MHB How Does Trigonometric Substitution Simplify the Integral Calculation?

[karush]

206.8.8.49
$a=0 \ \ b=12$
$\displaystyle I_{49}=\int_{a}^{b} \frac{dx}{\sqrt[]{144-x^2}} \, dx = \arcsin{\left[\frac{1}{12}\right]} \\$
$ \text{use identity} $
$\sin^2\theta+\cos^2\theta = 1
\Rightarrow 1-\cos^2\theta=\sin^2\theta
\\$
$\text{x substituion} $
$\displaystyle
x=12\sin{\theta}
\therefore dx=12\cos{\theta}
\therefore \theta=\arcsin\left[\frac{x}{12}\right]
\\$
$\displaystyle I_{49}=\int_{a}^{b} \frac{12\cos {\theta}}{\sqrt[]{144-144cos^2\theta}} \, d\theta
= \int_{a}^{b} \frac{\cos\theta}{\cos{\theta} }\, d\theta =
\int_{a}^{b} 1 \,d\theta = \left[\theta\right]_a^b
\\ $
$\text{back substitute } \\
\displaystyle
a=0 \ \ b=12 \ \ \theta=\arcsin\left[\frac{x}{12}\right]$
$\arcsin\left[\frac{12}{12}\right]
-\arcsin\left[\frac{0}{12}\right]=\frac{\pi}{2}$

 

[MarkFL]

Here's how I'd work this problem:

$$I=\int_{0}^{12}\frac{1}{\sqrt{12^2-x^2}}\,dx$$

Let:

$$x=12\sin(\theta)\implies dx=12\cos(\theta)\,d\theta$$

$$I=\int_{0}^{\frac{\pi}{2}}\,d\theta=\frac{\pi}{2}$$

 

[karush]

Ok, after drying my years
Why would the domain be $0$ to $\frac{\pi}{2}$ ?

 

[MarkFL]

Ok, after drying my years
Why would the domain be $0$ to $\frac{\pi}{2}$ ?

Originally, the limits were in terms of $x$, but with the substitution were using, we need them to be in terms of $\theta$. So we use:

$$\theta(x)=\arcsin\left(\frac{x}{12}\right)$$

So, we find the new limits to be:

$$\theta(0)=\arcsin\left(\frac{0}{12}\right)=0$$

$$\theta(12)=\arcsin\left(\frac{12}{12}\right)=\frac{\pi}{2}$$

 

[karush]

Ok got it

 

[MarkFL]

The only real issue I would find with what you originally posted is this line:

$$I_{49}=\int_{a}^{b} \frac{dx}{\sqrt{144-x^2}}\,dx=\arcsin\left(\frac{1}{12}\right)$$

You have two differentials in the integral, and the result should be:

$$I_{49}=\int_{a}^{b} \frac{1}{\sqrt{144-x^2}}\,dx=\arcsin\left(\frac{b}{12}\right)-\arcsin\left(\frac{a}{12}\right)$$