Mathematics

# $\int^{\pi/3}_0 \dfrac{tan x}{\sqrt{2k sec x}} dx = 1-\dfrac{1}{\sqrt{2}}$, then the value of k is

$2$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

#### Realted Questions

Q1 Single Correct Hard
$\displaystyle \lim _{ n\rightarrow \infty }{ \sum _{ r=1 }^{ n }{ \dfrac { 1 }{ n } } } \sqrt { \left[ \dfrac { n+r }{ n-r } \right] }$
• A. $\dfrac { \pi }{ 2 }$
• B. $2\pi$
• C. $\dfrac { \pi }{ 2 } -1$
• D. $\dfrac { \pi }{ 2 } +1$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Medium
Solve:
$\int \frac{dx}{\sqrt{4-x^{2}}}$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
$\displaystyle \int_0^{2a} f(x) dx = 0$ is
• A. $\displaystyle \int_0^{2a} f(2a - x) =\int_0^{2a} f(x)$
• B. $\displaystyle \int_0^{2a} f(x) = -\int_0^{2a} f (x)$
• C. $\displaystyle \int_0^{2a} f(-x) = \int_0^{2a} f(x)$
• D. $\displaystyle \int_0^{2a} f(2a - x) = - \int_0^{2a} f(x)$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Hard
lf $I_{m,n}=\displaystyle \int x^{m}(\log x)^{n}dx$, then$I_{m,n}-\displaystyle \frac{x^{m+1}}{(m+1)}(\log x)^{n}=$
• A. $\displaystyle \frac{n}{m+1},I_{mn-1}$
• B. $\displaystyle \frac{m}{n+1},I_{mn-1}$
• C. $\displaystyle \frac{n}{m+1}.I_{m-1}, n-1$
• D. $-\displaystyle \frac{n}{m+1},I_{mn-1}$

Evaluate $\int \dfrac{{1 - {x^2}}}{{x\left( {1 - 2x} \right)}}$