Mathematics

# $\displaystyle\int { \sqrt { 4-{ x }^{ 2 } } } dx$ is equals to

##### SOLUTION
$I=\displaystyle\int{\sqrt{4-{x}^{2}}dx}$

Take $x=2\sin{\theta}\Rightarrow dx=2\cos{\theta}d{\theta}$

$\sqrt{4-{x}^{2}}=\sqrt{4-4{\sin}^{2}{\theta}}=\sqrt{4\left(1-{\sin}^{2}{\theta}\right)}=2\sqrt{{\cos}^{2}{\theta}}=2\cos{\theta}$

$\displaystyle\int{\sqrt{4-{x}^{2}}dx}$

$=\displaystyle\int{2\cos{\theta}.2\cos{\theta}d{\theta}}$

$=2 \displaystyle \int{2{\cos}^{2}{\theta}d{\theta}}$

$=2 \displaystyle \int{\left(1+\cos{2\theta}\right)d{\theta}}$ since $1+\cos{2\theta}=2{\cos}^{2}{\theta}$

$=2\left(\theta+\dfrac{\sin{2\theta}}{2}\right)+c$

$=2\left({\sin}^{-1}{x/2}+\dfrac{\sin{2{\sin}^{-1}{x/2}}}{2}\right)+c$ since $x=2\sin{\theta}\Rightarrow \theta={\sin}^{-1}{x/2}$

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

#### Realted Questions

Q1 Single Correct Medium
$\int_{0}^{\pi /2}sin2xtan^{-1}\left ( sinx \right )dx=$

• A.  $\dfrac{\pi }{2}$+1
• B.  $\dfrac{3\pi }{2}$+1
• C.  $\dfrac{3\pi }{2}$-1
• D.  $\dfrac{\pi }{2}$-1

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
If $f(y)=e^y, g(y)-y; y > 0$ and $F(t)=\displaystyle\int^t_0f(t-y)g(y)dy$, then?
• A. $F(t)=t e^t$
• B. $F(t)=t e^{-t}$
• C. $F(t)=1-e^{-t}(1+t)$
• D. $F(t)=e^t-(1+t)$

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Medium
Find $\int \dfrac {x^{3} - 1}{x^{2}}dx$.

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Hard
Evaluate: $\displaystyle \int \dfrac{\left ( \sec x\:co\sec \right )}{\left ( \log \tan x \right )} dx$
• A. $\displaystyle \log \left ( \tan x \right )$
• B. $\displaystyle \cot \left ( \log x \right )$
• C. $\displaystyle \tan \left ( \log x \right )$
• D. $\displaystyle \log \log \left ( \tan x \right )$

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 $\displaystyle f(x)>0 \forall x \in R$