Question

Evaluate the integral int(cos²x dx)
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Answer 1

Step-by-step explanation:

We have,

$i = \int \cos^{2} (x) dx$

$i = \int \cos(x) . \cos(x) dx$

$= > i = \cos(x) . \int \cos(x) dx - \int( \frac{d}{dx} ( \cos(x) ). \int \cos(x) dx)dx$

$= > i = \cos(x) \sin(x) - \int - \sin(x) . \sin(x) dx$

$= >i = \sin(x) \cos(x) + \int \sin^{2} (x) dx$

$= > i = \sin(x) \cos(x) + \int(1 - \cos^{2} (x) )dx$

$= > i = \sin(x) \cos(x) + \int1 \times dx - \int \cos^{2} (x) dx$

$= > i = \sin(x) \cos(x) + x - i$

$= > 2i = \sin(x) \cos(x) + x$

$= > i = \frac{ \sin(x) \cos(x ) }{2} + \frac{x}{2}$