MHB Solving an Integral Using Trig Substitution

[shamieh]

$$\int \frac{81/100}{100x^2 + 81}$$

so I know this is $$1 + tan^2\theta$$

but how do I implement it? isn't it $$a^2 + x$$

so should i let $$x = 100tan\theta$$?

 

[MarkFL]

I think I would choose to write the integral as:

$$I=\frac{1}{100}\int\frac{1}{\left(\dfrac{10x}{9}\right)^2+1}\,dx$$

Now, what substitution do you think you should make?

 

[Prove It]

$$\int \frac{81/100}{100x^2 + 81}$$

so I know this is $$1 + tan^2\theta$$

but how do I implement it? isn't it $$a^2 + x$$

so should i let $$x = 100tan\theta$$?

$\displaystyle \begin{align*} \int{\frac{\frac{81}{100}}{100x^2 + 81}\,\mathrm{d}x} &= \frac{81}{100} \int{ \frac{1}{ \left( 10 x \right) ^2 + 9^2 } \,\mathrm{d}x } \end{align*}$

So now make the substitution $\displaystyle \begin{align*} 10x = 9\tan{(\theta)} \implies \mathrm{d}x = \frac{9}{10}\sec^2{(\theta)}\,\mathrm{d}\theta \end{align*}$ and the integral becomes

$\displaystyle \begin{align*} \frac{81}{100} \int{ \frac{1}{ \left[ 9\tan{(\theta)} \right] ^2 + 9^2 } \, \frac{9}{10}\sec^2{(\theta)}\,\mathrm{d}\theta } \end{align*}$

You should be able to do something with this now...