calculating

These dudes are mose easily solved by drawing a right triangle. Label an angle and build the two parts suggested by the inner-most expression. Use the pythagorean theorem for the third part and you are done.

Of course, if you are feeling adventurous, you may wish to apply certain identities that would take you down a different path. sin(2a) = 2sin(a)cos(a) comes to mind.
 
i am not seeing it tkhunny can you explain more please.

"These dudes are mose easily solved by drawing a right triangle. Label an angle and build the two parts suggested by the inner-most expression. Use the pythagorean theorem for the third part and you are done."
 
Note by the diagram:

\(\displaystyle sec(y)=\frac{x}{1}\)

\(\displaystyle y=sec^{-1}(x)\)

\(\displaystyle sin(sec^{-1}(x))=sin(y)\)

So, we get \(\displaystyle \frac{\sqrt{x^{2}-1}}{x}\)

Same way for cos.

Technically, the one above should be |x| in the denominator.
 
couldnt sec^(-1)(-7) = cos(-7)

with that said u get

1/2*sin(cos(-7))*cos(cos(-7))

the example that my book gave me an answer of (-2squareroot3)/49


uh?
 
No, \(\displaystyle sec^{-1}(x)\neq cos(x)\)

Why not use what I gave you and just plug in x=-7?.

\(\displaystyle \frac{1}{2}\frac{\sqrt{x^{2}-1}}{|x|}\cdot \frac{1}{x}\)
 
here is the original problem and my work. got tired of trying to figure out how to calculate that one part i was having trouble with so i put it into my hp 49g+ and got it. here is my work from the very beginning.
 
Making your subs, this whittles down to:

\(\displaystyle \int sin^{2}(t)dt\)

To integrate, use the identity \(\displaystyle sin^{2}(t)=\frac{1}{2}(1-cos(2t))\)

\(\displaystyle \int \frac{\sqrt{sec^{2}(t)-1}}{sec^{3}(t)}\cdot sec(t)tan(t)dt\)

\(\displaystyle \frac{tan^{2}(t)}{sec^{2}(t)}dt=\int sin^{2}(t)dt\)
 
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