Question

Hi there, I am having trouble solving these two integrals as I continue to get the answers wrong:

Answer 1

Answer:

663.84 degrees

Explanations:

Given the integral function

$\int_{-\infty}^8\frac{8}{x^2+4}dx$

Since 8 is a constant, it will be outside the integral function to have:

$\begin{gathered} 8\int_{-\infty}^8\frac{1}{x^2+4}dx \\ 8\int_{-\infty}^8\frac{1}{x^2+2^2}dx \end{gathered}$

Using the general rule shown below;

$\int\frac{1}{x^2+a^2}dx=\frac{1}{a}tan^{-1}(\frac{x}{a})+C$

Comparing with the given integral, a = 2, such that;

$\begin{gathered} 8\int_{-\infty}^8\frac{1}{x^2+2^2}=[8(\frac{1}{2}tan^{-1}(\frac{x}{2}))]_{-\infty}^8 \\ \int_{-\infty}^8\frac{8}{x^2+2^2}=[4tan^{-1}(\frac{x}{2})]_{-\infty}^8 \end{gathered}$

Substitute the limits

$\begin{gathered} \int_{-\infty}^8\frac{8}{x^2+2^2}=[4tan^{-1}(\frac{8}{2})-4tan^{-1}(-\frac{\infty}{2})] \\ \int_{-\infty}^8\frac{8}{x^2+2^2}=[4(75.96)-4(-90)] \\ \int_{-\infty}^8\frac{8}{x^2+2^2}=303.84+360=663.84^0 \\ \end{gathered}$