What is the integral of (cos(x/2)-sin(x/2))^2 ?

The integral of (cos(x/2)-sin(x/2))^2 is cos(x)+x+C

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integral of (cos(x/2)-sin(x/2))^2

\int (cos (\frac{x}{2}) - sin (\frac{x}{2}))^{2} d x

cos (x) + x + C

Solution steps

\int (cos (\frac{x}{2}) - sin (\frac{x}{2}))^{2} d x

Factor

cos (\frac{x}{2}) - sin (\frac{x}{2}) : - (- cos (\frac{x}{2}) + sin (\frac{x}{2}))

= \int (- (- cos (\frac{x}{2}) + sin (\frac{x}{2})))^{2} d x

Simplify

(- (- cos (\frac{x}{2}) + sin (\frac{x}{2})))^{2} : (- cos (\frac{x}{2}) + sin (\frac{x}{2}))^{2}

= \int (- cos (\frac{x}{2}) + sin (\frac{x}{2}))^{2} d x

Expand

(- cos (\frac{x}{2}) + sin (\frac{x}{2}))^{2} : cos^{2} (\frac{x}{2}) - 2 cos (\frac{x}{2}) sin (\frac{x}{2}) + sin^{2} (\frac{x}{2})

= \int cos^{2} (\frac{x}{2}) - 2 cos (\frac{x}{2}) sin (\frac{x}{2}) + sin^{2} (\frac{x}{2}) d x

Rewrite using trig identities

= \int - 2 cos (\frac{x}{2}) sin (\frac{x}{2}) + 1 d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int 2 cos (\frac{x}{2}) sin (\frac{x}{2}) d x + \int 1 d x

\int 2 cos (\frac{x}{2}) sin (\frac{x}{2}) d x = - cos (x)

\int 1 d x = x

= - (- cos (x)) + x

Simplify

= cos (x) + x

Add a constant to the solution

= cos (x) + x + C

What is the integral of (cos(x/2)-sin(x/2))^2 ?
The integral of (cos(x/2)-sin(x/2))^2 is cos(x)+x+C