Question

Evaluate the integral of the quotient of the sine of x and the square root of the quantity 1 plus cosine x, dx.

Answer 1

$\bf \displaystyle \int~\cfrac{sin(x)}{\sqrt{1+cos(x)}}\cdot dx\\\\ -------------------------------\\\\ u=1+cos(x)\implies \cfrac{du}{dx}=-sin(x)\implies \cfrac{du}{-sin(x)}=dx\\\\ -------------------------------\\\\ \displaystyle \int~\cfrac{sin(x)}{\sqrt{u}}\cdot \cfrac{du}{-sin(x)}\implies -\int~\cfrac{1}{\sqrt{u}}\cdot du\implies -\int~u^{-\frac{1}{2}}\cdot du \\\\\\ -\cfrac{u^{\frac{1}{2}}}{\frac{1}{2}}\implies -2u^{\frac{1}{2}}\implies -2\sqrt{1+cos(x)}+C$

Answer 2

Answer:

∫((cos(x)*dx)/(√(1+sin(x)))) = 2√(1 + sin(x)) + c.

Step-by-step explanation:

In order to solve this question, it is important to notice that the derivative of the expression (1 + sin(x)) is present in the numerator, which is cos(x). This means that the question can be solved using the u-substitution method.

Let u = 1 + sin(x).

This means du/dx = cos(x). This implies dx = du/cos(x).

Substitute u = 1 + sin(x) and dx = du/cos(x) in the integral.

∫((cos(x)*dx)/(√(1+sin(x)))) = ∫((cos(x)*du)/(cos(x)*√(u))) = ∫((du)/(√(u)))

= ∫(u^(-1/2) * du). Integrating:

(u^(-1/2+1))/(-1/2+1) + c = (u^(1/2))/(1/2) + c = 2u^(1/2) + c = 2√u + c.

Put u = 1 + sin(x). Therefore, 2√(1 + sin(x)) + c. Therefore:

∫((cos(x)*dx)/(√(1+sin(x)))) = 2√(1 + sin(x)) + c!!