1. ## Definite integration

Using the substitution $t=\tan x$, or otherwise, find
$\int^{\frac{\pi}{3}}_0\frac{1}{9-8\sin^2x}dx$

$\int^{\frac{\pi}{3}}_0\frac{1}{5+4\cos2x}dx$
when $x=\frac{\pi}{3}\rightarrow t=\sqrt{3}$ and $x=0\rightarrow t=0$
$t=\tan x\rightarrow dt=sec^2xdx$
$\int^{\frac{\pi}{3}}_0\frac{1}{5+4(\frac{1-t^2}{1+t^2})}dx$
$\int^{\frac{\pi}{3}}_0\frac{1+t^2}{9+t^2}dx$
$\int^{\sqrt{3}}_0\frac{1}{9+t^2}dt$
$[\frac{1}{t}\ln(9+t^2)]^{\sqrt{3}}_0$
now then the answer i would get would be
$\frac{1}{\sqrt{3}}\ln(12)$
but the answer is supposed to be $\frac{1}{18}\pi$ which cannot be arrived at from $[\frac{1}{t}\ln(9+t^2)]^{\sqrt{3}}_0$ as an inverse trig is needed
thanks

2. Originally Posted by arze
[snip]

Using the substitution $t=\tan x$, or otherwise, find
$\int^{\frac{\pi}{3}}_0\frac{1}{9-8\sin^2x}dx$

$\int^{\frac{\pi}{3}}_0\frac{1}{5+4\cos2x}dx$
when $x=\frac{\pi}{3}\rightarrow t=\sqrt{3}$ and $x=0\rightarrow t=0$
$t=\tan x\rightarrow dt=sec^2xdx$
$\int^{\frac{\pi}{3}}_0\frac{1}{5+4(\frac{1-t^2}{1+t^2})}dx$
$\int^{\frac{\pi}{3}}_0\frac{1+t^2}{9+t^2}dx$
$\int^{\sqrt{3}}_0\frac{1}{9+t^2}dt$
$[\frac{1}{t}\ln(9+t^2)]^{\sqrt{3}}_0$

[/snip]
Everything seems ok, except you integrated the last integral incorrectly.

You should have gotten $\left.\left[\frac{1}{3}\arctan\left(\frac{t}{3}\right)\right]\right|_0^{\sqrt{3}}$

3. integration of (1/9+t^2) is not ln(9+t^2)....you need to use this formula for that purpose...

integral of [1/ (a^2 + x^2) ] = (1/a) . arctan (x/a)

hopefully that will help!

ohh and arctan is the same as (inverse tan) ...

Good luck!