Question

please answer.............. ​

Answer 1

GIVEN :-

$\\ \\ \bigstar \: \displaystyle \sf \int_{-a} ^{a} \sqrt{ \frac{a - x}{a + x} }$

TO FIND :-

$\\ \\ \bigstar \: \displaystyle \sf \: value \: of \: \int_{-a} ^{a} \sqrt{ \frac{a - x}{a + x} }$

SOLUTION :-

$\\ \\ : \implies\displaystyle \sf \int_{-a} ^{a} \sqrt{ \frac{a - x}{a + x} }$

$: \implies\displaystyle \sf Let \: I = \int_{-a} ^{a} \sqrt{ \frac{a - x}{a + x} }$

$\bullet \: \displaystyle \sf Let \: x = a \cos2 \theta$

$: \implies\displaystyle \sf \frac{dx}{d \theta} = - 2a \sin2 \theta$

$: \implies\displaystyle \sf \: dx = - 2a \sin2 \theta.d \theta$

$\therefore\displaystyle \sf \: I = \int \sqrt{ \frac{a - a \cos2 \theta}{a + a \cos2 \theta} } \: . \: - 2a \sin2 \theta \: d \theta$

$: \implies\displaystyle \sf I = \int \: \sqrt{ \frac{ \cancel{a}(1 - cos2 \theta)}{ \cancel{a}(1 + cos2 \theta)} } \: \: . - 2a \sin2 \theta \: d \theta$

$: \implies\displaystyle \sf I = - 2a \int \sqrt{ \dfrac{ \sin ^{2} \theta}{ \cos ^{2} \theta } } \: \: . \sin2 \theta \: d \theta$

$: \implies\displaystyle \sf I = - 2a \int \sqrt{ \frac{ \sin \theta}{ \cancel{\cos \theta}} } \: \: . 2 \sin\theta. \cancel{ \cos \theta}.d \theta$

$: \implies\displaystyle \sf I = - 4a \int \sin ^{2}\theta \: d \theta$

$: \implies\displaystyle \sf I = - 4a \int \frac{1 - \cos2 \theta}{2} \: d \theta$

$: \implies\displaystyle \sf I = 2a \int \bigg( \cos2 \theta - 1 \bigg) \: d \theta$

$: \implies\displaystyle \sf I = 2a \Bigg[ \int \cos2 \theta \: d \theta - \int d \theta\Bigg]$

$: \implies\displaystyle \sf I = 2a\Bigg[ \dfrac{ \sin2 \theta}{2} - a \theta\Bigg] + c$

$: \implies\displaystyle \sf I =a \sin2 \theta - 2a \theta + c$

$: \implies\displaystyle \sf I =a \sqrt{1 - \frac{x ^{2} }{a ^{2} } } - a \cos ^{ - 1} \frac{x}{a} + c$

$: \implies\displaystyle \sf I = \sqrt{a ^{2} - x ^{2} } - a \cos ^{ - 1} \dfrac{x}{a} + c$

$: \implies\displaystyle \sf I \bigg | ^{a} _{-a} = \sqrt{a ^{2} - x ^{2} } \bigg | ^{a} _{-a} - a \cos ^{ - 1} \frac{x}{a} + c$

$: \implies\displaystyle \sf I = \big \{0 - 0 \big \} - a \bigg \{ \cos ^{ - 1} 1 - \cos ^{ - 1} ( - 1) \bigg \}$

$: \implies\displaystyle \sf I = - a \bigg(0 - \pi \: \bigg)$

$: \implies \underline{ \boxed{\displaystyle \sf I = a \pi}}$

$\therefore \underline{\displaystyle \sf Required \ answer \ of \ this \ question \ is \ a \pi.}$