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Ma poate ajuta cineva prin metoda integrarii prin parti integrala din( cosx+cos la a 2 de x )dx?​

Răspuns :

[tex]I=\displaystyle\int\cos x+\cos^2x\,dx=\int\cos xdx+\int\cos^2x dx=I_1+I_2\\\\I_1=\displaystyle\int\cos x dx=\sin x+C\\\\I_2=\displaystyle\int\cos^2 x dx=\int\cos x\cdot cos x dx=\int\cos x(\sin x)'dx=\\=\sin x\cos x-\int\sin x(\cos x)dx=\sin x\cos x+\int\sin^2xdx=\\=\sin x\cos x+\int1-\cos^2 xdx=\sin x\cos x+\int1dx-\int\cos^2xdx=\\=\sin x\cos x+x-I_2\\\\2I_2=x+\sin x \cos x\\\\\displaystyle I_2=\frac{x+\sin x\cos x}{2}+C\\\\I=I_1+I_2=\sin x+\frac{x+\sin x\cos x}{2}+C[/tex]