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

Answer:

$\int{\frac{cos(x)}{\sqrt{1+sin(x)}}dx = 2\sqrt{1+sin(x)}+c$

Step-by-step explanation:

We have the following integral:

$\int{\frac{cos(x)}{\sqrt{1+sin(x)}}dx$

Use the following substitution:

$u = 1 + sin (x)$

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

So:

$du = cos(x) dx$

$\frac{du}{cos(x)} = dx$

When making this substitution, the integral remains as follows:

$\int{\frac{cos(x)}{\sqrt{u}}}*\frac{du}{cos(x)}\\\\\\\int{\frac{1}{\sqrt{u}}}du\\\\\int u^{-\frac{1}{2}}du\\\\\int u^{-\frac{1}{2}}du=\frac{u^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}=2\sqrt{u}+c$

Then we have to:

$\int{\frac{cos(x)}{\sqrt{1+sin(x)}}dx = 2\sqrt{1+sin(x)}+c$

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!!!