Evaluate the integral of $ cos(x)sin(x) $ from $ 0 $ to $ frac{pi}{2} $

To evaluate the integral of $ \cos(x)\sin(x) $ from $ 0 $ to $ \frac{\pi}{2} $, we can use the substitution method. Let:

$ u = \sin(x) $

Then,

$ du = \cos(x) dx $

The integral transforms to:

$ \int_0^{\frac{\pi}{2}} \cos(x)\sin(x) dx = \int_0^1 u du $

Evaluating this integral:

$ \int_0^1 u du = \frac{u^2}{2} \Bigg|_0^1 = \frac{1}{2} $

Thus, the value of the integral is $ \frac{1}{2} $.

To evaluate the integral of $ cos(x)sin(x) $ from $ 0 $ to $ frac{pi}{2} $, we use the substitution method:

$ u = sin(x) $

$ du = cos(x) dx $

Then:

$ int_0^{frac{pi}{2}} cos(x)sin(x) dx = int_0^1 u du = frac{1}{2} $

Using substitution:

$ u = sin(x) $

Then:

$ int cos(x)sin(x) dx = frac{1}{2} $

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