integral of 1/(cos (x) -1)
can we rewrite this as secx-1
\[\int\limits_{}^{}\frac{1}{\cos(x)-1} \cdot \frac{\cos(x)+1}{\cos(x)+1} dx=\int\limits_{}^{}\frac{\cos(x)+1}{\cos^2(x)-1} dx\] \[\int\limits_{}^{}\frac{\cos(x)+1}{-(1-\cos^2(x))}dx=\int\limits_{}^{}\frac{\cos(x)+1}{-\sin^2(x)} dx\] \[\int\limits_{}^{}\frac{\cos(x)}{-\sin^2(x)} dx +\int\limits_{}^{}\frac{1}{-\sin^2(x)} dx\]
\[-\int\limits_{}^{}\frac{\cos(x)}{\sin^2(x)} dx-\int\limits_{}^{}\frac{1}{\frac{1}{2}(1-\cos(2x)} dx\]
for the first one let u=sin(x) let me know if you need more help
oh, nice job myin, multiplying by the conjugate
how do you know this???
so what is the actual answer?
and just to show off a little if you want \[\int\frac{1}{-\sin^2(x)}dx\] i would just recall that this is \[-\int\csc^2(x)dx\] so that the integrand is instantly the derivative of \[\cot(x)\]
i get an actual answer of \[\csc(x)+\cot(x)\]
that is the actual answer
in actuality
how do i know this?
that was my question. it was a serious one. i would never know to do this
well i wanted to get the bottom as one term so i can break it up
although i have to say i like my method of solving \[-\int\frac{1}{\sin^2(x)}dx\]
yes i wasn't thinking there
ah that makes sense.
also might explain why we see all those trig problems that say "show \[\frac{\sin(x)}{1+\cos(x)}=\text{whatever}\]
oh yes identities
it can be useful in trig
i mean in calc
guess they prove useful later. only problem is it is two semesters later so they forget
right
some people don't even take trig
same as those problems that say find \[\tan(\sin^{-1}(x))\] by the time they see it again it is in calc 2
i love those problems
i know you do. actually i do too
i was like telling my trig students make sure you know how to do these i love these
try to explain as best i can that is it exactly the same as those problems that say "find the other trig functions of theta if sin(theta) = whatever
thats key for its gonna be on the test
ti 89 does it symbolically.
lol ok gnight
goodnight
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