Hi, Can someone help me with the integral Cot/sq. 1+2sinx, please? I tried doing u substitution I set u=1+sinx du=2cosx Will this work or should I try integration by parts? I also know that cot=1/tan Thank you
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Originally Posted by chocolatelover Hi, Can someone help me with the integral Cot/sq. 1+2sinx, please? I tried doing u substitution I set u=1+sinx du=2cosx Will this work or should I try integration by parts? I also know that cot=1/tan Thank you you cannot have cot by itself, you have to have cot(something). please type your question properly
Sorry Cotx/sq. 1+2sinx
$\displaystyle \displaystyle\int\frac{\cot x}{(1+2\sin x)^2}dx=\int\frac{\cos x}{\sin x(1+2\sin x)^2}dx$ Substitute $\displaystyle \sin x=u$.
I'm trying to integrate "Cotx/sq. 1+2sinx" Can someone please help me? Thank you
Originally Posted by chocolatelover Sorry Cotx/sq. 1+2sinx and by that i suppose you mean $\displaystyle \frac {\cot x}{\sqrt{1 + 2sin(x)}}$ ? parentheses are important too. change cot(x) into cos(x)/sin(x), then make the substitution $\displaystyle u = \sqrt {1 + 2 \sin x}$
Would the final answer be 1/2ln sq. 1+2sinx+c I did: (1/2) int du/sq. u dx= 1/2 int. 1/sq. u dx 1/2 int. ln|sq. u|+c= 1/2 ln sq. 1+2sinx+c Thank you
Originally Posted by chocolatelover Would the final answer be 1/2ln sq. 1+2sinx+c I did: (1/2) int du/sq. u dx= 1/2 int. 1/sq. u dx 1/2 int. ln|sq. u|+c= 1/2 ln sq. 1+2sinx+c Thank you please read my instructions in my last post, and use parentheses to clarify what you mean
Originally Posted by chocolatelover Can someone help me with the integral Cot/sq. 1+2sinx, please? Does this make sense? ¬¬
Originally Posted by Krizalid Does this make sense? ¬¬ Hey, it's the same question (asked by the same person). I saw that thread but i didn't remember it, since i never posted in it. i made the same substitution as you did
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