trig substitution: int (sec^2(x)) / (4-(tan^2(x))^(3/2)

[T_TEngineer_AdamT_T]

Hi again
this is still trigonometric substitution and i got stuck with this problem
\(\displaystyle \int \frac{(\sec^2(x))}{(4-\tan^2(x))^{\frac{3}{2}}dx\)

i know that this is the pattern:
let x = 2sin(theta)d(theta)
and dx = 2cos(theta)d(theta)

but i dont know how to because of the tan and the sec..
i really cant move on
help me pls just 1 clue

 

[galactus]

What was your original problem, Adam?. That may help to see if there are any errors. Is this the original problem?. Or have you made the trig substitutions?.

 

[T_TEngineer_AdamT_T]

i need to get rid of the tangent and secant
but i'll try...
let tan(x) = 2sin(theta)
d(tan(x)) = 2cos(theta)d(theta)

then integrate:
\(\displaystyle \int \frac{\sec^2(x)dx}{(4-4\sin(\theta))^{\frac{3}{2}}}\)
\(\displaystyle \int \frac{\sec^2(x)(2\cos(\theta)d\theta)}{2(1-\sin^2{\theta})^{\frac{3}{2}}\)

cancel the cosine

then i get a int sec^2(x)dx

but the answer is
\(\displaystyle \frac{\tan(x)}{4\sqrt{4-\tan^2(x)}} + C\)

 

[galactus]

Let \(\displaystyle \L\\u=tan(x), \;\ du=sec^{2}(x)dx\)

\(\displaystyle \L\\\int\frac{1}{(4-u^{2})^{\frac{3}{2}}}du\)

\(\displaystyle \L\\=\frac{u}{4\sqrt{4-u^{2}}}du\)

Now, resub and you have your solution.

 

[soroban]

Hello, T_TEngineer_AdamT_T!

I think you've got it!

. . . cancel the cosine

then i get: \(\displaystyle \:\frac{1}{4}\L\int\)\(\displaystyle \sec^2\theta\,d\theta\;\;\) . . . Good!

but the answer is: \(\displaystyle \:\frac{\tan(x)}{4\sqrt{4\,-\,\tan^2(x)}}\, +\, C\;\;\) . . . Yes!

You have: \(\displaystyle \L\:\frac{1}{4}\int\sec^2\theta\,d\theta\;=\;\frac{1}{4}\cdot\tan\theta \,+\,C\;\) [1]


Now we must back-substitute . . .

We began with: \(\displaystyle \tan x \:=\:2\sin\theta\;\;\Rightarrow\;\;\sin\theta\:=\:\frac{\tan x}{2}\:=\:\frac{opp}{hyp}\)

\(\displaystyle \theta\) is an angle in a right triangle with: \(\displaystyle \,opp\,=\,\tan x\) and \(\displaystyle hyp\,=\,2\)

. . Using Pythagorus, we find that: \(\displaystyle \,adj\,=\,\sqrt{4\,-\,\tan^2x}\)

Hence: \(\displaystyle \:\tan\theta \:=\:\frac{opp}{adj} \:=\:\frac{\tan x}{\sqrt{4\,-\,\tan^2x}}\)


Substitute into [1]: \(\displaystyle \L\:\frac{1}{4}\cdot\frac{\tan x}{\sqrt{4\,-\,\tan^2x}} \,+\,C\;\;\) . . . see?

 

[T_TEngineer_AdamT_T]

ohh lucky...
so i did it correctly thanks