Date: 02/17/99 at 17:48:47 From: crys Subject: Integration Can you help me with (integral sign) x tan^2 x dx ? I think you have to use some kind of substitution or maybe trigonometry identities Thank you.
Date: 02/18/99 at 01:28:50 From: Doctor Luis Subject: Re: Integration The first thing that comes to mind is integration by parts. My primary motivation for this method is the factor of x that appears on the integrand, because I know that differentiating that factor will give me a constant. Quick review: integration by parts is essentially based on the following formula / / | | | u dv = u*v - | v du | | / / In this case the substitutions should be u = x -----> du = dx dv = tan^2(x) dx -----> v = ? Note that there is a small problem here. We need to integrate the function tan^2(x) in order to obtain v from dv. But this does not really present that much of a problem, if you remember that tan(x) = sin(x)/cos(x) and that cos^2(x)+sin^2(x) = 1. / / | sin^2(x) | 1 - cos^2(x) v = | --------- dx = | ------------- dx | cos^2(x) | cos^2(x) / / / | = | (sec^2(x)-1)dx = tan(x)-x | / Therefore, v = tan(x)-x. Now, substituting into our "integration by parts formula," you get / / | | | (x)*(tan^2(x)dx) = (x)*(tan(x)-x) - | (tan(x)-x)*(dx) | | / / where I have enclosed the substitutions we made in parentheses. Now, the second term on the integral on the righthand side of the equation presents no problem. It's just (1/2)x^2, but the first term seems a bit more difficult. However, if you rewrite tan(x) as sin(x)cos(x)/cos^2(x), you can see that we almost have a complete differential; the derivative of cos^2(x) is 2*cos(x)*(-sin(x)). So, concentrating on the integral of tan(x), we have: / / | | sin(x)cos(x) | tan(x)dx = | ------------- dx | | cos^2(x) / / If you make the substitution u = cos^2(x), which implies that du = 2cos(x)*(-sin(x))dx or cos(x)sin(x)dx = -du/2), then this last integral becomes / / | sin(x)cos(x) 1 | du | ------------- = - --- | ----- = -(1/2)ln(u) = -(1/2)ln(cos^2(x)) | cos^2(x) 2 | u / / Therefore the integral of tan(x) is just -(1/2)ln(cos^2(x)) and so our original integral becomes: / / | | | (x)*(tan^2(x)dx) = (x)*(tan(x)-x) - | (tan(x)-x)*(dx) | | / / = x*tan(x)-x^2 - (-(1/2)ln(cos^2(x))-(1/2)x^2) = x*tan(x) - (1/2)x^2 + (1/2)ln(cos^2(x)) I hope this explanation helps, although it would have been better if you had included an explanation of how you attempted this problem and where you got stuck. (I did not know how or why you got stuck, so I tried to explain everything.) Feel free to reply if you have any questions about this answer. Sincerely, - Doctor Luis, The Math Forum http://mathforum.org/dr.math/
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