Evaluate the integral sinx/cos^2x dx from 0 to pi/3. So I've gotten it to tanxcosx, but I don't know the indefinite integral of these, and I couldn't find it in my book. Can anyone help?
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Originally Posted by bobsanchez Evaluate the integral sinx/cos^2x dx from 0 to pi/3. So I've gotten it to tanxcosx, but I don't know the indefinite integral of these, and I couldn't find it in my book. Can anyone help? you should get under integral .....
Last edited by Plato; Aug 11th 2010 at 01:37 PM.
Ah okay, let me work the problem. Thanks.
-1?
i got result 1
-1/cos(60) = -2 -1/cos(0) = -1 -2 - (-1) = -2 + 1 = -1 Did I mess up somewhere in there or enter it wrong?
Let to get Plato's result (without the typo minus).
Originally Posted by bobsanchez -1/cos(60) = -2 -1/cos(0) = -1 -2 - (-1) = -2 + 1 = -1 Did I mess up somewhere in there or enter it wrong? and now the after that ... i just got and with limits you get "1"
Evaluate the integral sec2xtan2x dx. sec2x + C This is another good one. My first try resulted in sec2x + C, but was wrong.
Originally Posted by bobsanchez Evaluate the integral sinx/cos^2x dx from 0 to pi/3. So I've gotten it to tanxcosx, but I don't know the indefinite integral of these, and I couldn't find it in my book. Can anyone help? Just use a substitution... Let so that and the integral becomes . Therefore .
Originally Posted by bobsanchez Evaluate the integral sec2xtan2x dx. sec2x + C This is another good one. My first try resulted in sec2x + C, but was wrong. Can someone take a quick look at this one please?
Originally Posted by bobsanchez Can someone take a quick look at this one please? You forgot to use the chain rule when finding du. Can you take it from here?
...wow...forgot the chain rule? Really? *sigh* Anyways, thanks for the help. Alright, so...I just integrate and multiply it by du, correct? So (1/2)sec2x + C?
I could read that as , but I know that's not what you meant.
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