While doing some exercises in my cal II book i stumbled on these 2 questions.. Any help will be appreciated. Thanks (1) The integral of tan3x --> -1/3ln|cos3x|? (2) The integral of x/(x^(2)-1)^2 --> not sure
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Originally Posted by Bure03 While doing some exercises in my cal II book i stumbled on these 2 questions.. Any help will be appreciated. Thanks (1) The integral of tan3x --> -1/3ln|cos3x|? (2) The integral of x/(x^(2)-1)^2 --> not sure (1) Substitute $\displaystyle u = \cos 3x$ (2) Substitute $\displaystyle u = x^2 - 1$
Originally Posted by Bure03 While doing some exercises in my cal II book i stumbled on these 2 questions.. Any help will be appreciated. Thanks (1) The integral of tan3x --> -1/3ln|cos3x|? (2) The integral of x/(x^(2)-1)^2 --> not sure #1 $\displaystyle \int\tan(3x)\,dx=\int\frac{\sin(3x)}{\cos(3x)}\,dx$ Make the substitution $\displaystyle z=\cos(3x)$ #2 $\displaystyle \int\frac{x\,dx}{(x^2-1)^2}$ Make the substitution $\displaystyle z=x^2-1$ Try taking it from here. Hope this makes sense!
Originally Posted by Bure03 While doing some exercises in my cal II book i stumbled on these 2 questions.. Any help will be appreciated. Thanks (1) The integral of tan3x --> -1/3ln|cos3x|? Yes, that's correct. It can also be written as $\displaystyle \frac{1}{3}\ln|\sec(3x)|$.
thanks guys just wanted to double check my answers.
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