Can anyone help me to solve these two indefinite integrals? 1) Find int[dx/(sin(x)*sqrt{1+cos(x)})]. 2) Find int [cos(x)/(sin(x+1))^2]dx. For me this integrals were too complicated. Thanks in advance!
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Originally Posted by Dext91 1) Find int[dx/(sin(x)*sqrt{1+cos(x)})]. Substitue and you arrive at . Now use the subtitution . It should be quite managable from here on, you might want to consider using partial fractions for the next step. Good luck, pomp.
Last edited by pomp; Jul 18th 2009 at 09:31 AM.
Originally Posted by Dext91 1) Find int[dx/(sin(x)*sqrt{1+cos(x)})]. Rewrite: Let , then Therefore Not sure if this is correct.
Originally Posted by VonNemo19 Rewrite: Let , then Therefore Not sure if this is correct. Slightly wrong, you multiplied through by sin instead of dividing when you subbed in. . Subbing in gives: So the integral becomes
Is that second one supposed to be: or I am thinking the latter since that would be easier. Do you have a typo?. Just checking.
Originally Posted by galactus Is that second one supposed to be: or I am thinking the latter since that would be easier. Do you have a typo?. Just checking. I hope you're right. After putting the former through matlab I'm afraid to even attempt it.
Yes indeed, Pomp. If it is the latter, the simple sub will suffice. But the former is a booger.
Not quite the sin(x) term does not cancel but rather you have 1/[sin^2(x)sqrt(u)] See attachment we can avoid sqr roots altogether
Originally Posted by Dext91 Can anyone help me to solve these two indefinite integrals? 1) Find int[dx/(sin(x)*sqrt{1+cos(x)})]. 2) Find int [cos(x)/(sin(x+1))^2]dx. For me this integrals were too complicated. Thanks in advance! For number one Finally
Last edited by DeMath; Jul 18th 2009 at 11:21 AM. Reason: typos
Originally Posted by Dext91 Can anyone help me to solve these two indefinite integrals? 1) Find int[dx/(sin(x)*sqrt{1+cos(x)})]. 2) Find int [cos(x)/(sin(x+1))^2]dx. For me this integrals were too complicated. Thanks in advance! For number 2
My original attachment contains a mistake--see correct version
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