Question

integrala din sin patrat x

Answer 1

$\int\limits sin^2 {x} \, dx = \int\limits sin x sin x \, dx = \int\limits (-cos x)' sinx \, dx =$

$-cosxsinx- \int\limits (-cos x) (sinx)' \, dx = -cosx sinx + \int\limits cosx *cosx \, dx$

[tex]= -cosxsinx+ \int\limits cos^2x \, dx = -cosxsinx+ \int\limits (1-sin^2x) \, dx = [/tex]

$=-cosxsinx+ \int\limits 1dx \, dx - \int\limits sin^2x \, dx =$

Notam $sin^2x$ cu I

[tex]-cosxsinx+x-I I=-cosx sinx+x-I 2I=-cosx sinx + x [/tex]

$I= \frac{-cosxsinx+x}{2} +C$

[tex]Sin^2x+Cos^2x=1 Sin^2x=1-cos^2x Cos^2x=1-Sin^2x[/tex]