Question

Calculus! Evalutate the deff. integral

Answer 1

\[\int\limits_{\pi/2}^{\pi}\frac{ Sin x }{ 1+Cos ^{2} x }dx \]

Answer 2

I got to Sinxarctan(cosx), But when I put in the pi and pi/2 i kept getting the wrong answere

Answer 3

If you let u = cos x, then du = -sin x dx, so your integral becomes:\[\int\limits\frac{-du}{1+u^2}\]the sin x isnt there anymore.

Answer 4

ohhhhh! xD ok, got it thank you!

Answer 5

In the interest of Calculus.... Could you finish the question on here so I can see how the rest of it's done please.

Answer 6

From there, its pretty much something you either know, or don't. It turns out that:\[\frac{d}{dx}\tan ^{-1}x=\frac{1}{1+x^2}\], so we have that:\[\int\limits\frac{1}{1+u^2}du=\tan^{-1}u\]Putting back our substitution, we get:\[\tan^{-1}(\cos x)\]Now we just need to evaluate the limits.\[\tan^{-1}(\cos \pi)-\tan^{-1}(\cos \frac{\pi}{2})=\tan^{-1}(-1)-\tan^{-1}(0)\]\[=-\frac{\pi}{4}-0=-\frac{\pi}{4}\]

Answer 7

Would it not be negative? \[-\int\limits\frac{1}{1+u^2}du=-\tan^{-1}u\] Because you had negative du after you changed it to 'u' and 'du'.

Answer 8

yeah,its supp oased to be posetive

Answer 9

So the answer would be \(\dfrac{\pi}{4}\) not \(-\dfrac{\pi}{4}\) ?

Answer 10

YUP! \[\frac{ \pi }{ 4 }\]