Trig sub problem

[masteroc]

So the problem I am having is that I know (u)=arcsin(x/2), but then I have to substitute that into: (1/2)ln(sec(u)+tan(u)) and im not sure how to go about that. Someone told me that comes out to (1/2)ln((x/2)/sqrt(1-(x/2)^2) + 1/sqrt(1-(x/2)^2)) but im not sure how they got that.


Thanks

 

[tkhunny]

sec(arcsin(x/2))

Draw a right triangle.
Lable an acute angle 'a'.

Consider a = arcsin(x/2), or sin(a) = x/2

Lable the side opposite 'a'. Call the length 'x'.
Lable the hypotenuse. Call the length '2'.
The Pythagorean Theorem tells us that the length of the adjacent side is sqrt(2^2 - x^2)

sec(arcsin(x/2)) = sec(a) = [sqrt(2^2 - x^2)]/2

Do the same with the tangent.

 

[soroban]

Hello, masteroc!

A slightly different explanation . . .

I know: .\(\displaystyle u \,=\,\arcsin\left(\frac{x}{2}\right)\)
Then I have to substitute that into: .\(\displaystyle \frac{1}{2}\ln|\sec u + \tan u |\)

We have: .\(\displaystyle u\:=\:\arcsin\left(\frac{x}{2}\right)\)

. . Then: .\(\displaystyle \sin u \:=\:\frac{x}{2} \:=\:\frac{opp}{hyp}\)

That is, \(\displaystyle u\) is in a right triangle with \(\displaystyle opp = x,\:hyp = 2.\)

. . Pythagorus tell us: .\(\displaystyle adj = \sqrt{4-x^2}\)

Hence: .\(\displaystyle \begin{Bmatrix}\sec u &=& \frac{hyp}{adj} &=& \frac{2}{\sqrt{4-x^2}} \\ \tan u &=& \frac{opp}{adj} &=& \frac{x}{\sqrt{4-x^2}} \end{Bmatrix}\)


Substitute: .\(\displaystyle \frac{1}{2}|\sec u + \tan u| \;=\;\frac{1}{2}\left|\dfrac{2}{\sqrt{4-x^2}} + \dfrac{x}{\sqrt{4-x^2}}\right| \;=\;\frac{1}{2}\ln\left|\dfrac{2+x}{\sqrt{4-x^2}}\right| \)


. . . . . . \(\displaystyle =\;\frac{1}{2}\ln\left|\dfrac{2+x}{\sqrt{(2-x)(2+x)}}\right| \;=\;\frac{1}{2}\ln\left|\dfrac{2+x}{(2-x)^{\frac{1}{2}}(2+x)^{\frac{1}{2}}}\right| \)

. . . . . . \(\displaystyle =\;\frac{1}{2}\ln\left|\dfrac{(2+x)^{\frac{1}{2}}}{(2-x)^{\frac{1}{2}}}\right| \;=\;\frac{1}{2}\ln\left|\dfrac{2+x}{2-x}\right|^{\frac{1}{2}} \;=\;\frac{1}{4}\ln\left|\dfrac{2+x}{2-x}\right|\)