Question

For each integral in Exercises 5–8, write down three integrals that will have that form after a substitution of variables.

$\int \frac{1}{\sqrt{u}}du$

Short answer

The three integrals $\int \frac{2x+1}{\sqrt{x^2+x}}dx,\int \frac{\sin x}{\sqrt{\cos x}}dx\mathrm{and}\int \frac{1}{x\sqrt{\ln x}}dx$will have form $\int \frac{1}{\sqrt{u}}du$after a substitution of variables.

Step-by-step solution

Step 1. Given Information 

For given integral we have to write down three integrals that will have that form after a substitution of variables.
$\int \frac{1}{\sqrt{u}}du$

Step 2. The first integrals that will have that form after a substitution of variables.

Let

$$\begin{gathered} u=x^2+x \\ \frac{du}{dx}=2x+1 \\ du=(2x+1)dx \end{gathered}$$

This substitution changes the integral into

localid="1649155815093" $\int \frac{2x+1}{\sqrt{x^2+x}}dx$

Step 3. The first integrals that will have that form after a substitution of variables.

Let

$$\begin{gathered} u=\cos x \\ \frac{du}{dx}=\sin x \\ du=\sin xdx \end{gathered}$$

This substitution changes the integral into

role="math" localid="1648747414306" $\int \frac{\sin x}{\sqrt{\cos x}}dx$

Step 4. The first integrals that will have that form after a substitution of variables.

Let

$$\begin{gathered} u=\ln x \\ \frac{du}{dx}=\frac{1}{x} \\ du=\frac{1}{x}dx \end{gathered}$$

This substitution changes the integral into

$\int \frac{1}{x\sqrt{\ln x}}dx$