What is the integral of bsqrt(1-(x^2)/(a^2)) ?

The integral of bsqrt(1-(x^2)/(a^2)) is 1/2 ba(arcsin(x/a)+1/2 sin(2arcsin(x/a)))+C

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integral of bsqrt(1-(x^2)/(a^2))

\int b 1 - \frac{x ^{2}}{a ^{2}} d x

\frac{1}{2} ba (arcsin (\frac{x}{a}) + \frac{1}{2} sin (2 arcsin (\frac{x}{a}))) + C

Solution steps

\int b 1 - \frac{x ^{2}}{a ^{2}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= b \cdot \int 1 - \frac{x ^{2}}{a ^{2}} d x

Apply Trigonometric Substitution

= b \cdot \int a cos^{2} (u) d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= ba \cdot \int cos^{2} (u) d u

Rewrite using trig identities

= ba \cdot \int \frac{1 + cos ( 2 u )}{2} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= ba \frac{1}{2} \cdot \int 1 + cos (2 u) d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= ba \frac{1}{2} (\int 1 d u + \int cos (2 u) d u)

\int 1 d u = u

\int cos (2 u) d u = \frac{1}{2} sin (2 u)

= ba \frac{1}{2} (u + \frac{1}{2} sin (2 u))

Substitute back

u = arcsin (\frac{1}{a} x)

= ba \frac{1}{2} (arcsin (\frac{1}{a} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{a} x)))

Simplify

ba \frac{1}{2} (arcsin (\frac{1}{a} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{a} x))) : \frac{1}{2} ba (arcsin (\frac{x}{a}) + \frac{1}{2} sin (2 arcsin (\frac{x}{a})))

= \frac{1}{2} ba (arcsin (\frac{x}{a}) + \frac{1}{2} sin (2 arcsin (\frac{x}{a})))

Add a constant to the solution

= \frac{1}{2} ba (arcsin (\frac{x}{a}) + \frac{1}{2} sin (2 arcsin (\frac{x}{a}))) + C

What is the integral of bsqrt(1-(x^2)/(a^2)) ?
The integral of bsqrt(1-(x^2)/(a^2)) is 1/2 ba(arcsin(x/a)+1/2 sin(2arcsin(x/a)))+C