Question

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

Answer 1

`\int{(cos(x))/(√(1+sin(x)))dx = 2√(1+sin(x))+c`

Explanation:

We have the following integral:

`\int{(cos(x))/(√(1+sin(x)))dx`

Use the following substitution:

`u = 1 + sin (x)`

Remember that the derivative of
`sin(x)` is
`cos(x)`

So:

`du = cos(x) dx`

`(du)/(cos(x)) = dx`

When making this substitution, the integral remains as follows:

`\int{(cos(x))/(√(u))}*(du)/(cos(x))\\\\\\\int{(1)/(√(u))}du\\\\\int u^{-(1)/(2)}du\\\\\int u^{-(1)/(2)}du=\frac{u^{-(1)/(2)+1}}{-(1)/(2)+1}=2√(u)+c`

Then we have to:

`\int{(cos(x))/(√(1+sin(x)))dx = 2√(1+sin(x))+c`

Answer 2

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

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!!!