Question

Calcule a integral com x variando de 0 a 2pi de raiz quadrada de 1+cos(X) dx. Como resolver?

Answer 1

Devemos saber que:

$\cos(2u)=2\cos^2(u)-1\\\\ \Longrightarrow\cos(x)=2\cos^2\left(\frac{x}{2}\right)-1$

Com isso, podemos escrever:

$I=\displaystyle\int\sqrt{1+\cos x}\,dx\\\\ I=\displaystyle\int\sqrt{1+(2\cos^2\left(\frac{x}{2}\right)-1)}\,dx\\\\ I=\displaystyle\int\sqrt{2\cos^2\left(\frac{x}{2}\right)}\,dx\\\\ I=\displaystyle\int\sqrt2|\cos\left(\frac{x}{2}\right)|\,dx\\\\ \text{Para cos(x/2)≥0:}\\\\ I=\sqrt2\displaystyle\int\cos\left(\frac{x}{2}\right)\,dx$

Fazendo uma substituição:

$y=\dfrac{x}{2}\\\\ dy=\dfrac{dx}{2}\iff dx=2dy$

Usando na integral:

$I=\sqrt2\displaystyle\int\cos\left(\frac{x}{2}\right)\,dx\\\\ I=\sqrt2\displaystyle\int\cos(y)\,2dy\\\\ I=2\sqrt2\displaystyle\int\cos(y)\,dy\\\\ I=2\sqrt2(\sin(y))+C$

Desfazendo a substituição:

$I=2\sqrt2\sin\left(\frac{x}{2}\right)+C$

Com os limites de integração dados, devemos prestar atenção. Veja que a integral foi definida acima colocando cos(x/2)≥0. No intervalo em que cos(x/2)≤0, o sinal inverte. Vamos dividir a integral:

$I_1=\displaystyle\int^{2\pi}_{0}\sqrt{1+\cos x}\,dx\\\\ I_1=\displaystyle\int^{\pi}_{0}\sqrt{1+\cos x}\,dx-\displaystyle\int^{2\pi}_{\pi}\sqrt{1+\cos x}\,dx\\\\ I_1=[2\sqrt2\sin\left(\frac{x}{2}\right)]^{\pi}_0-[2\sqrt2\sin\left(\frac{x}{2}\right)]^{2\pi}_{\pi}\\\\ I_1=[(2\sqrt2\sin\left(\frac{\pi}{2}\right))-(2\sqrt2\sin\left(\frac{0}{2}\right))]-[(2\sqrt2\sin\left(\frac{2\pi}{2}\right))-(2\sqrt2\sin\left(\frac{\pi}{2}\right))]\\\\ I_1=[(2\sqrt2\sin(\pi/2))-(2\sqrt2\sin(0))]_[(2\sqrt2\sin(\pi))-(2\sqrt2\sin(\pi/2))]\\\\ I_1=[2\sqrt2\cdot1-0]-[0-2\sqrt2\cdot1]\\\\ \boxed{\displaystyle\int^{2\pi}_{0}\sqrt{1+\cos x}\,dx=4\sqrt2}$