Integration with Trigonometric Substitution

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[SOLVED] Integration with Trigonometric Substitution

Homework Statement



Given integral (I):
I[(x)sqrt(9-x^2)dx]

by words:
Integral of "X" times square root of "9-X(squared)

Use proper trigonometric substitution to solve this problem.

Homework Equations





The Attempt at a Solution

 

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You don't even need Trig substitution.

\int x\sqrt{9-x^2}dx

u=9-x^2
du=-2xdx \rightarrow xdx=-\frac 1 2 du
 
I know that I don't need that.
But the problem is, I have to use it.
The exercise require it.
 
Well you posted the solution to it? I don't know what else to tell you. Just analyze what they did. Work it yourself a couple times if you have to.
 
You forgot a term when you first did the substitution.

x = 3sin(u)
dx = 3cos(u)du

(9 - x^2)^(1/2) = 3cos(u)

So the integral becomes 27sin(u)cos^2(u)du
 
\int x^2\sqrt{9-x^2}dx
apropo
u=x^2\sqrt{9-x^2}dx
 
Last edited:
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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