Can't solve these two. Apparently you're NOT supposed to use a trig substitution for the second one. I don't know how you can for the first since 5 isn't a prime number.
First one:
∫ dx/[x*√(5-x^2)]
Second one:
∫ x^3*√(9-x^2) dx
Thanks
Can't solve these two. Apparently you're NOT supposed to use a trig substitution for the second one. I don't know how you can for the first since 5 isn't a prime number.
First one:
∫ dx/[x*√(5-x^2)]
Second one:
∫ x^3*√(9-x^2) dx
Thanks
For ∫ x^3*√(9-x^2) dx
Mr. Fantastic told you what to substitute. If you take $\displaystyle u = \sqrt{9-x^2}$.
$\displaystyle du = \frac{-\not{2}x}{\not{2}\sqrt{9-x^2}} dx$
$\displaystyle du = -\frac{x}{\sqrt{9-x^2}} dx$
$\displaystyle dx = -\frac{\sqrt{9-x^2}}{x} du$
Now replace! you'll see that you can substitute $\displaystyle 9-x^2$ and $\displaystyle x^2$ (remember Mr. F said $\displaystyle u^2 = 9 - x^2$).