How do you find the indefinite integral of #int sin2xcos2xdx #?

Question

William Abdalla · Accepted Answer

Here are three solutions.

Substitution 1 #int sin2xcos2xdx # Let #u = sin2x#, so that #du = 2cos2x dx#and the integral becomes #1/2intudu = 1/4u^2 +C# #int sin2xcos2xdx = 1/4sin^2 2x +C # Substitution 2 #int sin2xcos2xdx # Let #u = cos2x#, so that #du = -2sin2x dx#and the integral becomes #-1/2intudu = -1/4u^2 +C# #int sin2xcos2xdx = -1/4cos^2 2x +C # Solution 3 #int sin2xcos2xdx = int 1/2sin4x dx# Let #u = 4x#, so #du = 4dx# and the integral becomes #1/8 int sinu du = -1/8 cos u +C# #int sin2xcos2xdx = -1/8cos4x +C# The challenge is to see why these answers are the same. Hint find the difference between the answers. Hint 2 "difference" means subtract.

Emma Aldridge · Answer

undefined To find the indefinite integral of $ \int \sin^2(2x) \cos^2(2x) \, dx $, you can use the trigonometric identity $ \sin^2(2x) = \frac{1 - \cos(4x)}{2} $ and $ \cos^2(2x) = \frac{1 + \cos(4x)}{2} $. Then, you can substitute these identities into the integral and integrate term by term. The result will be:

$$ \int \sin^2(2x) \cos^2(2x) \, dx = \frac{1}{4} \int (1 - \cos(4x))(1 + \cos(4x)) \, dx $$

$$ = \frac{1}{4} \int (1 - \cos^2(4x)) \, dx $$

$$ = \frac{1}{4} \int (1 - \frac{1 + \cos(8x)}{2}) \, dx $$

$$ = \frac{1}{4} \left( \int 1 \, dx - \frac{1}{2} \int (1 + \cos(8x)) \, dx \right) $$

$$ = \frac{1}{4} \left( x - \frac{x}{2} - \frac{1}{16} \sin(8x) \right) + C $$

$$ = \frac{1}{4} \left( \frac{x}{2} - \frac{1}{16} \sin(8x) \right) + C $$

Where $ C $ is the constant of integration.