MHB You have only 3 minutes for a better solution to this integral

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SUMMARY

The integral of $\cos^3(2x)$ can be simplified using the substitution $u = \sin(2x)$, leading to the expression $\frac{1}{2} \int_0^1 (1 - u^2) \, du$. This method effectively transforms the integral into a more manageable form, allowing for straightforward evaluation. The discussion emphasizes the importance of recognizing trigonometric identities and substitution techniques in calculus.

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$\cos^3(2x) = [1-\sin^2(2x)]\cos(2x)$

use the substitution $u = \sin(2x) \implies du = 2\cos(2x) \, dx$

$\displaystyle \dfrac{1}{2} \int_0^1 1-u^2 \, du$

you can finish up from here
 
skeeter said:
$\cos^3(2x) = [1-\sin^2(2x)]\cos(2x)$

use the substitution $u = \sin(2x) \implies du = 2\cos(2x) \, dx$

$\displaystyle \dfrac{1}{2} \int_0^1 1-u^2 \, du$

you can finish up from here
Thank you!
 

Thread 'Ambiguity of the term "indefinite integral"'

I've encountered a few different definitions of "indefinite integral," denoted ##\int f(x) \, dx##. any particular antiderivative ##F:\mathbb{R} \to \mathbb{R}, F'(x) = f(x)## the set of all antiderivatives ##\{F:\mathbb{R} \to \mathbb{R}, F'(x) = f(x)\}## a "canonical" antiderivative any expression of the form ##\int_a^x f(x) \, dx##, where ##a## is in the domain of ##f## and ##f## is continuous Sometimes, it becomes a little unclear which definition an author really has in mind, though...
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