Mathematics

# If $I_{1}=\displaystyle \int{\sin^{-1}xdx}$ and $I_{2}=\displaystyle \int{\sin^{-1}\sqrt{1-x^{2}}xdx}$ then

$I_{1}=I_{2}$

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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 Medium
The value of${\smallint _{100}}{1000}\frac{{dx}}{x}is$
• A. $10$
• B. $1$
• C. $2.303$
• D. $4.606$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Hard
Evaluate the following definite integrals as limit of sums.
$\displaystyle\int^{1}_{-1}e^xdx$.
• A. $e^2-1$
• B. $\dfrac{e^2-1}{2}$
• C. $\dfrac{e^2-1}{e^2}$
• D. $\dfrac{e^2-1}{e}$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Hard
Evaluate $\displaystyle\int{\frac{dx}{{(x^2+a^2)}^3}}$.
• A. $\displaystyle I=\frac{x}{2a^2{(x^2+a^2)}^2}+\frac{3}{4a^2}\left\{\frac{x}{a^2(x^2+a^2)}+\frac{1}{a^3}\tan^{-1}{\left(\frac{x}{a}\right)}\right\}+C$
• B. $\displaystyle I=\frac{x}{4a^2{(x^2+a^2)}^2}+\frac{1}{4a^2}\left\{\frac{x}{2a^2(x^2+a^2)}+\frac{1}{2a^3}\sec^{-1}{\left(\frac{x}{a}\right)}\right\}+C$
• C. $\displaystyle I=\frac{x}{a^2{(x^2+a^2)}^2}+\frac{3}{2a^2}\left\{\frac{x}{2a^2(x^2+a^2)}+\frac{1}{2a^3}\tan^{-1}{\left(\frac{x}{a}\right)}\right\}+C$
• D. $\displaystyle I=\frac{x}{4a^2{(x^2+a^2)}^2}+\frac{3}{4a^2}\left\{\frac{x}{2a^2(x^2+a^2)}+\frac{1}{2a^3}\tan^{-1}{\left(\frac{x}{a}\right)}\right\}+C$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium

$\displaystyle \int_{0}^\frac{\pi}{2}\frac{dx}{1+\tan x}=$
• A. $\displaystyle \frac{\pi}{2}$
• B. $1$
• C. $\log 2$
• D. $\displaystyle \frac{\pi}{4}$

Let $n \space\epsilon \space N$ & the A.M., G.M., H.M. & the root mean square of $n$ numbers $2n+1, 2n+2, ...,$ up to $n^{th}$ number are $A_{n}$, $G_{n}$, $H_{n}$ and $R_{n}$ respectively.