Mastering Integration Techniques: Solving Intx2(1-5x2)10dx with Expert Help

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need urgent help with this

any ideas will be appreciated

\intx2(1-5x2)10dx

tried all kinds of things, what i now think needs to be done is to change the differential to d(1-5x2) so that i won't have to open the (1-5x2)10 which i really don't want to nor do i think i am expected to.

d(1-5x2)=-10xdx... not really helping me :(
 
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I tried several approaches, among them substitution and integration by parts, but didn't seem to get anywhere with either of these.

Another approach that seems promising is a trig substitution, using x = 1/sqrt(5) * cos(theta).

Using this approach I was able to rewrite the integral as
-1/25 \int cos^2(\theta) sin^{21}(\theta)d\theta
This could be rewritten as two integrals, both in powers of sin(theta). At that point, I would use a table of integrals.

This approach might be beyond your present capabilities, but it's the only one I can think of that leads anywhere.
 


Dell said:
any ideas will be appreciated

\intx2(1-5x2)10dx

tried all kinds of things, what i now think needs to be done is to change the differential to d(1-5x2) so that i won't have to open the (1-5x2)10 which i really don't want to nor do i think i am expected to.

d(1-5x2)=-10xdx... not really helping me :(
You could just expand the integrand using the binomial theorem.
 
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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