16 cos^2 x in terms of sec^2 x
- Im in the middle of solving an integration and just need to know what is
- what is the integral of cosx/sin^2 x
- integral of x^2/(x^2 +1)
thank you
appreciate any help
1. $\displaystyle \sec{x} = \frac{1}{\cos{x}}$
2. You can try $\displaystyle u = \sin{x}$, or maybe it would be more clear if you see it as $\displaystyle \csc{x}\cot{x}$.
3. Simply add 0 using 1. This is a basically a quick way around long division:
$\displaystyle \frac{x^2+1-1}{x^2+1} = \frac{x^2+1}{x^2+1} - \frac{1}{x^2+1} = \ldots$
Thank you for replying.
the problem is
1- the integral is (sec^2 x +8secx cosx +16cos^x dx) so i was trying to find the relation or form of 16cos^2x in terms of the sec^2 x)
2-the integral of this question was (1/(sin^2x) - cosx/sin^2x )
3- i got the same result: integral 1+ 1/(x^2+1)
so im looking for the integral of 1/(x^+1)
THANK YOU AGAIN