1. trigonometric integration

I have this example

$\int \frac{1}{\sqrt{x^2+25}}dx$

let $x=tan(u)$, then $dx=5sec^2(u)du$

$\Raightarrow \int \frac{5sec^2(u)}{5\sqrt{x^2+1}}du \Rightarrow \int \frac{5sec^2(u)}{5sec(u)}du$

$\Rightarrow \int sec (u) du = ln|sec(u)+tan(u)|+C$

now here is where the problem for me lies.

$tan(u)=\frac{x}{5}$ I understand this but something I am not seeing is how $sec(u)=\frac{\sqrt{x^2+5}}{5}$

$ln\Bigg|\frac{\sqrt{x^2+5}}{5} + \frac{x}{5}\Bigg|+C$

2. Re: trigonometric integration

Originally Posted by Jonroberts74
I have this example
$\int \frac{1}{\sqrt{x^2+25}}dx$
let $x=tan(u)$, then $dx=5sec^2(u)du$
$\Raightarrow \int \frac{5sec^2(u)}{5\sqrt{x^2+1}}du \Rightarrow \int \frac{5sec^2(u)}{5sec(u)}du$
$\Rightarrow \int sec (u) du = ln|sec(u)+tan(u)|+C$
now here is where the problem for me lies.
$tan(u)=\frac{x}{5}$ I understand this but something I am not seeing is how $sec(u)=\frac{\sqrt{x^2+5}}{5}$
for the answer $ln\Bigg|\frac{\sqrt{x^2+5}}{5} + \frac{x}{5}\Bigg|+C$
Draw a right triangle in which there is an acute angle $u$ and $\tan(u)=\frac{x}{5}$

Now what is the $sec(u)~?$

3. Re: trigonometric integration

Originally Posted by Jonroberts74
I have this example

$\int \frac{1}{\sqrt{x^2+25}}dx$

let $x=tan(u)$, then $dx=5sec^2(u)du$

$\Raightarrow \int \frac{5sec^2(u)}{5\sqrt{x^2+1}}du \Rightarrow \int \frac{5sec^2(u)}{5sec(u)}du$

$\Rightarrow \int sec (u) du = ln|sec(u)+tan(u)|+C$

now here is where the problem for me lies.

$tan(u)=\frac{x}{5}$ I understand this but something I am not seeing is how $sec(u)=\frac{\sqrt{x^2+5}}{5}$

$ln\Bigg|\frac{\sqrt{x^2+5}}{5} + \frac{x}{5}\Bigg|+C$

$\tan^2(x) + 1 = \sec^2(x)$

$\tan(u)=\frac{x}{5}$

$\tan^2(u)+1=\sec^2(u)=\frac{x^2}{25}+1$

$\sec(u)=\sqrt{\frac{x^2}{25}+1}=\frac{\sqrt{x^2+25 }}{5}$

I think if you check you'll see that it's 25 not 5. My table shows

$\int\frac{1}{\sqrt{x^2+a^2}}dx=\log\left|x+\sqrt{x ^2+a^2}\right|$

4. Re: trigonometric integration

Originally Posted by Jonroberts74
I have this example

$\int \frac{1}{\sqrt{x^2+25}}dx$

let $x=tan(u)$, then $dx=5sec^2(u)du$

$\Raightarrow \int \frac{5sec^2(u)}{5\sqrt{x^2+1}}du \Rightarrow \int \frac{5sec^2(u)}{5sec(u)}du$

$\Rightarrow \int sec (u) du = ln|sec(u)+tan(u)|+C$

now here is where the problem for me lies.

$tan(u)=\frac{x}{5}$ I understand this but something I am not seeing is how $sec(u)=\frac{\sqrt{x^2+5}}{5}$

$ln\Bigg|\frac{\sqrt{x^2+5}}{5} + \frac{x}{5}\Bigg|+C$
Actually, you let \displaystyle \begin{align*} x = 5\tan{(u)} \end{align*}. Also if \displaystyle \begin{align*} \tan{(u)} = \frac{x}{5} \end{align*} then \displaystyle \begin{align*} \sec{(u)} = \frac{\sqrt{x^2 + 25}}{5} \end{align*}.

As for your other question, by Pythagoras we have

\displaystyle \begin{align*} \sin^2{(u)} + \cos^2{(u)} &\equiv 1 \\ \frac{\sin^2{(u)} + \cos^2{(u)}}{\cos^2{(u)}} &\equiv \frac{1}{\cos^2{(u)}} \\ \frac{\sin^2{(u)}}{\cos^2{(u)}} + \frac{\cos^2{(u)}}{\cos^2{(u)}} &\equiv \frac{1}{\cos^2{(u)}} \\ \tan^2{(u)} + 1 &= \sec^2{(u)} \end{align*}

From here you should be able to see why \displaystyle \begin{align*} \tan{(u)} = \frac{x}{5} \implies \sec{(u)} = \frac{\sqrt{x^2 + 25}}{5} \end{align*}.

5. Re: trigonometric integration

oops, type. meant $x=5tan(u)$

I was over-thinking that one. thanks