<p>Hey everyone, I've been stuck doing this stupid thing for almost half an hour, I've tried trig identities and different substitutions by manipulating u, but nothing is coming out!</p>
<p>integrate cosx / 1 + sinx^2, from 0 to pi/2</p>
<p>thanks if anyone can help..</p>
<p>It will be impossible to help you on questions unless you're more careful at stating equations unambiguously. Assuming, that what you really meant was cos(x) / (1 + Sin^2(x)), you can use simple substitution with u=1+sin^2(x) to solve the integral. If you've been workin on it for half an hour, it's probably not that simple, but the actual problem posted is rather unclear.</p>
<p>I don't see how else he could mean it, because cos x / 1 would just be cos x.</p>
<p>The only thing hard about that one is the ambiguity of the equation you posted.</p>
<p>Assuming you meant the integral of cos(x)/(1+((sin(x))^2), make u = sinx, then you have the integral from 0 to pi/2 of du/(1+u^2), which is an arctangent. The integral is then arctan(sinx) from 0 to pi/2. At pi/2, sinx = 1, and arctan of 1 = pi/4. At 0, sin = 0, and arctan of 0 is 0. The answer is pi/4.</p>
<p>yeah, maybe I should fix it, it is what you wrote:</p>
<p>integrate:</p>
<p>cos(x) / [1 + sin^2(x)], from 0 to pi/2</p>
<p>I did a substitution, du = 2sinxcosx dx, but there is no way to get rid of 2sinx. After pulling it out as 1/2sinx, the rest of the function is in terms of u while 1/2sinx is in x... so I tried putting everything back to replace u and then once you start evaluating, it comes out indefined.</p>
<p>Sorry, I made a really stupid mistake in my post. The substitution I wrote does not work, as you pointed out. I think obsessedAndre is right, but it's been a while since I've done integration with inverse trigonometric functions.</p>
<p>And sorry I was kind of snappish in my post. Obnoxious English assignments that look like they're going to keep me up late into the night put me in a really bad mood.</p>
<p>I don't feel like solving it but 1/(1+sin^2(x)) is actually a trig identity for the derivitive of arctan(sinx). See if that helps at all.</p>
<p>yep, thanks andre, i was confused becuase my teacher probably accidentally assigned this when we aren't actually on the chapter yet, but yes, arctan does work. thanks!</p>