MHB What's the best strategy to solving this Integral in 3 minutes?

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To solve the integral of cos^3(2x), it is effective to separate it into (cos^2(2x))cos(2x) and use the identity cos^2(2x) = 1 - sin^2(2x). A substitution of u = sin(2x) simplifies the integral, leading to du = 2cos(2x)dx, which allows for the transformation of cos(2x)dx into (1/2)du. The integral then evaluates to (1/2)∫(1 - u^2)du from 0 to sin(8), resulting in the expression sin(8) - (sin^3(8)/3). This method demonstrates a clear strategy to solve the integral efficiently within a limited time frame.
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Please do not double post here, either.

-Dan
 
Separate cos^3(2x) as (cos^2(2x))cos(2x)= (1- sin^2(2x))cos(2x). Now use the substitution u= sin(2x) so that du= 2cos(2x)dx, cos(2x)dx= (1/2)du. When x= 0, u= sin(0)= 0 and when x= 4, u= sin(8). The integral becomes $$\frac{1}{2}\int_0^{sin(8)}(1- u^2)du=\left[u- \frac{u^3}{3}\right]_0^{sin(8)}$$$$= sin(8)- \frac{sin^3(8)}{3}$$.
 
$x = \dfrac{\pi}{4} \implies \text{ upper limit }, u= 1$
 
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Yes, and thank you!
 

Thread 'Area of Overlapping Squares'

Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?
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