Integral of x(sqrt(1-x^2)) + 1/2 -x^2/2 IS IT RIGHT?

Homework Statement

integral of [ x*sqrt[1-x2] + 1/2
- x2/2] dx - with the limits of 1 and 0

the last two bits are easy but i just wanted to confirm what i have done for the first bit is correct...ive tried using a method that takes less time but i dont know if its right...

Homework Equations

sin(x)^2 + cos(x)^2 = 1

cos(arcsinx) = sqrt(1-x2)

The Attempt at a Solution

for the first bit:

x = sinu
dx = cosu du
1-x2 = cos(u)^2

so it becomes: sinu(cos(u)2)cosu du

sinu du = -dcosu

so :- -cos(u)3 dcosu
integrate this beacomes: -cos(u)4/4

which then becomes
-cos(arcsinx)4/4 which is just: -(1-x2)2/4

so putting this together with the 2nd and 3rd part gives

-(1-x2)2/4 + x/2 - x3/6 with limts 1&0

which equals 1/3

so is this right?

Cyosis
Homework Helper
Use the right tool for the right situation. Trigonometric substitution certainly works, but it takes you on quite a detour. To check if your answer is correct you simply differentiate it so lets do that.

$$\frac{d}{dx} \frac{-(1-x^2)^2}{4}=x(1-x^2) \neq x \sqrt{1-x^2}$$

You make an error very early on after you state that 1-x^2=cos(u)^2 you conclude in the next line that $\sqrt{1-x^2}=\cos^2u$. In other words you forgot to take the square root. The rest seems to be correct.

There is however a method that is way easier take u=1-x^2 and you're done.

as soon as you pointed out that error early on i felt like kicking myself yeah i see what to do now

(do you still get 1/3 as the answer cause ive just redone it and im still getting the same answer with the limits in)

thanks

Last edited:
Cyosis
Homework Helper
For the first term I get 1/3 for the second 1/2 and for the third -1/6. So 1/3+1/2-1/6=2/3

tiny-tim