# just hints in integrals

• Feb 12th 2009, 07:07 PM
alex83
just hints in integrals
• Im in the middle of solving an integration and just need to know what is
16 cos^2 x in terms of sec^2 x
• what is the integral of cosx/sin^2 x
• integral of x^2/(x^2 +1)

thank you
appreciate any help
• Feb 12th 2009, 07:11 PM
Chop Suey
1. $\sec{x} = \frac{1}{\cos{x}}$

2. You can try $u = \sin{x}$, or maybe it would be more clear if you see it as $\csc{x}\cot{x}$.

3. Simply add 0 using 1. This is a basically a quick way around long division:
$\frac{x^2+1-1}{x^2+1} = \frac{x^2+1}{x^2+1} - \frac{1}{x^2+1} = \ldots$
• Feb 12th 2009, 07:36 PM
alex83
Chop Suey
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)(Thinking)

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 (Bow)
• Apr 24th 2009, 03:32 PM
mr fantastic
Quote:

Originally Posted by alex83
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)(Thinking)

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 (Bow)

Maths is not a spectator sport. Review the methods you have been taught. What have you tried? Where do you get stuck? You should realise that $\int \frac{1}{x^2 + 1} \, dx$ is a standard form.