Question

Inverse trigonometric functions,

Can you please help me where am I wrong in this problem ?
http://screencast.com/t/mQtujojz4F

Answer 1

except for the fact that I already used the limit for the second sin while on the first I did not yet

Answer 2

\[\int\limits_{}^{}\frac{ 1 }{ \sqrt{1 - x^2} }dx = \sin^{-1}(x)\]
what you did seems illegal

Answer 3

You'll have to complete the square on the denominator in the square root i think....

Answer 4

that is right

4x - x^2 = 4 - (x-2)^2

Answer 5

let me know if you'd like me to solve it @Christos

Answer 6

Please can you? @Euler271

Answer 7

btw look at this part of the solution: http://screencast.com/t/5ivVVg6RLSS results are almost the same thing

Answer 8

@Christos that's different cos it's only a constant and an x^2. Not an x term and an x-squared. Try it the same way tho, use u-sub for the 4 - (x-2)^2... let u=x-2

Answer 9

\[\int\limits_{1}^{2}\frac{ 1 }{ \sqrt{4 - (x-2)^2} }dx\]
let u = x - 2; du = dx
\[\int\limits_{-1}^{0}\frac{ 1 }{ \sqrt{4 - u^2} }du = \int\limits_{-1}^{0}\frac{ 1 }{ 2 }\frac{ 1 }{ \sqrt{1 - \frac{ u^2 }{ 4 }} }du \]
let v = u/2; dv = du/2
\[\int\limits_{-\frac{ 1 }{ 2 }}^{0}\frac{ 1 }{ \sqrt{1 - v^2} }dv = \sin^{-1} (v) = \sin^{-1}(0) - \sin^{-1}(-1/2) = \frac{ \pi }{ 6 }\]