intergration check working please?

Question

s (sqrt)(1 − x^2) dx

were s is the integral sign

is the answer
(x-1/3x^3)* (1-x^2)^3/2 + c

if not y not?

Answer 1

If x = sin t

dx/dt = cos t


int sqrt(1 - x^2) dx = int sqrt (1 - sin^2 t)* dx/dt * dt =

= int cos t * cos t dt = int cos^2 t dt

Now, cos^2 t = (1 + cos 2t )/2

Take u = 2t

dt/du = 1/2

int sqrt(1 - x^2) dx = = int (1 + cos 2t)/2 dt =
= int (1/2) dt + int (cos 2t)/2 dt =
= t/2 + int (cos u)/2 * dt/du * du =

= t/2 + int (cos u)/2 * 1/2 du = t/2 + 1/4 int cos u du
= t/2 + 1/4 sin u + c = t/2 + (sin 2t)/4 + c =

= (arcsin x)/2 + sin (2 arcsin x)/4 + c =

= (arcsin x)/2 + 2sin(arcsin x) cos(arcsin x)/4 + c =

= (arcsin x)/2 + 2x * sqrt(1 - x^2)/4 + c =

= (arcsin x)/2 + x * sqrt(1 - x^2)/2 + c

Answer 2

The answer is (1/2)*arc sin(x) + (1/2)*x*sqrt(1 - x^2) + C.

Use trig substitution: take x = sin A , sqrt(1 - x^2) = cos A , and dx = cos A dx . The integral becomes

Integ((cos A)^2) dA = (1/2)*A + (1/4)*2*(cos A)*(sin A) + C ,

and the reverse substitutions give you the answer stated above.

The answer to your "y not" is because the derivative of your "answer" is not the integrand.

Answer 3

the objective of formal substitution (for indispensable calculus) is to make the integrand "less complicated" -- plenty the way, back in algebra you found out some 4th degree equations have been a disguised form of quadratics. So given the hint which you ought to substitute u = cosx you persist with with removing any strains of x interior the integrand. unique indispensable (e^cosx) [sinx dx] u = cosx du = - sinx dx [which very actually is damaging one situations the bracketed quantity in the present day above!] so unique indispensable turns into - (e^u) du [no x's showing!] it particularly is easier: its indispensable is basically itself: -e^u + ok Now "back substitute" - e^u + ok = - e^cosx + ok and evaluate as a diverse indispensable if needed.

Answer 4

the answer is wrong.
u can check it by differentiating the answer if u get the question then it is right.
correct answer is 1/2*x*sqrt(1-x^2) + 1/2*arcsinx
u can get it by substituting x=sinz

Answer 5

If you want to check whether your answer is correct, differenciate (find dy/dx) your answer. You must end up with sqrt(1-x^2).

Answer 6

An easy check for you to do in general is to take the derivative of the answer you got. If it matches the integrand...then hooray...if not...better go at it again.