Question

Q1.integration of √ ( a² - x² ) dx is?​

Answer 1

Answer:

$∫dx√a2−x2=arcsin(xa)+C Explanation:∫dx√a2−x2After using x=asiny and dx=acosy⋅dy transforms, this integral became∫acosy⋅dyacosy=∫dy=y+cAfter using x=asiny and y=arcsin(xa) inverse transforms, I found∫dx√a2−x2=arcsin(xa)+C$

Explanation:

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Answer 2

Answer:

$\orange{ \sf\displaystyle \int \sqrt{ {a}^{2} - {x}^{2} } = \sf = \frac{ {a}^{2} }{2} {sin}^{ - 1} \frac{x}{a} + \frac{x}{2} \sqrt{ {a}^{2} - {x}^{2} } + c}$

Step-by-step explanation:

$\bf \: given \\ \displaystyle \: \int \sqrt{ {a}^{2} - {x}^{2} } \: dx \\ \sf substitute \\ \sf\: \: x = asin \theta \: \implies \: \theta = {sin}^{ - 1} \frac{x}{a} \: \: \: and \\ \sf \: \pink{dx = acos\theta \: d\theta \: } \\ \bf \: now \\ \\ \displaystyle \: \int \sqrt{ {a}^{2} - {x}^{2} } \: dx \\ = \sf \int \sqrt{ {a}^{2} - {a}^{2} {sin}^{2} \theta} \: \: acos\theta \: d\theta \\ = \sf \int \sqrt{ {a}^{2} (1 - {sin}^{2} \theta)} \: \: acos\theta \: d\theta \\ = \sf \int \: a \sqrt{ {1 - {sin}^{2} \theta} } \: a cos\theta \: d\theta \\ = \sf \int {a}^{2} . {cos}^{2} \theta \: d\theta \: \\ = \sf {a}^{2} \int \: \frac{1}{2} (1 + cos2\theta) \: d\theta \: \: \: \: \sf \green{\{ \because \: 2 {cos}^{2} \theta = 1 + cos2\theta \}} \\ \sf= \frac{ {a}^{2} }{2} (\theta + \frac{sin2\theta}{2} ) + c \: \: \: \: \: \: \: \ \sf \red{ \{c = integral \: constant \}} \\ \\ \sf = \frac{ {a}^{2} }{2} (\theta + \frac{2sin\theta \: cos\theta}{2} ) + c \\ \\ \sf = \frac{ {a}^{2} }{2}(\theta + sin \theta \: \sqrt{1 - {sin}^{2}\theta } ) + c \\ \\ \sf = \frac{ {a}^{2} }{2} ( {sin}^{ - 1} \frac{x}{a} + \frac{x}{a} \sqrt{1 - \frac{ {x}^{2} }{ {a}^{2} } } ) + c \\ \\ \sf = \frac{ {a}^{2} }{2} {sin}^{ - 1} \frac{x}{a} + \frac{ {a}^{2} }{2} . \frac{x}{ {a}^{2} } \sqrt{ {a}^{2} - {x}^{2} } + c \\ \\ \sf = \frac{ {a}^{2} }{2} {sin}^{ - 1} \frac{x}{a} + \frac{x}{2} \sqrt{ {a}^{2} - {x}^{2} } + c$