Mathematics

# A curve is represented parametrically by the equation $x=e^t \cos t$ and $y=e^t \sin t$, where $t$ is a parameter. Then,If $F(t)=\int (x+y)dt$, then the value of $F\left (\displaystyle \frac {\pi}{2}\right )-F(0)$ is

$e^{\tfrac{\pi}{2}}$

##### SOLUTION
Given,

$F(t)=\int (x+y)dt$

$=\int (e^t \cos t+e^t \sin t)dt$

$=\dfrac{e^t\sin \left(t\right)}{2}+\dfrac{e^t\cos \left(t\right)}{2}-\dfrac{e^t\cos \left(t\right)}{2}+\dfrac{e^t\sin \left(t\right)}{2}$

$=e^t\sin t+c$

Now,

$F\left ( \dfrac{\pi}{2} \right )-F(0)=e^{\frac{\pi}{2}}\sin \dfrac{\pi}{2}-(e^0\sin 0)$

$=e^{\frac{\pi}{2}}(1)-1 (0)$

$=e^{\frac{\pi}{2}}$

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

#### Realted Questions

Q1 Subjective Medium
Evaluate :
$\int x^2 (1-\dfrac{1}{x^2})dx$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
$\int \sqrt {\dfrac {x - 1}{x + 1}}dx$ is equal to
• A. $2\sqrt {x^{2} + 1} + \sin^{-1} x + c$
• B. $2\sqrt {x^{2} - 1} + \sin^{-1} x + c$
• C. $\dfrac {\sqrt {x^{2} + 1}}{2} + \sin^{-1} x + c$
• D. $\sqrt {x^{2} + 1} - \sin^{-1} x + c$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Hard
$\displaystyle \int \cos^{-1}\sqrt{\frac{1-x}{2}}dx=$
• A. $\displaystyle \frac{\pi}{2}x-\frac{1}{2}xcos^{-1}x+c$
• B. $\dfrac{\pi}{2}x-\dfrac{1}{2}xcos^{-1}x-\sqrt{1-x^{2}}+c$
• C. $\displaystyle \dfrac{\pi}{2}x+\dfrac{1}{2}xcos^{-1}x+c$
• D. $\dfrac{\pi}{2}x-\dfrac{1}{2}xcos^{-1}x+\dfrac{1}{2}\sqrt{1-x^{2}}+c$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
What is $\int \dfrac {xe^{x}dx}{(x + 1)^{2}}$ equal to?
where $c$ is the constant of integration.
• A. $(x + 1)^{2} e^{x} + c$
• B. $(x + 1)e^{x} + c$
• C. $\dfrac {e^{x}}{(x + 1)^{2}} + c$
• D. $\dfrac {e^{x}}{x + 1} + c$

Consider two differentiable functions $f(x), g(x)$ satisfying $\displaystyle 6\int f(x)g(x)dx=x^{6}+3x^{4}+3x^{2}+c$ & $\displaystyle 2 \int \frac {g(x)dx}{f(x)}=x^{2}+c$. where $\displaystyle f(x)>0 \forall x \in R$