Răspuns:
i) Se face schimbarea de variabilă [tex]\sin^2x=t\Rightarrow 2\sin x\cos x=dt\Rightarrow \sin 2x=dt[/tex]
[tex]I(t)=\displaystyle\int\dfrac{dt}{\sqrt{t^2+1}}=\ln(t+\sqrt{t^2+1})+\mathcal{C}[/tex]
Atunci
[tex]I(x)=\ln\left(\sin^2x+\sqrt{\sin^4 x+1}\right)+\mathcal{C}[/tex]
j) Integrala se mai scrie
[tex]\displaystyle\int \sin x(1-\cos^2x)\cos^2xdx[/tex]
Se face schimbarea de variabilă
[tex]\cos x=t\Rightarrow -\sin xdx=dt\Rightarrow\sin xdx=-dt[/tex]
[tex]I(t)=-\displaystyle\int(1-t^2)t^2dt=\int(t^4-t^2)dt=\dfrac{t^5}{5}-\dfrac{t^3}{3}+\mathcal{C}[/tex]
[tex]I(x)=\dfrac{\cos^5x}{5}-\dfrac{\cos^3x}{3}+\mathcal{C}[/tex]
Explicație pas cu pas:
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Răspuns:
i) Se face schimbarea de variabilă [tex]\sin^2x=t\Rightarrow 2\sin x\cos x=dt\Rightarrow \sin 2x=dt[/tex]
[tex]I(t)=\displaystyle\int\dfrac{dt}{\sqrt{t^2+1}}=\ln(t+\sqrt{t^2+1})+\mathcal{C}[/tex]
Atunci
[tex]I(x)=\ln\left(\sin^2x+\sqrt{\sin^4 x+1}\right)+\mathcal{C}[/tex]
j) Integrala se mai scrie
[tex]\displaystyle\int \sin x(1-\cos^2x)\cos^2xdx[/tex]
Se face schimbarea de variabilă
[tex]\cos x=t\Rightarrow -\sin xdx=dt\Rightarrow\sin xdx=-dt[/tex]
[tex]I(t)=-\displaystyle\int(1-t^2)t^2dt=\int(t^4-t^2)dt=\dfrac{t^5}{5}-\dfrac{t^3}{3}+\mathcal{C}[/tex]
[tex]I(x)=\dfrac{\cos^5x}{5}-\dfrac{\cos^3x}{3}+\mathcal{C}[/tex]
Explicație pas cu pas: