4
$\begingroup$

What's the value of it?I hope you can help me.

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

You are right.Actually what really wanted is a definite integral .In fact,it's $$\int_{-\pi}^r\frac{x(1+\sin{x})}{1+\cos^2{x}}dx$$ But I still don't know how to value it. Thank Q for your reply.

$\endgroup$
3
  • 5
    $\begingroup$ See this . Maybe it's what u want or it maybe a definite integral. $\endgroup$
    – ABC
    Apr 23, 2013 at 7:55
  • $\begingroup$ If you want some value, you need a definite integral... $\endgroup$
    – V-X
    Apr 23, 2013 at 13:23
  • $\begingroup$ Is the $r$ in the upper limit $\pi$? $\endgroup$
    – user17762
    Apr 24, 2013 at 7:08

1 Answer 1

4
$\begingroup$

We have $$I=\int_{-\pi}^{\pi} \dfrac{x(1+\sin(x))}{1+\cos^2(x)} dx$$ Replacing $x$ by $-x$, we get that $$I=\int_{-\pi}^{\pi} \dfrac{-x(1-\sin(x))}{1+\cos^2(x)} dx$$ Adding both and dividing by $2$, we get that $$I = \int_{-\pi}^{\pi} \dfrac{x \sin(x)}{1+\cos^2(x)}dx = 2\int_{0}^{\pi} \dfrac{x \sin(x)}{1+\cos^2(x)}dx = 2\sum_{k=0}^{\infty}(-1)^k\int_0^{\pi}x\sin(x) \cos^{2k}(x) dx$$ Now integral $$\int_0^{\pi}x\sin(x) \cos^{2k}(x) dx = \dfrac{\pi}{2k+1}$$ Hence, we get that $$I = \sum_{k=0}^{\infty}(-1)^k \dfrac{2\pi}{2k+1} = \dfrac{\pi^2}2$$

$\endgroup$
2
  • $\begingroup$ Thank Q !I always think about integral but not defenite. $\endgroup$
    – 07216
    Apr 24, 2013 at 10:12
  • $\begingroup$ $\int_0^{\pi}x\sin(x) \cos^{2k}(x) \,\mathrm dx = \dfrac{\pi}{2k+1}$ can be proven using the substitution $x\to\pi -x$ $\endgroup$ Oct 7, 2019 at 20:45

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.