Solving Integrals with Substitution Method - Step-by-Step Guide

[Nyasha]

Homework Statement

In each of the following cases use the given substitution in order to evaluate the given integral:

$$\int\frac{2+\sqrt x}{1-\sqrt x}dx=2(4(1+\sqrt x)^\frac{3}{2}-\frac{(1-\sqrt x)^2}{2}-3In(1-\sqrt x)$$

Homework Equations

For the substitution
$$u=1-\sqrt x$$

$$x=(1-u)^2$$

The Attempt at a Solution

$$du=\frac{-1}{2\sqrt x}$$

$$-2\sqrt x du=dx$$

$$-2\sqrt(1-u)^2du=dx$$


$$-2\int\frac{2+\sqrt(1-u)^2}{u}(\sqrt(1-u)^2$$

$$-2\int 2\sqrt(1-u^2)(u^-1)du+ \sqrt(1-u^2)(u^-1)$$

Guys how come when l simplify this integration l do not get the answer correct answer which is shown in the problem statement ? Where have l done my mistake ?

 

[gabbagabbahey]

$$du=\frac{-1}{2\sqrt x}$$

Surely you mean $$\frac{du}{dx}=\frac{-1}{2\sqrt x}$$...right?

$$-2\sqrt x du=dx$$

$$-2\sqrt(1-u)^2du=dx$$

Technically, this is incorrect; since sqrt() always reteurns the positive root,

$$\sqrt{(1-u)^2}=\left\{\begin{array}{lr}1-u, & 1-u \geq 0 \\ u-1, & 1-u\leq 0\end{array}$$

But it should be clear from your definition of u that $u=1-\sqrt{x}\implies \sqrt{x}=1-u$ always. So $\sqrt{x}\neq\sqrt{(1-u)^2}$ in general.

$$-2\int 2\sqrt(1-u^2)(u^-1)du+ \sqrt(1-u^2)(u^-1)$$

Now you seem to be claiming $\sqrt{(1-u)^2}=\sqrt{1-u^2}$...surely you didn't mean to do that!

Use $\sqrt{x}=1-u$ instead.

 

[Nyasha]

Surely you mean $$\frac{du}{dx}=\frac{-1}{2\sqrt x}$$...right?



Technically, this is incorrect; since sqrt() always reteurns the positive root,

$$\sqrt{(1-u)^2}=\left\{\begin{array}{lr}1-u, & 1-u \geq 0 \\ u-1, & 1-u\leq 0\end{array}$$

But it should be clear from your definition of u that $u=1-\sqrt{x}\implies \sqrt{x}=1-u$ always. So $\sqrt{x}\neq\sqrt{(1-u)^2}$ in general.




Now you seem to be claiming $\sqrt{(1-u)^2}=\sqrt{1-u^2}$...surely you didn't mean to do that!

Use $\sqrt{x}=1-u$ instead.

Maybe l should take a new approach since this one was a complete disaster.

$$x=(1-u)^2$$

$$dx= -2(1-u)du$$

$$dx=2u-2du$$

$$\int\frac{2+\sqrt (1-u)^2}{u}.(2u-2)du$$

I would like to know if l am on the right path before l take another wild goose chase

 

[gabbagabbahey]

Maybe l should take a new approach since this one was a complete disaster.

$$x=(1-u)^2$$

$$dx= -2(1-u)du$$

$$dx=2u-2du$$

$$\int\frac{2+\sqrt (1-u)^2}{u}.(2u-2)du$$

I would like to know if l am on the right path before l take another wild goose chase

Again, $\sqrt{(1-u)^2}$ depends on whether $1-u$ is positive or negative.

To avoid having to analyze the two different cases, stick to your original substitution; $u=1-\sqrt{x}\implies \sqrt{x}=1-u$...

 

[Nyasha]

Again, $\sqrt{(1-u)^2}$ depends on whether $1-u$ is positive or negative.

To avoid having to analyze the two different cases, stick to your original substitution; $u=1-\sqrt{x}\implies \sqrt{x}=1-u$...

Ohhh, l now get it. This means my integrand will become:


$$\int\frac{2+ (1-u)}{u}(2u-2)du=\int\frac{4u-2}{u}du + \int\frac{(2u-2)(1-u)}{u}du$$

 

[gabbagabbahey]

Ohhh, l now get it. This means my integrand will become:


$$\int\frac{2+ (1-u)}{u}(2u-2)du=\int\frac{4u-2}{u}du + \int\frac{(2u-2)(1-u)}{u}du$$

Actually, $$\int\frac{2+ (1-u)}{u}(2u-2)du=\int\frac{4u-4}{u}du + \int\frac{(2u-2)(1-u)}{u}du$$

But there is a much easier way to simplify this:

$$\int\frac{2+ (1-u)}{u}(2u-2)du=2\int\frac{(3-u)(u-1)}{u}du$$

And $(3-u)(u-1)=$___?

 

[Nyasha]

Actually, $$\int\frac{2+ (1-u)}{u}(2u-2)du=\int\frac{4u-4}{u}du + \int\frac{(2u-2)(1-u)}{u}du$$

But there is a much easier way to simplify this:

$$\int\frac{2+ (1-u)}{u}(2u-2)du=2\int\frac{(3-u)(u-1)}{u}du$$

And $(3-u)(u-1)=$___?

Thanks very much for the help