Verified answer

Ответ:

[tex]\boxed{ \boldsymbol{ \displaystyle \int {\sqrt{a^{2} - x^{2} } } \, dx = \frac{a^{2}\arcsin \bigg(\dfrac{x}{a} \bigg)}{2} + \frac{a^{2} \sin \bigg( 2\arcsin \bigg(\dfrac{x}{a} \bigg) \bigg)}{4} + C} }[/tex]

Примечание:

Формула понижения степени:

[tex]\boxed{\cos^{2} \alpha = \frac{1 + \cos 2\alpha }{2} }[/tex]

По таблице интегралов:

[tex]\boxed{\displaystyle \int x^{n} \ dx = \frac{x^{n + 1}}{n + 1} + C; n \neq -1, x > 0}[/tex]

[tex]\boxed{\displaystyle \int \cos x \ dx = \sin x + C}[/tex]

По свойствам интегралов:

[tex]\boxed{ \displaystyle \int \sum\limits_{i=1}^n {C_{i}f_{i}(x)} \, dx = \sum\limits_{i=1}^nC_{i} \int {f_{i}(x)} \, dx}[/tex]

Пошаговое объяснение:

2.2

[tex]\displaystyle I = \int {\sqrt{a^{2} - x^{2} } } \, dx , \ - a < x < a[/tex]

[tex]\displaystyle \int {\sqrt{a^{2} - x^{2} } } \, dx =[/tex]

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Замена:

[tex]x = a \sin t \Longrightarrow dx = (a \sin t)' \ dt = a \cos t \ dt[/tex]

[tex]x = a \sin t \Longrightarrow t = \arcsin \bigg(\dfrac{x}{a} \bigg)[/tex]

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[tex]\displaystyle \int { a \cos t \sqrt{a^{2} - (a \sin t)^{2} } } \, dt = \int {a \cos t \sqrt{a^{2}(1 - \sin^{2} t)} } \, dt = \int {a^{2} \cos t \sqrt{\cos^{2} t} } \, dt=[/tex]

[tex]\displaystyle = a^{2} \int { \cos^{2} t } \, dt = a^{2} \int { \frac{1 + \cos 2t}{2} } \, dt = \frac{a^{2}}{2} \int { (1 + \cos 2t) } \, dt =[/tex]

[tex]\displaystyle = \frac{a^{2}}{2} \Bigg( \int { 1 } \, dt + \int { \cos 2t } \, dt \Bigg) = \frac{a^{2}}{2} \Bigg(t+ \frac{1}{2} \int { \cos 2t } \, d(2t) \Bigg) =[/tex]

[tex]\displaystyle = \frac{a^{2}}{2} \Bigg(t+ \frac{\sin 2t}{2} \Bigg) + C = \frac{a^{2}t}{2} + \frac{a^{2} \sin 2t}{4} + C =[/tex]

[tex]\displaystyle = \frac{a^{2}\arcsin \bigg(\dfrac{x}{a} \bigg)}{2} + \frac{a^{2} \sin \bigg( 2\arcsin \bigg(\dfrac{x}{a} \bigg) \bigg)}{4} + C[/tex]