Question

Integrate the following functions.

$$ \int \sqrt{\sin 2x} \cos 2x , dx $$

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Answer

To integrate the given function

$$ \int \sqrt{\sin 2x} \cos 2x , dx $$

first, we perform a substitution. Let $$ \sin 2x = t $$.

This implies: $$ 2 \cos 2x = \frac{dt}{dx} $$ or $$ dx = \frac{dt}{2 \cos 2x} $$.

Substituting into the integral:

$$ \int \sqrt{\sin 2x} \cos 2x , dx = \int \sqrt{t} \cos 2x , \frac{dt}{2 \cos 2x} $$

The ( \cos 2x ) terms cancel out:

$$ = \frac{1}{2} \int \sqrt{t} , dt $$

Next, integrate ( \sqrt{t} ):

$$ \frac{1}{2} \int t^{\frac{1}{2}} , dt = \frac{1}{2} \cdot \frac{2}{3} t^{\frac{3}{2}} + C = \frac{1}{3} t^{\frac{3}{2}} + C $$

Finally, substitute ( t ) back for ( \sin 2x ):

$$ = \frac{1}{3} (\sin 2x)^{\frac{3}{2}} + C $$

Thus, the integral is:

$$ \boxed{\frac{1}{3} (\sin 2x)^{\frac{3}{2}} + C} $$


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