Integral: Solve with Substitution

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Homework Statement



Integrate[Cos[(2x+3)^(1/3)]


Homework Equations





The Attempt at a Solution



Would I use simple substitution?

Would it just simply be 1/2*Sin((2x+3)^(1/3))??
 
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You will have to use a substitution yes. No your answer is wrong. When you're integrating always differentiate your final answer to see if it gives the correct result.

That said do you have any ideas about the kind of substitution you want to use?
 
Your proposed solution completely disregards the 1/3 power.

I would suggest integration by parts.
 
Indeed.

U=(2x+3)^1/3

Du= ((2x+3)^(-2/3))/(2/3)

dv=Cos(X)

V= Sin(x)

then use U*du=d*v-integral[du*v] ?
 
The substitution is correct, but your du is not. The expression should be multiplied by 2/3, not divided.

So we have the following:

<br /> du=\frac{2}{3} \left(\frac{1}{(2x+3)^{\frac{1}{3}}}\right)^2 dx<br />

Now write the bracket expression in terms of u.
 
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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