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

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Answer 1 · 2015-10-20T01:22:09

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.

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