QBManager

Integral Calculus

86429 $\int_{0}^{\frac{π}{2}} \frac{\sin^{n} θ}{\sin^{n} θ + \cos^{n} θ} d θ$ is equal to

1 1

2

\frac{π}{2}

3 0

4

\frac{π}{4}

Explanation:

(D) : Let,
$I = \int_{0}^{\frac{π}{2}} \frac{\sin^{n} θ}{\sin^{n} θ + \cos^{n} θ} d θ$
$= \int_{0}^{\frac{π}{2}} \frac{\sin^{n} (π / 2 - θ)}{\sin^{n} (\frac{π}{2} - θ) + \cos^{n} (\frac{π}{2} - θ)} d θ$
$I = \int_{0}^{\frac{π}{2}} \frac{\cos^{n} θ}{\cos^{n} θ + \sin^{n} θ} d θ$
Adding Equation (i) and (ii)
$2 I = \int_{0}^{\frac{π}{2}} \frac{\sin^{n} θ + \cos^{n} θ}{\cos^{n} θ + \sin^{n} θ}$
$2 I = \int_{0}^{\frac{π}{2}} 1 d θ \Rightarrow 2 I = [θ]_{0}^{\frac{π}{2}}$
$2 I = [\frac{π}{2} - 0] \Rightarrow I = \frac{π}{4}$

Assam CEE-2021

Integral Calculus

86430 $\int_{0}^{π / 2} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} d x$

1

\frac{π}{2}

2 0

3

\frac{π}{4}

4

π

Explanation:

(C) : Let,
$I = \int_{0}^{π / 2} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} d x$
We know that,
$\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x$
$I = \int_{0}^{π / 2} \frac{\sqrt{\sin (\frac{π}{2} - x)}}{\sqrt{\sin (\frac{π}{2} - x)} + \sqrt{\cos (\frac{π}{2} - x)}} d x$
$I = \int_{a}^{π / 2} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} d x$
Adding equation (i) and (ii)
$2 I = \int_{0}^{π / 2} (\frac{\sqrt{\sin x} + \sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\cos x}}) d x$
$2 I = \int_{0}^{π / 2} 1 dx$
$2 I = [x]_{0}^{\frac{π}{2}} = \frac{π}{2} - 0 = \frac{π}{4}$

Assam CEE-2021

Integral Calculus

86365 For the curve $4 x^{5} = 5 y^{4}$ , the ratio of the cube of the subtangent at a point on the curve to the square of the subnormal at the same point $i$

1

y {(\frac{5}{4})}^{4}

2

{(\frac{5}{4})}^{4}

3

{(\frac{5}{4})}^{3}

4

{(\frac{4}{5})}^{4}

Explanation:

(D) : Given,Curve $4 x^{5} = 5 y^{4}$
$20 x^{4} = 20 y^{3} \cdot \frac{d y}{d x} \Rightarrow \frac{d y}{d x} = \frac{x^{4}}{y^{3}}$
We know that,
Length of subnormal $(S N) = (y \cdot \frac{d x}{d y}) = \frac{y^{4}}{x^{4}}$
Length of subtangent $(ST) = (y \cdot \frac{dy}{dx}) = \frac{x^{4}}{y^{2}}$
But given condition,
$\frac{(SN)^{3}}{(ST)^{2}} = \frac{{(\frac{y^{4}}{x^{4}})}^{3}}{{(\frac{x^{4}}{y^{2}})}^{2}} = {(\frac{y^{4}}{x^{4}})}^{3} \times {(\frac{y^{2}}{x^{4}})}^{2}$
$= \frac{y^{12}}{x^{12}} \times \frac{y^{4}}{x^{8}} = \frac{y^{16}}{x^{20}} = {(\frac{y^{4}}{x^{5}})}^{4}$
$= {(\frac{4}{5})}^{4}$

Karnataka CET-2010

Integral Calculus

86366 The value of $\int e^{\sin x} \sin 2 x d x$ is

1

2 e^{\sin x} (\sin x + 1) + C

2

2 e^{\sin x} (\cos x + 1) + C

3

2 e^{\sin x} (\cos x - 1) + C

4

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

Explanation:

(D) :Let, $I = \int e^{\sin x} \cdot \sin 2 x \cdot dx$
$= \int e^{\sin x} \cdot (2 \sin x \cdot \cos x) dx$
Let, $\sin x = t$
$\cos x . dx = dt$
$∴ I = 2 \int e^{t} . t d t$
$= 2 [t . e^{t} - \int e^{t} dt]$
$= 2 t \cdot e^{t} - 2 e^{t} + C$
$= 2 \sin x . e^{\sin x} - 2 e^{\sin x}$
$= 2 e^{\sin x} (\sin x - 1) + C$

Karnataka CET-2020

Integral Calculus

86431 $\int_{0}^{1} \frac{d x}{(3 x + 2) + \sqrt{3 x + 2}} =$

1

- \frac{2}{3} \log | \frac{\sqrt{5 + 1}}{\sqrt{2} + 1} |

2

\frac{2}{3} \log | \frac{\sqrt{5 + 1}}{\sqrt{2} + 1} |

3

2 \log | \sqrt{5 + 1} |

4

\frac{2}{3} \log | \sqrt{5 + 1} |

Explanation:

(B): $I = \int_{0}^{1} \frac{d x}{(3 x + 2) + \sqrt{3 x + 2}}$
$Unknown environment 'tabular'$
$I = \frac{2}{3} \int_{\sqrt{2}}^{\sqrt{5}} \frac{t d t}{t^{2} + t} = \frac{2}{3} \int_{\sqrt{2}}^{\sqrt{5}} \frac{t}{t (t + 1)} d t$
$I = \frac{2}{3} \int_{\sqrt{2}}^{\sqrt{5}} \frac{1}{(t + 1)} d t$
$I = \frac{2}{3} [\log (t + 1)]_{\sqrt{2}}^{\sqrt{5}}$
$I = \frac{2}{3} [\log (\sqrt{5} + 1) - \log (\sqrt{2} + 1)]$
$I = \frac{2}{3} \log | \frac{\sqrt{5} + 1}{\sqrt{2} + 1} |$

GUJCET-2021

Integral Calculus

86429 $\int_{0}^{\frac{π}{2}} \frac{\sin^{n} θ}{\sin^{n} θ + \cos^{n} θ} d θ$ is equal to

1 1

2

\frac{π}{2}

3 0

4

\frac{π}{4}

Explanation:

(D) : Let,
$I = \int_{0}^{\frac{π}{2}} \frac{\sin^{n} θ}{\sin^{n} θ + \cos^{n} θ} d θ$
$= \int_{0}^{\frac{π}{2}} \frac{\sin^{n} (π / 2 - θ)}{\sin^{n} (\frac{π}{2} - θ) + \cos^{n} (\frac{π}{2} - θ)} d θ$
$I = \int_{0}^{\frac{π}{2}} \frac{\cos^{n} θ}{\cos^{n} θ + \sin^{n} θ} d θ$
Adding Equation (i) and (ii)
$2 I = \int_{0}^{\frac{π}{2}} \frac{\sin^{n} θ + \cos^{n} θ}{\cos^{n} θ + \sin^{n} θ}$
$2 I = \int_{0}^{\frac{π}{2}} 1 d θ \Rightarrow 2 I = [θ]_{0}^{\frac{π}{2}}$
$2 I = [\frac{π}{2} - 0] \Rightarrow I = \frac{π}{4}$

Assam CEE-2021

Integral Calculus

86430 $\int_{0}^{π / 2} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} d x$

1

\frac{π}{2}

2 0

3

\frac{π}{4}

4

π

Explanation:

(C) : Let,
$I = \int_{0}^{π / 2} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} d x$
We know that,
$\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x$
$I = \int_{0}^{π / 2} \frac{\sqrt{\sin (\frac{π}{2} - x)}}{\sqrt{\sin (\frac{π}{2} - x)} + \sqrt{\cos (\frac{π}{2} - x)}} d x$
$I = \int_{a}^{π / 2} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} d x$
Adding equation (i) and (ii)
$2 I = \int_{0}^{π / 2} (\frac{\sqrt{\sin x} + \sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\cos x}}) d x$
$2 I = \int_{0}^{π / 2} 1 dx$
$2 I = [x]_{0}^{\frac{π}{2}} = \frac{π}{2} - 0 = \frac{π}{4}$

Assam CEE-2021

Integral Calculus

86365 For the curve $4 x^{5} = 5 y^{4}$ , the ratio of the cube of the subtangent at a point on the curve to the square of the subnormal at the same point $i$

1

y {(\frac{5}{4})}^{4}

2

{(\frac{5}{4})}^{4}

3

{(\frac{5}{4})}^{3}

4

{(\frac{4}{5})}^{4}

Explanation:

(D) : Given,Curve $4 x^{5} = 5 y^{4}$
$20 x^{4} = 20 y^{3} \cdot \frac{d y}{d x} \Rightarrow \frac{d y}{d x} = \frac{x^{4}}{y^{3}}$
We know that,
Length of subnormal $(S N) = (y \cdot \frac{d x}{d y}) = \frac{y^{4}}{x^{4}}$
Length of subtangent $(ST) = (y \cdot \frac{dy}{dx}) = \frac{x^{4}}{y^{2}}$
But given condition,
$\frac{(SN)^{3}}{(ST)^{2}} = \frac{{(\frac{y^{4}}{x^{4}})}^{3}}{{(\frac{x^{4}}{y^{2}})}^{2}} = {(\frac{y^{4}}{x^{4}})}^{3} \times {(\frac{y^{2}}{x^{4}})}^{2}$
$= \frac{y^{12}}{x^{12}} \times \frac{y^{4}}{x^{8}} = \frac{y^{16}}{x^{20}} = {(\frac{y^{4}}{x^{5}})}^{4}$
$= {(\frac{4}{5})}^{4}$

Karnataka CET-2010

Integral Calculus

86366 The value of $\int e^{\sin x} \sin 2 x d x$ is

1

2 e^{\sin x} (\sin x + 1) + C

2

2 e^{\sin x} (\cos x + 1) + C

3

2 e^{\sin x} (\cos x - 1) + C

4

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

Explanation:

(D) :Let, $I = \int e^{\sin x} \cdot \sin 2 x \cdot dx$
$= \int e^{\sin x} \cdot (2 \sin x \cdot \cos x) dx$
Let, $\sin x = t$
$\cos x . dx = dt$
$∴ I = 2 \int e^{t} . t d t$
$= 2 [t . e^{t} - \int e^{t} dt]$
$= 2 t \cdot e^{t} - 2 e^{t} + C$
$= 2 \sin x . e^{\sin x} - 2 e^{\sin x}$
$= 2 e^{\sin x} (\sin x - 1) + C$

Karnataka CET-2020

Integral Calculus

86431 $\int_{0}^{1} \frac{d x}{(3 x + 2) + \sqrt{3 x + 2}} =$

1

- \frac{2}{3} \log | \frac{\sqrt{5 + 1}}{\sqrt{2} + 1} |

2

\frac{2}{3} \log | \frac{\sqrt{5 + 1}}{\sqrt{2} + 1} |

3

2 \log | \sqrt{5 + 1} |

4

\frac{2}{3} \log | \sqrt{5 + 1} |

Explanation:

(B): $I = \int_{0}^{1} \frac{d x}{(3 x + 2) + \sqrt{3 x + 2}}$
$Unknown environment 'tabular'$
$I = \frac{2}{3} \int_{\sqrt{2}}^{\sqrt{5}} \frac{t d t}{t^{2} + t} = \frac{2}{3} \int_{\sqrt{2}}^{\sqrt{5}} \frac{t}{t (t + 1)} d t$
$I = \frac{2}{3} \int_{\sqrt{2}}^{\sqrt{5}} \frac{1}{(t + 1)} d t$
$I = \frac{2}{3} [\log (t + 1)]_{\sqrt{2}}^{\sqrt{5}}$
$I = \frac{2}{3} [\log (\sqrt{5} + 1) - \log (\sqrt{2} + 1)]$
$I = \frac{2}{3} \log | \frac{\sqrt{5} + 1}{\sqrt{2} + 1} |$

GUJCET-2021

Integral Calculus

86429 $\int_{0}^{\frac{π}{2}} \frac{\sin^{n} θ}{\sin^{n} θ + \cos^{n} θ} d θ$ is equal to

1 1

2

\frac{π}{2}

3 0

4

\frac{π}{4}

Explanation:

(D) : Let,
$I = \int_{0}^{\frac{π}{2}} \frac{\sin^{n} θ}{\sin^{n} θ + \cos^{n} θ} d θ$
$= \int_{0}^{\frac{π}{2}} \frac{\sin^{n} (π / 2 - θ)}{\sin^{n} (\frac{π}{2} - θ) + \cos^{n} (\frac{π}{2} - θ)} d θ$
$I = \int_{0}^{\frac{π}{2}} \frac{\cos^{n} θ}{\cos^{n} θ + \sin^{n} θ} d θ$
Adding Equation (i) and (ii)
$2 I = \int_{0}^{\frac{π}{2}} \frac{\sin^{n} θ + \cos^{n} θ}{\cos^{n} θ + \sin^{n} θ}$
$2 I = \int_{0}^{\frac{π}{2}} 1 d θ \Rightarrow 2 I = [θ]_{0}^{\frac{π}{2}}$
$2 I = [\frac{π}{2} - 0] \Rightarrow I = \frac{π}{4}$

Assam CEE-2021

Integral Calculus

86430 $\int_{0}^{π / 2} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} d x$

1

\frac{π}{2}

2 0

3

\frac{π}{4}

4

π

Explanation:

(C) : Let,
$I = \int_{0}^{π / 2} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} d x$
We know that,
$\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x$
$I = \int_{0}^{π / 2} \frac{\sqrt{\sin (\frac{π}{2} - x)}}{\sqrt{\sin (\frac{π}{2} - x)} + \sqrt{\cos (\frac{π}{2} - x)}} d x$
$I = \int_{a}^{π / 2} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} d x$
Adding equation (i) and (ii)
$2 I = \int_{0}^{π / 2} (\frac{\sqrt{\sin x} + \sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\cos x}}) d x$
$2 I = \int_{0}^{π / 2} 1 dx$
$2 I = [x]_{0}^{\frac{π}{2}} = \frac{π}{2} - 0 = \frac{π}{4}$

Assam CEE-2021

Integral Calculus

86365 For the curve $4 x^{5} = 5 y^{4}$ , the ratio of the cube of the subtangent at a point on the curve to the square of the subnormal at the same point $i$

1

y {(\frac{5}{4})}^{4}

2

{(\frac{5}{4})}^{4}

3

{(\frac{5}{4})}^{3}

4

{(\frac{4}{5})}^{4}

Explanation:

(D) : Given,Curve $4 x^{5} = 5 y^{4}$
$20 x^{4} = 20 y^{3} \cdot \frac{d y}{d x} \Rightarrow \frac{d y}{d x} = \frac{x^{4}}{y^{3}}$
We know that,
Length of subnormal $(S N) = (y \cdot \frac{d x}{d y}) = \frac{y^{4}}{x^{4}}$
Length of subtangent $(ST) = (y \cdot \frac{dy}{dx}) = \frac{x^{4}}{y^{2}}$
But given condition,
$\frac{(SN)^{3}}{(ST)^{2}} = \frac{{(\frac{y^{4}}{x^{4}})}^{3}}{{(\frac{x^{4}}{y^{2}})}^{2}} = {(\frac{y^{4}}{x^{4}})}^{3} \times {(\frac{y^{2}}{x^{4}})}^{2}$
$= \frac{y^{12}}{x^{12}} \times \frac{y^{4}}{x^{8}} = \frac{y^{16}}{x^{20}} = {(\frac{y^{4}}{x^{5}})}^{4}$
$= {(\frac{4}{5})}^{4}$

Karnataka CET-2010

Integral Calculus

86366 The value of $\int e^{\sin x} \sin 2 x d x$ is

1

2 e^{\sin x} (\sin x + 1) + C

2

2 e^{\sin x} (\cos x + 1) + C

3

2 e^{\sin x} (\cos x - 1) + C

4

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

Explanation:

(D) :Let, $I = \int e^{\sin x} \cdot \sin 2 x \cdot dx$
$= \int e^{\sin x} \cdot (2 \sin x \cdot \cos x) dx$
Let, $\sin x = t$
$\cos x . dx = dt$
$∴ I = 2 \int e^{t} . t d t$
$= 2 [t . e^{t} - \int e^{t} dt]$
$= 2 t \cdot e^{t} - 2 e^{t} + C$
$= 2 \sin x . e^{\sin x} - 2 e^{\sin x}$
$= 2 e^{\sin x} (\sin x - 1) + C$

Karnataka CET-2020

Integral Calculus

86431 $\int_{0}^{1} \frac{d x}{(3 x + 2) + \sqrt{3 x + 2}} =$

1

- \frac{2}{3} \log | \frac{\sqrt{5 + 1}}{\sqrt{2} + 1} |

2

\frac{2}{3} \log | \frac{\sqrt{5 + 1}}{\sqrt{2} + 1} |

3

2 \log | \sqrt{5 + 1} |

4

\frac{2}{3} \log | \sqrt{5 + 1} |

Explanation:

(B): $I = \int_{0}^{1} \frac{d x}{(3 x + 2) + \sqrt{3 x + 2}}$
$Unknown environment 'tabular'$
$I = \frac{2}{3} \int_{\sqrt{2}}^{\sqrt{5}} \frac{t d t}{t^{2} + t} = \frac{2}{3} \int_{\sqrt{2}}^{\sqrt{5}} \frac{t}{t (t + 1)} d t$
$I = \frac{2}{3} \int_{\sqrt{2}}^{\sqrt{5}} \frac{1}{(t + 1)} d t$
$I = \frac{2}{3} [\log (t + 1)]_{\sqrt{2}}^{\sqrt{5}}$
$I = \frac{2}{3} [\log (\sqrt{5} + 1) - \log (\sqrt{2} + 1)]$
$I = \frac{2}{3} \log | \frac{\sqrt{5} + 1}{\sqrt{2} + 1} |$

GUJCET-2021

Integral Calculus

86429 $\int_{0}^{\frac{π}{2}} \frac{\sin^{n} θ}{\sin^{n} θ + \cos^{n} θ} d θ$ is equal to

1 1

2

\frac{π}{2}

3 0

4

\frac{π}{4}

Explanation:

(D) : Let,
$I = \int_{0}^{\frac{π}{2}} \frac{\sin^{n} θ}{\sin^{n} θ + \cos^{n} θ} d θ$
$= \int_{0}^{\frac{π}{2}} \frac{\sin^{n} (π / 2 - θ)}{\sin^{n} (\frac{π}{2} - θ) + \cos^{n} (\frac{π}{2} - θ)} d θ$
$I = \int_{0}^{\frac{π}{2}} \frac{\cos^{n} θ}{\cos^{n} θ + \sin^{n} θ} d θ$
Adding Equation (i) and (ii)
$2 I = \int_{0}^{\frac{π}{2}} \frac{\sin^{n} θ + \cos^{n} θ}{\cos^{n} θ + \sin^{n} θ}$
$2 I = \int_{0}^{\frac{π}{2}} 1 d θ \Rightarrow 2 I = [θ]_{0}^{\frac{π}{2}}$
$2 I = [\frac{π}{2} - 0] \Rightarrow I = \frac{π}{4}$

Assam CEE-2021

Integral Calculus

86430 $\int_{0}^{π / 2} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} d x$

1

\frac{π}{2}

2 0

3

\frac{π}{4}

4

π

Explanation:

(C) : Let,
$I = \int_{0}^{π / 2} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} d x$
We know that,
$\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x$
$I = \int_{0}^{π / 2} \frac{\sqrt{\sin (\frac{π}{2} - x)}}{\sqrt{\sin (\frac{π}{2} - x)} + \sqrt{\cos (\frac{π}{2} - x)}} d x$
$I = \int_{a}^{π / 2} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} d x$
Adding equation (i) and (ii)
$2 I = \int_{0}^{π / 2} (\frac{\sqrt{\sin x} + \sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\cos x}}) d x$
$2 I = \int_{0}^{π / 2} 1 dx$
$2 I = [x]_{0}^{\frac{π}{2}} = \frac{π}{2} - 0 = \frac{π}{4}$

Assam CEE-2021

Integral Calculus

86365 For the curve $4 x^{5} = 5 y^{4}$ , the ratio of the cube of the subtangent at a point on the curve to the square of the subnormal at the same point $i$

1

y {(\frac{5}{4})}^{4}

2

{(\frac{5}{4})}^{4}

3

{(\frac{5}{4})}^{3}

4

{(\frac{4}{5})}^{4}

Explanation:

(D) : Given,Curve $4 x^{5} = 5 y^{4}$
$20 x^{4} = 20 y^{3} \cdot \frac{d y}{d x} \Rightarrow \frac{d y}{d x} = \frac{x^{4}}{y^{3}}$
We know that,
Length of subnormal $(S N) = (y \cdot \frac{d x}{d y}) = \frac{y^{4}}{x^{4}}$
Length of subtangent $(ST) = (y \cdot \frac{dy}{dx}) = \frac{x^{4}}{y^{2}}$
But given condition,
$\frac{(SN)^{3}}{(ST)^{2}} = \frac{{(\frac{y^{4}}{x^{4}})}^{3}}{{(\frac{x^{4}}{y^{2}})}^{2}} = {(\frac{y^{4}}{x^{4}})}^{3} \times {(\frac{y^{2}}{x^{4}})}^{2}$
$= \frac{y^{12}}{x^{12}} \times \frac{y^{4}}{x^{8}} = \frac{y^{16}}{x^{20}} = {(\frac{y^{4}}{x^{5}})}^{4}$
$= {(\frac{4}{5})}^{4}$

Karnataka CET-2010

Integral Calculus

86366 The value of $\int e^{\sin x} \sin 2 x d x$ is

1

2 e^{\sin x} (\sin x + 1) + C

2

2 e^{\sin x} (\cos x + 1) + C

3

2 e^{\sin x} (\cos x - 1) + C

4

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

Explanation:

(D) :Let, $I = \int e^{\sin x} \cdot \sin 2 x \cdot dx$
$= \int e^{\sin x} \cdot (2 \sin x \cdot \cos x) dx$
Let, $\sin x = t$
$\cos x . dx = dt$
$∴ I = 2 \int e^{t} . t d t$
$= 2 [t . e^{t} - \int e^{t} dt]$
$= 2 t \cdot e^{t} - 2 e^{t} + C$
$= 2 \sin x . e^{\sin x} - 2 e^{\sin x}$
$= 2 e^{\sin x} (\sin x - 1) + C$

Karnataka CET-2020

Integral Calculus

86431 $\int_{0}^{1} \frac{d x}{(3 x + 2) + \sqrt{3 x + 2}} =$

1

- \frac{2}{3} \log | \frac{\sqrt{5 + 1}}{\sqrt{2} + 1} |

2

\frac{2}{3} \log | \frac{\sqrt{5 + 1}}{\sqrt{2} + 1} |

3

2 \log | \sqrt{5 + 1} |

4

\frac{2}{3} \log | \sqrt{5 + 1} |

Explanation:

(B): $I = \int_{0}^{1} \frac{d x}{(3 x + 2) + \sqrt{3 x + 2}}$
$Unknown environment 'tabular'$
$I = \frac{2}{3} \int_{\sqrt{2}}^{\sqrt{5}} \frac{t d t}{t^{2} + t} = \frac{2}{3} \int_{\sqrt{2}}^{\sqrt{5}} \frac{t}{t (t + 1)} d t$
$I = \frac{2}{3} \int_{\sqrt{2}}^{\sqrt{5}} \frac{1}{(t + 1)} d t$
$I = \frac{2}{3} [\log (t + 1)]_{\sqrt{2}}^{\sqrt{5}}$
$I = \frac{2}{3} [\log (\sqrt{5} + 1) - \log (\sqrt{2} + 1)]$
$I = \frac{2}{3} \log | \frac{\sqrt{5} + 1}{\sqrt{2} + 1} |$

GUJCET-2021

Integral Calculus

86429 $\int_{0}^{\frac{π}{2}} \frac{\sin^{n} θ}{\sin^{n} θ + \cos^{n} θ} d θ$ is equal to

1 1

2

\frac{π}{2}

3 0

4

\frac{π}{4}

Explanation:

(D) : Let,
$I = \int_{0}^{\frac{π}{2}} \frac{\sin^{n} θ}{\sin^{n} θ + \cos^{n} θ} d θ$
$= \int_{0}^{\frac{π}{2}} \frac{\sin^{n} (π / 2 - θ)}{\sin^{n} (\frac{π}{2} - θ) + \cos^{n} (\frac{π}{2} - θ)} d θ$
$I = \int_{0}^{\frac{π}{2}} \frac{\cos^{n} θ}{\cos^{n} θ + \sin^{n} θ} d θ$
Adding Equation (i) and (ii)
$2 I = \int_{0}^{\frac{π}{2}} \frac{\sin^{n} θ + \cos^{n} θ}{\cos^{n} θ + \sin^{n} θ}$
$2 I = \int_{0}^{\frac{π}{2}} 1 d θ \Rightarrow 2 I = [θ]_{0}^{\frac{π}{2}}$
$2 I = [\frac{π}{2} - 0] \Rightarrow I = \frac{π}{4}$

Assam CEE-2021

Integral Calculus

86430 $\int_{0}^{π / 2} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} d x$

1

\frac{π}{2}

2 0

3

\frac{π}{4}

4

π

Explanation:

(C) : Let,
$I = \int_{0}^{π / 2} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} d x$
We know that,
$\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x$
$I = \int_{0}^{π / 2} \frac{\sqrt{\sin (\frac{π}{2} - x)}}{\sqrt{\sin (\frac{π}{2} - x)} + \sqrt{\cos (\frac{π}{2} - x)}} d x$
$I = \int_{a}^{π / 2} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} d x$
Adding equation (i) and (ii)
$2 I = \int_{0}^{π / 2} (\frac{\sqrt{\sin x} + \sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\cos x}}) d x$
$2 I = \int_{0}^{π / 2} 1 dx$
$2 I = [x]_{0}^{\frac{π}{2}} = \frac{π}{2} - 0 = \frac{π}{4}$

Assam CEE-2021

Integral Calculus

86365 For the curve $4 x^{5} = 5 y^{4}$ , the ratio of the cube of the subtangent at a point on the curve to the square of the subnormal at the same point $i$

1

y {(\frac{5}{4})}^{4}

2

{(\frac{5}{4})}^{4}

3

{(\frac{5}{4})}^{3}

4

{(\frac{4}{5})}^{4}

Explanation:

(D) : Given,Curve $4 x^{5} = 5 y^{4}$
$20 x^{4} = 20 y^{3} \cdot \frac{d y}{d x} \Rightarrow \frac{d y}{d x} = \frac{x^{4}}{y^{3}}$
We know that,
Length of subnormal $(S N) = (y \cdot \frac{d x}{d y}) = \frac{y^{4}}{x^{4}}$
Length of subtangent $(ST) = (y \cdot \frac{dy}{dx}) = \frac{x^{4}}{y^{2}}$
But given condition,
$\frac{(SN)^{3}}{(ST)^{2}} = \frac{{(\frac{y^{4}}{x^{4}})}^{3}}{{(\frac{x^{4}}{y^{2}})}^{2}} = {(\frac{y^{4}}{x^{4}})}^{3} \times {(\frac{y^{2}}{x^{4}})}^{2}$
$= \frac{y^{12}}{x^{12}} \times \frac{y^{4}}{x^{8}} = \frac{y^{16}}{x^{20}} = {(\frac{y^{4}}{x^{5}})}^{4}$
$= {(\frac{4}{5})}^{4}$

Karnataka CET-2010

Integral Calculus

86366 The value of $\int e^{\sin x} \sin 2 x d x$ is

1

2 e^{\sin x} (\sin x + 1) + C

2

2 e^{\sin x} (\cos x + 1) + C

3

2 e^{\sin x} (\cos x - 1) + C

4

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

Explanation:

(D) :Let, $I = \int e^{\sin x} \cdot \sin 2 x \cdot dx$
$= \int e^{\sin x} \cdot (2 \sin x \cdot \cos x) dx$
Let, $\sin x = t$
$\cos x . dx = dt$
$∴ I = 2 \int e^{t} . t d t$
$= 2 [t . e^{t} - \int e^{t} dt]$
$= 2 t \cdot e^{t} - 2 e^{t} + C$
$= 2 \sin x . e^{\sin x} - 2 e^{\sin x}$
$= 2 e^{\sin x} (\sin x - 1) + C$

Karnataka CET-2020

Integral Calculus

86431 $\int_{0}^{1} \frac{d x}{(3 x + 2) + \sqrt{3 x + 2}} =$

1

- \frac{2}{3} \log | \frac{\sqrt{5 + 1}}{\sqrt{2} + 1} |

2

\frac{2}{3} \log | \frac{\sqrt{5 + 1}}{\sqrt{2} + 1} |

3

2 \log | \sqrt{5 + 1} |

4

\frac{2}{3} \log | \sqrt{5 + 1} |

Explanation:

(B): $I = \int_{0}^{1} \frac{d x}{(3 x + 2) + \sqrt{3 x + 2}}$
$Unknown environment 'tabular'$
$I = \frac{2}{3} \int_{\sqrt{2}}^{\sqrt{5}} \frac{t d t}{t^{2} + t} = \frac{2}{3} \int_{\sqrt{2}}^{\sqrt{5}} \frac{t}{t (t + 1)} d t$
$I = \frac{2}{3} \int_{\sqrt{2}}^{\sqrt{5}} \frac{1}{(t + 1)} d t$
$I = \frac{2}{3} [\log (t + 1)]_{\sqrt{2}}^{\sqrt{5}}$
$I = \frac{2}{3} [\log (\sqrt{5} + 1) - \log (\sqrt{2} + 1)]$
$I = \frac{2}{3} \log | \frac{\sqrt{5} + 1}{\sqrt{2} + 1} |$

GUJCET-2021