Mathematics

# Let $I=\displaystyle \int _{1/\sqrt{2}}^{1}e^{\cos^{-1}{x}}\ dx$ then

$I < 1-\dfrac{1}{\sqrt{2}}$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

#### Realted Questions

Q1 Single Correct Medium
The integral $\displaystyle \int \dfrac {dx}{a\cos x + b\sin x}$ is of the form $\dfrac {1}{r} \ln \left [\tan \left (\dfrac {x + \alpha}{2}\right )\right ]$.
What is $\alpha$ equal to?
• A. $\tan^{-1}\left (\dfrac {b}{a}\right )$
• B. $\tan^{-1}\left (\dfrac {a + b}{a - b}\right )$
• C. $\tan^{-1}\left (\dfrac {a - b}{a + b}\right )$
• D. $\tan^{-1}\left (\dfrac {a}{b}\right )$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Medium
Solve $\displaystyle\int \dfrac {x^{2}}{(x^{2}+1)(x^{2}+4)}dx$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
The value of $\displaystyle\int_{-\pi}^{\pi}\sin^{3}x \cos^{2}x\ dx$ is equal to
• A. $1$
• B. $2$
• C. $3$
• D. $0$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Medium
Solve $\displaystyle\int \dfrac { \left( x ^ { 3 } + 8 \right) ( x - 1 ) } { x ^ { 2 } - 2 x + 4 }dx$

$\displaystyle\int^{\pi}_0\dfrac{dx}{(6-\cos x)}$.