The most direct solution for a definite integral

[Appleton]

Hi, I'm wondering if I have the most direct solution for this integral or if there is a more efficient way of solving this. I haven't seen a double substitution deployed on one of these problems yet, so I thought perhaps this was not necessary.

Homework Statement

Using the substitution t = tan x, or otherwise, evaluate:
$\int^{\frac{\pi}{3}}_{0} \frac{1}{9 - 8sin^2x} dx$

Homework Equations

The Attempt at a Solution

$\int^{\frac{\pi}{3}}_{0} \frac{1}{9 - 8sin^2x} dx$

$= \int^{\frac{\pi}{3}}_{0} \frac{1}{1 + 8cos^2x} dx$

$= \int^{\frac{\pi}{3}}_{0} \frac{1}{4cos2x + 5} dx$

$let\ t = tanx$

$cos2x = \frac{1 - t^2}{1 + t^2}$ (trig identity)

$so \int^{\frac{\pi}{3}}_{0} \frac{1}{4cos2x + 5} dx$ $= \int^{\frac{\pi}{3}}_{0} \frac{1}{4(\frac{1 - t^2}{1 + t^2}) + 5} dx$

$= \int^{\frac{\pi}{3}}_{0} \frac{1 + t^2}{t^2 +9} dx$

$= \int^{\frac{\pi}{3}}_{0} \frac{sec^2x}{tan^2x +9} dx$

$t = tanx → dt = sec^2x\ dx$

$so \int^{\frac{\pi}{3}}_{0} \frac{sec^2x}{tan^2x + 9} dx$ $= \int^{\sqrt 3}_{0} \frac{1}{t^2 +9} dt$

$let\ t = 3tanθ$ $so\ dt = 3sec^2θ\ dθ$

$so \int^{\sqrt 3}_{0} \frac{1}{t^2 +9} dt$ $= \int^{atan\frac{1}{\sqrt 3}}_{0} \frac{3sec^2θ}{9tan^2θ +9} dθ$

$= \int^{π/6}_{0} \frac{3sec^2θ}{9sec^2θ} dθ$

$= \int^{π/6}_{0} \frac{3}{9} dθ$

$= \frac{π}{18}$

 

[Tanya Sharma]

$\int^{\frac{\pi}{3}}_{0} \frac{1}{9 - 8sin^2x} dx$

$= \int^{\frac{\pi}{3}}_{0} \frac{1}{1 + 8cos^2x} dx$

$= \int^{\frac{\pi}{3}}_{0} \frac{1}{4cos2x + 5} dx$

$let\ t = tanx$

$cos2x = \frac{1 - t^2}{1 + t^2}$ (trig identity)

$so \int^{\frac{\pi}{3}}_{0} \frac{1}{4cos2x + 5} dx$ $= \int^{\frac{\pi}{3}}_{0} \frac{1}{4(\frac{1 - t^2}{1 + t^2}) + 5} dx$

$= \int^{\frac{\pi}{3}}_{0} \frac{1 + t^2}{t^2 +9} dx$

I haven't checked your entire work ,but what was quite evident is as follows-

You could have avoided the second substitution by properly performing the first substitution and changing the limits .When you changed the variable from 'x' to 't' ,you should also change the limits .When the variable is 'x' ,the limits correspond to 'x' .When the variable is 't' ,the limits should correspond to 't' .

Since t=tanx ,dt = sec²xdx or dx = dt/1+t²

When x=0 →t=0 and when x=π/3 → t = √3

The item in red should have been written as $= \int^{\sqrt{3}}_{0} \frac{1}{t^2 +9} dt$

You will find this integral quite easy to handle without making further substitutions.

Hope this helps

 

[ehild]

You result is correct, but you overcomplicated the integration.

Hi, I'm wondering if I have the most direct solution for this integral or if there is a more efficient way of solving this. I haven't seen a double substitution deployed on one of these problems yet, so I thought perhaps this was not necessary.

Homework Statement

Using the substitution t = tan x, or otherwise, evaluate:
$\int^{\frac{\pi}{3}}_{0} \frac{1}{9 - 8sin^2x} dx$

Homework Equations

The Attempt at a Solution

$\int^{\frac{\pi}{3}}_{0} \frac{1}{9 - 8sin^2x} dx$

$= \int^{\frac{\pi}{3}}_{0} \frac{1}{1 + 8cos^2x} dx$

very good... Substitute $\cos^2x=\frac{1}{1+\tan^2x}$

and then use the substitution

$let\ t = tanx$ x=arctan(t), $dx=\frac{1}{1+t^2}dt$. You directly arrive at

$= \int^{\sqrt 3}_{0} \frac{1}{t^2 +9} dt$

You can write the integral in form $= \frac{1}{9}\int^{\sqrt 3}_{0} \frac{1}{\left(\frac{t}{3}\right)^2 +1} dt$ which is a basic integral - do you recognize it?

Spoiler

∫(1/(1+x²)dx=arctan(x)

Edit: Oh, Tania beat me

ehild

 

[Appleton]

Thanks for your helpful comments, I can see I need to familiarize myself more thoroughly with some of these basic integrals.