The indefinite integral of a function f defined on an interval I is a function F (generally noted in uppercase), itself defined and differentiable on I, whose derivative is f, i.e. F ′ (x) = f (x)
The (unbounded) cosine primitive is equal to a positive sine, and the sine primitive is equal to a negative cosine. ∫ (cosine) = sine which gives: ∫ (cos (x)) dx = sin (x) ∫ (sine) = - cosine which gives: ∫ (sin (x)) dx = - cos (x)
You have to imagine being in the same story, but this time the scene takes place when SINUS arrived on dry land (he is positive and happy to have left the water) ! Now that he's safe, we put his head back (we put him in)! When SINUS is integrated, it finds its head (its "CO") and (re) transforms into negative COSINUS! (Negative because eventually he got used to his SINUS, and is not happy with this transformation)!
Use Pythagorean Identities: sin²x=1/2*(1 - cos(2x))/2
Use Sum Rule: ∫f(x)+g(x) dx=∫f(x) dx+ ∫g(x) dx.
∫1/2 dx - ∫((cos(2x))/2)dx
Use this rule:∫a dx = ax + C.
Use Constant Factor Rule ∫c f(x) dx = c ∫f(x) dx.
(x/2)-(1/2) ∫ cos2x dx
Use Integration by Substitution on ∫cos2x dx.
Let u=2x, du=2 dx, then dx=1/2 du
Using u and du above, rewrite ∫cos2x dx.
∫ (cos u)/2 du
Use Constant Factor Rule: ∫cf(x) dx = c∫f(x) dx.
1/2∫cos u du
Use Trigonometric Integration: the integral of cos u is sin u.
Substitute u=2x back into the original integral.
Rewrite the integral with the completed substitution.
(x/2)-(sin2x)/4 + C