1} \frac{t}{(b / a)}]}_{0}^{\infty} = \frac{1}{ab} \times \frac{π}{2} = \frac{π}{2 ab}$

COMEDK-2020

Integral Calculus

86262 Evaluate : $\int \frac{1}{1 + 3 \sin^{2} x + 8 \cos^{2} x} d x$

1

\frac{1}{6} \tan^{- 1} (2 \tan x) + C

2

\tan^{- 1} (2 \tan x) + C

3

\frac{1}{6} \tan^{- 1} (\frac{2 \tan x}{3}) + C

4 None of these

Explanation:

(C) : Let, $I = \int \frac{1}{1 + 3 \sin^{2} x + 8 \cos^{2} x} d x$
Dividing the numerator and denominator by $\cos^{2} x$ , we get
$I = \int \frac{\sec^{2} x}{\sec^{2} x + 3 \tan^{2} x + 8} d x = \int \frac{\sec^{2} x}{4 \tan^{2} x + 9} d x$
Putting $\tan x = t \Rightarrow \sec^{2} xdx = dt$ , we get
$I = \int \frac{dt}{4 t^{2} + 9} = \frac{1}{4} \int \frac{dt}{t^{2} + (3 / 2)^{2}} = \frac{1}{4} \times \frac{1}{3 / 2} \tan^{- 1} (\frac{t}{3 / 2}) + C$
$\Rightarrow I = \frac{1}{6} \tan^{- 1} (\frac{2 t}{3}) + C = \frac{1}{6} \tan^{- 1} (\frac{2 \tan x}{3}) + C$

BITSAT-2014

Integral Calculus

86256 If $x > 0, \int \frac{x^{3} - 7 x + 6}{x (x + 3)} d x = a x + b x^{2} + c \ln x + d$ , then $a + b + c =$

1 0

2

\frac{1}{2}

3

- \frac{1}{2}

4 -1

Explanation:

(C) : Given,
$\int \frac{x^{3} - 7 x + 6}{x (x + 3)} d x = a x + b x^{2} + c \ln x + d$
L.H.S., $I = \int \frac{x^{3} - 7 x + 6}{x (x + 3)} d x$
$= \int \frac{x^{3} - 7 x + 6}{x^{2} + 3 x} d x$
$= \int (x - 3 + \frac{2 x + 6}{x^{2} + 3 x}) d x$
$= \int x \cdot d x - \int 3 d x + \int \frac{2 x + 3}{x^{2} + 3 x} d x + 3 \int \frac{1}{x^{2} + 3 x} d x$
$= \frac{x^{2}}{2} - 3 x + \ln (x^{2} + 3 x) + 3 \int \frac{1}{x (x + 3)} d x$
$= \frac{x^{2}}{2} - 3 x + \ln (x^{2} + 3 x) + \frac{3 \times 1}{3} \int (\frac{1}{x} - \frac{1}{x + 3}) d x$
$= \frac{x^{2}}{2} - 3 x + \ln (x^{2} + 3 x) + \ln x - \ln (x + 3) + C$
$= \frac{x^{2}}{2} - 3 x + \ln (x^{2} + 3 x \times \frac{x}{x + 3}) + C$
$= \frac{x^{2}}{2} - 3 x + \ln (x (x + 3) \times \frac{x}{x + 3}) + C$
$= \frac{x^{2}}{2} - 3 x + \ln x^{2} + C$
$= \frac{x^{2}}{2} - 3 x + 2 \ln x + C$
On comparing L.H.S. and R.H.S., we get -
$a = - 3, b = \frac{1}{2}, c = 2$
Then, $a + b + c = - 3 + \frac{1}{2} + 2 = \frac{- 1}{2}$

COMEDK-2016

Integral Calculus

86257 $\int_{0}^{π / 4} \frac{\sin x \cos x}{\cos^{4} x + \sin^{4} x} d x$ is equal to

1

8 π

2

π / 4

3

4 π

4

π / 8

Explanation:

(D) : Let,
$I = \int_{0}^{π / 4} \frac{\sin x \cdot \cos x}{\cos^{4} x + \sin^{4} x} d x$
$= \int_{0}^{π / 4} \frac{\sin x \cdot \cos x}{\cos^{4} x + \sin^{4} x + 2 \sin^{2} \cdot \cos^{2} x - 2 \sin^{2} x \cdot \cos^{2} x} \cdot d x$
$= \int_{0}^{π / 4} \frac{\sin x \cdot \cos x}{{(\cos^{2} x + \sin^{2} x)}^{2} - 2 \sin^{2} x \cdot \cos^{2} x} d x$
$= \int_{0}^{π / 4} \frac{\sin x \cdot \cos x}{1 - \frac{4 \sin^{2} x \cdot \cos^{2} x}{2}} d x$
$= \int_{0}^{π / 4} \frac{2 \sin x \cdot \cos x}{2 - \sin^{2} 2 x} d x = \int_{0}^{π / 4} \frac{\sin 2 x}{1 + \cos^{2} 2 x} d x$
Let, $\cos 2 x = t$
$- 2 \sin 2 x . dx = dt$
$∴ I = \frac{- 1}{2} \int_{1}^{0} \frac{dt}{1 + t^{2}}$
$= \frac{- 1}{2} {[\tan^{- 1} (t)]}_{1}^{0} = \frac{- 1}{2} [\tan^{- 1} (0) - \tan^{- 1} (1)]$
$= \frac{- 1}{2} [0 - \frac{π}{4}] = \frac{π}{8}$

COMEDK-2019

Integral Calculus

86258 $\int \frac{d x}{1 - \cos x - \sin x}$ is equal to

1

\log | 1 + \cot \frac{x}{2} | + C

2

\log | 1 - \tan \frac{x}{2} | + C

3

\log | 1 - \cot \frac{x}{2} | + C

4

\log | 1 + \tan \frac{x}{2} | + C

Explanation:

(C) :Let, $I = \int \frac{d x}{1 - \cos x - \sin x}$
Put, $\cos x = \frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}$ and $\sin x = \frac{2 \tan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}$
$∴ I = \int \frac{d x}{1 - (\frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}) - \frac{2 \tan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}}$
$= \int \frac{\sec^{2} \frac{x}{2} d x}{2 \tan^{2} \frac{x}{2} - 2 \tan \frac{x}{2}} = \int \frac{\frac{1}{2} \sec^{2} \frac{x}{2} d x}{\tan^{2} \frac{x}{2} - \tan \frac{x}{2}}$
Put, $\tan \frac{x}{2} = t \Rightarrow \frac{1}{2} \sec^{2} \frac{x}{2} dx = dt$
$∴ I = \int \frac{dt}{t^{2} - t} = \int [\frac{1}{t - 1} - \frac{1}{t}] dt$
$= \log (t - 1) - \log t + C = \log | \frac{t - 1}{t} | + C$
$= \log | \frac{\tan \frac{x}{2} - 1}{\tan \frac{x}{2}} | + C = \log | 1 - \cot \frac{x}{2} | + C$

COMEDK-2020

Integral Calculus

86259 Evaluate : $\int_{0}^{π / 2} \frac{1}{a^{2} \sin^{2} x + b^{2} \cos^{2} x} d x$

1

\frac{π a}{4 b}

2

\frac{π a}{2 b}

3

\frac{π b}{4 a}

4

\frac{π}{2 ab}

Explanation:

(D) : Let,
$I = \int_{0}^{π / 2} \frac{1}{a^{2} \sin^{2} x + b^{2} \cos^{2} x} d x$
Dividing numerator and denominator by $\cos^{2} x$ we get
$I = \int_{0}^{π / 2} \frac{\sec^{2} x d x}{a^{2} \tan^{2} x + b^{2}}$
Put, $\tan x = t \Rightarrow \sec^{2} x d x = d t$
When, $x = 0 \Rightarrow t = 0$ and when $x = \frac{π}{2} \Rightarrow t = \infty$
$\Rightarrow I = \int_{0}^{\infty} \frac{dt}{a^{2} t^{2} + b^{2}} = \frac{1}{a^{2}} \int_{0}^{\infty} \frac{dt}{t^{2} + {(\frac{b}{a})}^{2}}$
$= \frac{1}{a^{2}} \times \frac{1}{(\frac{b}{a})} {[\tan^{- 1} \frac{t}{(b / a)}]}_{0}^{\infty} = \frac{1}{ab} \times \frac{π}{2} = \frac{π}{2 ab}$

COMEDK-2020

Integral Calculus

86262 Evaluate : $\int \frac{1}{1 + 3 \sin^{2} x + 8 \cos^{2} x} d x$

1

\frac{1}{6} \tan^{- 1} (2 \tan x) + C

2

\tan^{- 1} (2 \tan x) + C

3

\frac{1}{6} \tan^{- 1} (\frac{2 \tan x}{3}) + C

4 None of these

Explanation:

(C) : Let, $I = \int \frac{1}{1 + 3 \sin^{2} x + 8 \cos^{2} x} d x$
Dividing the numerator and denominator by $\cos^{2} x$ , we get
$I = \int \frac{\sec^{2} x}{\sec^{2} x + 3 \tan^{2} x + 8} d x = \int \frac{\sec^{2} x}{4 \tan^{2} x + 9} d x$
Putting $\tan x = t \Rightarrow \sec^{2} xdx = dt$ , we get
$I = \int \frac{dt}{4 t^{2} + 9} = \frac{1}{4} \int \frac{dt}{t^{2} + (3 / 2)^{2}} = \frac{1}{4} \times \frac{1}{3 / 2} \tan^{- 1} (\frac{t}{3 / 2}) + C$
$\Rightarrow I = \frac{1}{6} \tan^{- 1} (\frac{2 t}{3}) + C = \frac{1}{6} \tan^{- 1} (\frac{2 \tan x}{3}) + C$

BITSAT-2014

Integral Calculus

86256 If $x > 0, \int \frac{x^{3} - 7 x + 6}{x (x + 3)} d x = a x + b x^{2} + c \ln x + d$ , then $a + b + c =$

1 0

2

\frac{1}{2}

3

- \frac{1}{2}

4 -1

Explanation:

(C) : Given,
$\int \frac{x^{3} - 7 x + 6}{x (x + 3)} d x = a x + b x^{2} + c \ln x + d$
L.H.S., $I = \int \frac{x^{3} - 7 x + 6}{x (x + 3)} d x$
$= \int \frac{x^{3} - 7 x + 6}{x^{2} + 3 x} d x$
$= \int (x - 3 + \frac{2 x + 6}{x^{2} + 3 x}) d x$
$= \int x \cdot d x - \int 3 d x + \int \frac{2 x + 3}{x^{2} + 3 x} d x + 3 \int \frac{1}{x^{2} + 3 x} d x$
$= \frac{x^{2}}{2} - 3 x + \ln (x^{2} + 3 x) + 3 \int \frac{1}{x (x + 3)} d x$
$= \frac{x^{2}}{2} - 3 x + \ln (x^{2} + 3 x) + \frac{3 \times 1}{3} \int (\frac{1}{x} - \frac{1}{x + 3}) d x$
$= \frac{x^{2}}{2} - 3 x + \ln (x^{2} + 3 x) + \ln x - \ln (x + 3) + C$
$= \frac{x^{2}}{2} - 3 x + \ln (x^{2} + 3 x \times \frac{x}{x + 3}) + C$
$= \frac{x^{2}}{2} - 3 x + \ln (x (x + 3) \times \frac{x}{x + 3}) + C$
$= \frac{x^{2}}{2} - 3 x + \ln x^{2} + C$
$= \frac{x^{2}}{2} - 3 x + 2 \ln x + C$
On comparing L.H.S. and R.H.S., we get -
$a = - 3, b = \frac{1}{2}, c = 2$
Then, $a + b + c = - 3 + \frac{1}{2} + 2 = \frac{- 1}{2}$

COMEDK-2016

Integral Calculus

86257 $\int_{0}^{π / 4} \frac{\sin x \cos x}{\cos^{4} x + \sin^{4} x} d x$ is equal to

1

8 π

2

π / 4

3

4 π

4

π / 8

Explanation:

(D) : Let,
$I = \int_{0}^{π / 4} \frac{\sin x \cdot \cos x}{\cos^{4} x + \sin^{4} x} d x$
$= \int_{0}^{π / 4} \frac{\sin x \cdot \cos x}{\cos^{4} x + \sin^{4} x + 2 \sin^{2} \cdot \cos^{2} x - 2 \sin^{2} x \cdot \cos^{2} x} \cdot d x$
$= \int_{0}^{π / 4} \frac{\sin x \cdot \cos x}{{(\cos^{2} x + \sin^{2} x)}^{2} - 2 \sin^{2} x \cdot \cos^{2} x} d x$
$= \int_{0}^{π / 4} \frac{\sin x \cdot \cos x}{1 - \frac{4 \sin^{2} x \cdot \cos^{2} x}{2}} d x$
$= \int_{0}^{π / 4} \frac{2 \sin x \cdot \cos x}{2 - \sin^{2} 2 x} d x = \int_{0}^{π / 4} \frac{\sin 2 x}{1 + \cos^{2} 2 x} d x$
Let, $\cos 2 x = t$
$- 2 \sin 2 x . dx = dt$
$∴ I = \frac{- 1}{2} \int_{1}^{0} \frac{dt}{1 + t^{2}}$
$= \frac{- 1}{2} {[\tan^{- 1} (t)]}_{1}^{0} = \frac{- 1}{2} [\tan^{- 1} (0) - \tan^{- 1} (1)]$
$= \frac{- 1}{2} [0 - \frac{π}{4}] = \frac{π}{8}$

COMEDK-2019

Integral Calculus

86258 $\int \frac{d x}{1 - \cos x - \sin x}$ is equal to

1

\log | 1 + \cot \frac{x}{2} | + C

2

\log | 1 - \tan \frac{x}{2} | + C

3

\log | 1 - \cot \frac{x}{2} | + C

4

\log | 1 + \tan \frac{x}{2} | + C

Explanation:

(C) :Let, $I = \int \frac{d x}{1 - \cos x - \sin x}$
Put, $\cos x = \frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}$ and $\sin x = \frac{2 \tan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}$
$∴ I = \int \frac{d x}{1 - (\frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}) - \frac{2 \tan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}}$
$= \int \frac{\sec^{2} \frac{x}{2} d x}{2 \tan^{2} \frac{x}{2} - 2 \tan \frac{x}{2}} = \int \frac{\frac{1}{2} \sec^{2} \frac{x}{2} d x}{\tan^{2} \frac{x}{2} - \tan \frac{x}{2}}$
Put, $\tan \frac{x}{2} = t \Rightarrow \frac{1}{2} \sec^{2} \frac{x}{2} dx = dt$
$∴ I = \int \frac{dt}{t^{2} - t} = \int [\frac{1}{t - 1} - \frac{1}{t}] dt$
$= \log (t - 1) - \log t + C = \log | \frac{t - 1}{t} | + C$
$= \log | \frac{\tan \frac{x}{2} - 1}{\tan \frac{x}{2}} | + C = \log | 1 - \cot \frac{x}{2} | + C$

COMEDK-2020

Integral Calculus

86259 Evaluate : $\int_{0}^{π / 2} \frac{1}{a^{2} \sin^{2} x + b^{2} \cos^{2} x} d x$

1

\frac{π a}{4 b}

2

\frac{π a}{2 b}

3

\frac{π b}{4 a}

4

\frac{π}{2 ab}

Explanation:

(D) : Let,
$I = \int_{0}^{π / 2} \frac{1}{a^{2} \sin^{2} x + b^{2} \cos^{2} x} d x$
Dividing numerator and denominator by $\cos^{2} x$ we get
$I = \int_{0}^{π / 2} \frac{\sec^{2} x d x}{a^{2} \tan^{2} x + b^{2}}$
Put, $\tan x = t \Rightarrow \sec^{2} x d x = d t$
When, $x = 0 \Rightarrow t = 0$ and when $x = \frac{π}{2} \Rightarrow t = \infty$
$\Rightarrow I = \int_{0}^{\infty} \frac{dt}{a^{2} t^{2} + b^{2}} = \frac{1}{a^{2}} \int_{0}^{\infty} \frac{dt}{t^{2} + {(\frac{b}{a})}^{2}}$
$= \frac{1}{a^{2}} \times \frac{1}{(\frac{b}{a})} {[\tan^{- 1} \frac{t}{(b / a)}]}_{0}^{\infty} = \frac{1}{ab} \times \frac{π}{2} = \frac{π}{2 ab}$

COMEDK-2020

Integral Calculus

86262 Evaluate : $\int \frac{1}{1 + 3 \sin^{2} x + 8 \cos^{2} x} d x$

1

\frac{1}{6} \tan^{- 1} (2 \tan x) + C

2

\tan^{- 1} (2 \tan x) + C

3

\frac{1}{6} \tan^{- 1} (\frac{2 \tan x}{3}) + C

4 None of these

Explanation:

(C) : Let, $I = \int \frac{1}{1 + 3 \sin^{2} x + 8 \cos^{2} x} d x$
Dividing the numerator and denominator by $\cos^{2} x$ , we get
$I = \int \frac{\sec^{2} x}{\sec^{2} x + 3 \tan^{2} x + 8} d x = \int \frac{\sec^{2} x}{4 \tan^{2} x + 9} d x$
Putting $\tan x = t \Rightarrow \sec^{2} xdx = dt$ , we get
$I = \int \frac{dt}{4 t^{2} + 9} = \frac{1}{4} \int \frac{dt}{t^{2} + (3 / 2)^{2}} = \frac{1}{4} \times \frac{1}{3 / 2} \tan^{- 1} (\frac{t}{3 / 2}) + C$
$\Rightarrow I = \frac{1}{6} \tan^{- 1} (\frac{2 t}{3}) + C = \frac{1}{6} \tan^{- 1} (\frac{2 \tan x}{3}) + C$

BITSAT-2014

Integral Calculus

86256 If $x > 0, \int \frac{x^{3} - 7 x + 6}{x (x + 3)} d x = a x + b x^{2} + c \ln x + d$ , then $a + b + c =$

1 0

2

\frac{1}{2}

3

- \frac{1}{2}

4 -1

Explanation:

(C) : Given,
$\int \frac{x^{3} - 7 x + 6}{x (x + 3)} d x = a x + b x^{2} + c \ln x + d$
L.H.S., $I = \int \frac{x^{3} - 7 x + 6}{x (x + 3)} d x$
$= \int \frac{x^{3} - 7 x + 6}{x^{2} + 3 x} d x$
$= \int (x - 3 + \frac{2 x + 6}{x^{2} + 3 x}) d x$
$= \int x \cdot d x - \int 3 d x + \int \frac{2 x + 3}{x^{2} + 3 x} d x + 3 \int \frac{1}{x^{2} + 3 x} d x$
$= \frac{x^{2}}{2} - 3 x + \ln (x^{2} + 3 x) + 3 \int \frac{1}{x (x + 3)} d x$
$= \frac{x^{2}}{2} - 3 x + \ln (x^{2} + 3 x) + \frac{3 \times 1}{3} \int (\frac{1}{x} - \frac{1}{x + 3}) d x$
$= \frac{x^{2}}{2} - 3 x + \ln (x^{2} + 3 x) + \ln x - \ln (x + 3) + C$
$= \frac{x^{2}}{2} - 3 x + \ln (x^{2} + 3 x \times \frac{x}{x + 3}) + C$
$= \frac{x^{2}}{2} - 3 x + \ln (x (x + 3) \times \frac{x}{x + 3}) + C$
$= \frac{x^{2}}{2} - 3 x + \ln x^{2} + C$
$= \frac{x^{2}}{2} - 3 x + 2 \ln x + C$
On comparing L.H.S. and R.H.S., we get -
$a = - 3, b = \frac{1}{2}, c = 2$
Then, $a + b + c = - 3 + \frac{1}{2} + 2 = \frac{- 1}{2}$

COMEDK-2016

Integral Calculus

86257 $\int_{0}^{π / 4} \frac{\sin x \cos x}{\cos^{4} x + \sin^{4} x} d x$ is equal to

1

8 π

2

π / 4

3

4 π

4

π / 8

Explanation:

(D) : Let,
$I = \int_{0}^{π / 4} \frac{\sin x \cdot \cos x}{\cos^{4} x + \sin^{4} x} d x$
$= \int_{0}^{π / 4} \frac{\sin x \cdot \cos x}{\cos^{4} x + \sin^{4} x + 2 \sin^{2} \cdot \cos^{2} x - 2 \sin^{2} x \cdot \cos^{2} x} \cdot d x$
$= \int_{0}^{π / 4} \frac{\sin x \cdot \cos x}{{(\cos^{2} x + \sin^{2} x)}^{2} - 2 \sin^{2} x \cdot \cos^{2} x} d x$
$= \int_{0}^{π / 4} \frac{\sin x \cdot \cos x}{1 - \frac{4 \sin^{2} x \cdot \cos^{2} x}{2}} d x$
$= \int_{0}^{π / 4} \frac{2 \sin x \cdot \cos x}{2 - \sin^{2} 2 x} d x = \int_{0}^{π / 4} \frac{\sin 2 x}{1 + \cos^{2} 2 x} d x$
Let, $\cos 2 x = t$
$- 2 \sin 2 x . dx = dt$
$∴ I = \frac{- 1}{2} \int_{1}^{0} \frac{dt}{1 + t^{2}}$
$= \frac{- 1}{2} {[\tan^{- 1} (t)]}_{1}^{0} = \frac{- 1}{2} [\tan^{- 1} (0) - \tan^{- 1} (1)]$
$= \frac{- 1}{2} [0 - \frac{π}{4}] = \frac{π}{8}$

COMEDK-2019

Integral Calculus

86258 $\int \frac{d x}{1 - \cos x - \sin x}$ is equal to

1

\log | 1 + \cot \frac{x}{2} | + C

2

\log | 1 - \tan \frac{x}{2} | + C

3

\log | 1 - \cot \frac{x}{2} | + C

4

\log | 1 + \tan \frac{x}{2} | + C

Explanation:

(C) :Let, $I = \int \frac{d x}{1 - \cos x - \sin x}$
Put, $\cos x = \frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}$ and $\sin x = \frac{2 \tan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}$
$∴ I = \int \frac{d x}{1 - (\frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}) - \frac{2 \tan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}}$
$= \int \frac{\sec^{2} \frac{x}{2} d x}{2 \tan^{2} \frac{x}{2} - 2 \tan \frac{x}{2}} = \int \frac{\frac{1}{2} \sec^{2} \frac{x}{2} d x}{\tan^{2} \frac{x}{2} - \tan \frac{x}{2}}$
Put, $\tan \frac{x}{2} = t \Rightarrow \frac{1}{2} \sec^{2} \frac{x}{2} dx = dt$
$∴ I = \int \frac{dt}{t^{2} - t} = \int [\frac{1}{t - 1} - \frac{1}{t}] dt$
$= \log (t - 1) - \log t + C = \log | \frac{t - 1}{t} | + C$
$= \log | \frac{\tan \frac{x}{2} - 1}{\tan \frac{x}{2}} | + C = \log | 1 - \cot \frac{x}{2} | + C$

COMEDK-2020

Integral Calculus

86259 Evaluate : $\int_{0}^{π / 2} \frac{1}{a^{2} \sin^{2} x + b^{2} \cos^{2} x} d x$

1

\frac{π a}{4 b}

2

\frac{π a}{2 b}

3

\frac{π b}{4 a}

4

\frac{π}{2 ab}

Explanation:

(D) : Let,
$I = \int_{0}^{π / 2} \frac{1}{a^{2} \sin^{2} x + b^{2} \cos^{2} x} d x$
Dividing numerator and denominator by $\cos^{2} x$ we get
$I = \int_{0}^{π / 2} \frac{\sec^{2} x d x}{a^{2} \tan^{2} x + b^{2}}$
Put, $\tan x = t \Rightarrow \sec^{2} x d x = d t$
When, $x = 0 \Rightarrow t = 0$ and when $x = \frac{π}{2} \Rightarrow t = \infty$
$\Rightarrow I = \int_{0}^{\infty} \frac{dt}{a^{2} t^{2} + b^{2}} = \frac{1}{a^{2}} \int_{0}^{\infty} \frac{dt}{t^{2} + {(\frac{b}{a})}^{2}}$
$= \frac{1}{a^{2}} \times \frac{1}{(\frac{b}{a})} {[\tan^{- 1} \frac{t}{(b / a)}]}_{0}^{\infty} = \frac{1}{ab} \times \frac{π}{2} = \frac{π}{2 ab}$

COMEDK-2020

Integral Calculus

86262 Evaluate : $\int \frac{1}{1 + 3 \sin^{2} x + 8 \cos^{2} x} d x$

1

\frac{1}{6} \tan^{- 1} (2 \tan x) + C

2

\tan^{- 1} (2 \tan x) + C

3

\frac{1}{6} \tan^{- 1} (\frac{2 \tan x}{3}) + C

4 None of these

Explanation:

(C) : Let, $I = \int \frac{1}{1 + 3 \sin^{2} x + 8 \cos^{2} x} d x$
Dividing the numerator and denominator by $\cos^{2} x$ , we get
$I = \int \frac{\sec^{2} x}{\sec^{2} x + 3 \tan^{2} x + 8} d x = \int \frac{\sec^{2} x}{4 \tan^{2} x + 9} d x$
Putting $\tan x = t \Rightarrow \sec^{2} xdx = dt$ , we get
$I = \int \frac{dt}{4 t^{2} + 9} = \frac{1}{4} \int \frac{dt}{t^{2} + (3 / 2)^{2}} = \frac{1}{4} \times \frac{1}{3 / 2} \tan^{- 1} (\frac{t}{3 / 2}) + C$
$\Rightarrow I = \frac{1}{6} \tan^{- 1} (\frac{2 t}{3}) + C = \frac{1}{6} \tan^{- 1} (\frac{2 \tan x}{3}) + C$

BITSAT-2014

Integral Calculus

86256 If $x > 0, \int \frac{x^{3} - 7 x + 6}{x (x + 3)} d x = a x + b x^{2} + c \ln x + d$ , then $a + b + c =$

1 0

2

\frac{1}{2}

3

- \frac{1}{2}

4 -1

Explanation:

(C) : Given,
$\int \frac{x^{3} - 7 x + 6}{x (x + 3)} d x = a x + b x^{2} + c \ln x + d$
L.H.S., $I = \int \frac{x^{3} - 7 x + 6}{x (x + 3)} d x$
$= \int \frac{x^{3} - 7 x + 6}{x^{2} + 3 x} d x$
$= \int (x - 3 + \frac{2 x + 6}{x^{2} + 3 x}) d x$
$= \int x \cdot d x - \int 3 d x + \int \frac{2 x + 3}{x^{2} + 3 x} d x + 3 \int \frac{1}{x^{2} + 3 x} d x$
$= \frac{x^{2}}{2} - 3 x + \ln (x^{2} + 3 x) + 3 \int \frac{1}{x (x + 3)} d x$
$= \frac{x^{2}}{2} - 3 x + \ln (x^{2} + 3 x) + \frac{3 \times 1}{3} \int (\frac{1}{x} - \frac{1}{x + 3}) d x$
$= \frac{x^{2}}{2} - 3 x + \ln (x^{2} + 3 x) + \ln x - \ln (x + 3) + C$
$= \frac{x^{2}}{2} - 3 x + \ln (x^{2} + 3 x \times \frac{x}{x + 3}) + C$
$= \frac{x^{2}}{2} - 3 x + \ln (x (x + 3) \times \frac{x}{x + 3}) + C$
$= \frac{x^{2}}{2} - 3 x + \ln x^{2} + C$
$= \frac{x^{2}}{2} - 3 x + 2 \ln x + C$
On comparing L.H.S. and R.H.S., we get -
$a = - 3, b = \frac{1}{2}, c = 2$
Then, $a + b + c = - 3 + \frac{1}{2} + 2 = \frac{- 1}{2}$

COMEDK-2016

Integral Calculus

86257 $\int_{0}^{π / 4} \frac{\sin x \cos x}{\cos^{4} x + \sin^{4} x} d x$ is equal to

1

8 π

2

π / 4

3

4 π

4

π / 8

Explanation:

(D) : Let,
$I = \int_{0}^{π / 4} \frac{\sin x \cdot \cos x}{\cos^{4} x + \sin^{4} x} d x$
$= \int_{0}^{π / 4} \frac{\sin x \cdot \cos x}{\cos^{4} x + \sin^{4} x + 2 \sin^{2} \cdot \cos^{2} x - 2 \sin^{2} x \cdot \cos^{2} x} \cdot d x$
$= \int_{0}^{π / 4} \frac{\sin x \cdot \cos x}{{(\cos^{2} x + \sin^{2} x)}^{2} - 2 \sin^{2} x \cdot \cos^{2} x} d x$
$= \int_{0}^{π / 4} \frac{\sin x \cdot \cos x}{1 - \frac{4 \sin^{2} x \cdot \cos^{2} x}{2}} d x$
$= \int_{0}^{π / 4} \frac{2 \sin x \cdot \cos x}{2 - \sin^{2} 2 x} d x = \int_{0}^{π / 4} \frac{\sin 2 x}{1 + \cos^{2} 2 x} d x$
Let, $\cos 2 x = t$
$- 2 \sin 2 x . dx = dt$
$∴ I = \frac{- 1}{2} \int_{1}^{0} \frac{dt}{1 + t^{2}}$
$= \frac{- 1}{2} {[\tan^{- 1} (t)]}_{1}^{0} = \frac{- 1}{2} [\tan^{- 1} (0) - \tan^{- 1} (1)]$
$= \frac{- 1}{2} [0 - \frac{π}{4}] = \frac{π}{8}$

COMEDK-2019

Integral Calculus

86258 $\int \frac{d x}{1 - \cos x - \sin x}$ is equal to

1

\log | 1 + \cot \frac{x}{2} | + C

2

\log | 1 - \tan \frac{x}{2} | + C

3

\log | 1 - \cot \frac{x}{2} | + C

4

\log | 1 + \tan \frac{x}{2} | + C

Explanation:

(C) :Let, $I = \int \frac{d x}{1 - \cos x - \sin x}$
Put, $\cos x = \frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}$ and $\sin x = \frac{2 \tan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}$
$∴ I = \int \frac{d x}{1 - (\frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}) - \frac{2 \tan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}}$
$= \int \frac{\sec^{2} \frac{x}{2} d x}{2 \tan^{2} \frac{x}{2} - 2 \tan \frac{x}{2}} = \int \frac{\frac{1}{2} \sec^{2} \frac{x}{2} d x}{\tan^{2} \frac{x}{2} - \tan \frac{x}{2}}$
Put, $\tan \frac{x}{2} = t \Rightarrow \frac{1}{2} \sec^{2} \frac{x}{2} dx = dt$
$∴ I = \int \frac{dt}{t^{2} - t} = \int [\frac{1}{t - 1} - \frac{1}{t}] dt$
$= \log (t - 1) - \log t + C = \log | \frac{t - 1}{t} | + C$
$= \log | \frac{\tan \frac{x}{2} - 1}{\tan \frac{x}{2}} | + C = \log | 1 - \cot \frac{x}{2} | + C$

COMEDK-2020

Integral Calculus

86259 Evaluate : $\int_{0}^{π / 2} \frac{1}{a^{2} \sin^{2} x + b^{2} \cos^{2} x} d x$

1

\frac{π a}{4 b}

2

\frac{π a}{2 b}

3

\frac{π b}{4 a}

4

\frac{π}{2 ab}

Explanation:

(D) : Let,
$I = \int_{0}^{π / 2} \frac{1}{a^{2} \sin^{2} x + b^{2} \cos^{2} x} d x$
Dividing numerator and denominator by $\cos^{2} x$ we get
$I = \int_{0}^{π / 2} \frac{\sec^{2} x d x}{a^{2} \tan^{2} x + b^{2}}$
Put, $\tan x = t \Rightarrow \sec^{2} x d x = d t$
When, $x = 0 \Rightarrow t = 0$ and when $x = \frac{π}{2} \Rightarrow t = \infty$
$\Rightarrow I = \int_{0}^{\infty} \frac{dt}{a^{2} t^{2} + b^{2}} = \frac{1}{a^{2}} \int_{0}^{\infty} \frac{dt}{t^{2} + {(\frac{b}{a})}^{2}}$
$= \frac{1}{a^{2}} \times \frac{1}{(\frac{b}{a})} {[\tan^{- 1} \frac{t}{(b / a)}]}_{0}^{\infty} = \frac{1}{ab} \times \frac{π}{2} = \frac{π}{2 ab}$

COMEDK-2020

Integral Calculus

86262 Evaluate : $\int \frac{1}{1 + 3 \sin^{2} x + 8 \cos^{2} x} d x$

1

\frac{1}{6} \tan^{- 1} (2 \tan x) + C

2

\tan^{- 1} (2 \tan x) + C

3

\frac{1}{6} \tan^{- 1} (\frac{2 \tan x}{3}) + C

4 None of these

Explanation:

(C) : Let, $I = \int \frac{1}{1 + 3 \sin^{2} x + 8 \cos^{2} x} d x$
Dividing the numerator and denominator by $\cos^{2} x$ , we get
$I = \int \frac{\sec^{2} x}{\sec^{2} x + 3 \tan^{2} x + 8} d x = \int \frac{\sec^{2} x}{4 \tan^{2} x + 9} d x$
Putting $\tan x = t \Rightarrow \sec^{2} xdx = dt$ , we get
$I = \int \frac{dt}{4 t^{2} + 9} = \frac{1}{4} \int \frac{dt}{t^{2} + (3 / 2)^{2}} = \frac{1}{4} \times \frac{1}{3 / 2} \tan^{- 1} (\frac{t}{3 / 2}) + C$
$\Rightarrow I = \frac{1}{6} \tan^{- 1} (\frac{2 t}{3}) + C = \frac{1}{6} \tan^{- 1} (\frac{2 \tan x}{3}) + C$

BITSAT-2014

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