Mathematics

# $\underset{-a}{\overset{a}{\int}} \sqrt{\dfrac{a - x}{a + x}} dx$ is equal to

$a \pi$

##### SOLUTION
$I=\int^a_{-a}\sqrt{\dfrac{a-x}{a+x}}dx$
Put $x=a\cos\theta$
$dx=-a\sin d\theta$
$x\Rightarrow -a=a \cos\theta \Rightarrow \theta =\pi$
$x\Rightarrow a=\cos\theta\Rightarrow \theta =0$
$\displaystyle \int^{0}_{\pi}\sqrt{\frac{a-a \cos\theta}{a+a cos\theta}}\times (-a\sin\theta)d\theta$

$\displaystyle \int ^{\pi}_0\sqrt{\frac{a (1-\cos\theta}{a(1+cos\theta)}}\times a\sin\theta d\theta$
$\displaystyle \int^{\pi}_0\sqrt{\frac{2\sin^2\frac{\theta}{2}}{2\cos^2\frac{\theta}{2}}}\times a\sin\theta d\theta$
$\displaystyle a\int^{\pi}_0\tan\frac{\theta}{2}\times 2\sin\frac{\theta}{2}\cos\frac{\theta}{2}d\theta$.

$\displaystyle \Rightarrow 2a\int^{\pi}_0\sin^2\frac{\theta}{2}d\theta$
$\displaystyle \Rightarrow a\int^{\pi}_0(1-\cos\theta)d\theta$
$\displaystyle \Rightarrow a\,\,\,\theta]^{\pi}_0+a\sin]^{\pi}_0$
$\Rightarrow a( \pi-0)+a(0-0)$
$\Rightarrow \boxed{\pi a}$
$\therefore \boxed{ \int^a_{-a}\sqrt{\frac{a-x}{a+x}}dx=\pi a}$

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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 Hard
The integral $\int _{ \tfrac { \pi }{ 12 } }^{ \tfrac { \pi }{ 4 } }{ \dfrac { 8\cos { 2x } }{ { \left( \tan { x } +\cot { x } \right) }^{ 3 } } dx }$ equals:
• A. $\dfrac {5}{64}$
• B. $\dfrac {13}{32}$
• C. $\dfrac {13}{256}$
• D. $\dfrac {15}{128}$

1 Verified Answer | Published on 17th 09, 2020

Q2 Multiple Correct Medium
Let $f(x)=7tan^8x + 7tan^6x-3tan^4x-3tan^2x$ for all $x \in \left( \dfrac{-\pi}{2},\dfrac{\pi}{2} \right).$ Then the correct expression(s) is/are
• A. $\displaystyle \int_{0}^{\pi/4}xf(x)dx=\dfrac{1}{6}$
• B. $\displaystyle \int_{0}^{\pi/4}xf(x)dx=1$
• C. $\displaystyle \int_{0}^{\pi/4}xf(x)dx=\dfrac{1}{12}$
• D. $\displaystyle \int_{0}^{\pi/4}f(x)dx=0$

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Medium
Integrate $\int\limits_0^1 {{x \over {1 + {x^2}}}dx}$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
Evaluate $\displaystyle\int^{\pi/2}_0\sin^2xdx$
• A. $\dfrac{\pi}{3}$
• B. $\dfrac{\pi}{2}$
• C. $\dfrac{2\pi}{3}$
• D. $\dfrac{\pi}{4}$

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