Question

integral of sqrt(a^2-x^2)dx ... i got to the answer arcsin(x/a) + c ... but it appeared as wrong =\

Answer 1

It is wrong because
\[ \int \frac{1}{\sqrt{a^2 - x^2}} \ dx \ \ = \ \ \arcsin(x/a) + C \]

For your integral, integrate by parts with
du = 1 dx
v = sqrt(a^2 - x^2)

and take it from there.

Answer 2

i.e.,
\[ \int \sqrt{a^2 - x^2} \ dx \ = \ x\sqrt{a^2 - x^2} - \int \frac{-x^2}{\sqrt{a^2-x^2}} \ dx \]

Answer 3

You can solve it using trigonometric substitution. Substitute \(x=a\sin{u} \implies dx=a\cos{u}du\):
\[\int\limits_{}^{}\sqrt{a^2-x^2}dx=\int\limits \sqrt{a^2-a^2\sin^2{u}} a \cos{u} du=a^2 \int\limits \cos^2u du.\]
You can proceed from here.

Answer 4

You might need this triangle for your back substitution.