Question

(30 PONTOS) Calcule a integral indefinida:
$~$
$\displaystyle\int{\sqrt{1+\cos t}\,dt}$

Answer 1

Ola Lukyo 

  ∫ √(1 + cos(t)) dt 

  temos 1 + cos(t) = 2*cos²(t/2) 

  √2*∫√cos²(t/2) dt 

  fazendo u = t/2 e du = dx/2 

  2√2 ∫cos(u)*du = 2√2*sen(u) 

  por fim: 
  
  ∫ √(1 + cos(t)) dt = 2√2*sen(t) + C 

Answer 2

$\large\boxed{\begin{array}{l}\underline{\sf Substituic_{\!\!,}\tilde ao\,de\,Weierstrass}\\\rm t=tg\bigg(\dfrac{x}{2}\bigg)\\\rm sen(x)=\dfrac{2t}{t^2+1}\\\\\rm cos(x)=\dfrac{1-t^2}{t^2+1}\\\\\rm dx=\dfrac{2}{t^2+1}dt\end{array}}$

$\large\boxed{\begin{array}{l}\displaystyle\rm\int\sqrt{1+cos(t)}\,dt=\int\sqrt{1+\dfrac{1-x^2}{x^2+1}}\cdot\dfrac{2}{x^2+1}dx\\\\\rm1+\dfrac{1-x^2}{x^2+1}=\dfrac{\diagup\!\!\!\!\!x^2+1+1-\diagup\!\!\!\!x^2}{x^2+1}=\dfrac{2}{x^2+1}\\\\\displaystyle\rm\int\sqrt{\dfrac{2}{x^2+1}}\cdot\dfrac{2}{x^2+1}=2\sqrt{2}\int\dfrac{dx}{(x^2+1)\sqrt{x^2+1}}\end{array}}$

$\large\boxed{\begin{array}{l}\rm x= tg(\theta)\\\rm dx=sec^2(\theta)d\theta\\\rm \sqrt{x^2+1}=\sqrt{tg^2(\theta)+1}=\sqrt{sec^2(\theta)}=sec(\theta)\\\rm x^2+1=tg^2(\theta)+1=sec^2(\theta)\end{array}}$

$\large\boxed{\begin{array}{l}\displaystyle\rm2\sqrt{2}\int\dfrac{\diagup\!\!\!\!\!sec^2(\theta)}{\diagup\!\!\!\!sec^2(\theta)\cdot sec(\theta)}d\theta\\\\\displaystyle\rm2\sqrt{2}\int cos(\theta)d\theta=2\sqrt{2}sen(\theta)+k\end{array}}$

$\large\boxed{\begin{array}{l}\displaystyle\rm2\sqrt{2}\int\dfrac{dx}{(x^2+1)\sqrt{x^2+1}}=2\sqrt{2}sen\bigg(arctg\bigg(tg\bigg[\dfrac{t}{2}\bigg]\bigg)\bigg)+k\end{array}}$

$\large\boxed{\begin{array}{l}\rm 2\sqrt{2}sen\bigg(arctg\bigg(tg\bigg[\dfrac{t}{2}\bigg]\bigg)\bigg)=2\sqrt{2}tg\bigg(\dfrac{t}{2}\bigg)cos\bigg(\dfrac{t}{2}\bigg)\end{array}}$

$\large\boxed{\begin{array}{l}\displaystyle\rm\int\sqrt{1+cos(t)}dt=2\sqrt{2}\cdot tg\cdot\bigg(\dfrac{t}{2}\bigg)\cdot cos\bigg(\dfrac{t}{2}\bigg)+k\end{array}}$