6
$\begingroup$

$$\int \frac{1}{1+\cos ^2x} \,\mathrm dx$$

I have to integrate the expression above: I tried with substitutions $\cos x=t$ and $1+(\cos x)^2=t$, but those didn't work, and I couldn't find any useful way to use bisection and duplication formulas.

Any ideas?

$\endgroup$
4
16
$\begingroup$

Hint:

$$\int\frac{\mathrm dx}{1+\cos ^2x}=\int\frac{\sec^2x\,\mathrm dx}{\sec^2x+1}=\int\frac{\sec^2x\,\mathrm dx}{2+\tan^2x}$$

And then $$\tan x=t\iff \sec^2x\,\mathrm dx=\mathrm dt$$

$\endgroup$
1
  • 2
    $\begingroup$ Thanks! That really helped! $\endgroup$ Jan 8 '15 at 17:14
0
$\begingroup$

Maybe my soliution will be helful too.$$\int \frac{dx}{1+cos^2x} = \int \frac{dx}{cos^2x(\frac{1}{cos^2x}+1)} = [\frac{1}{cos^2x}dx=d(tgx)]= \int \frac{d(tgx)}{1+tg^2x+1} = \int \frac{d(tgx)}{tg^2x+2} = \frac{1}{\sqrt2}arctg(\frac{tgx}{\sqrt2})+C$$ Also we here remember trigonometry formula $\frac{1}{cos^2x}=1+tg^2x$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.