help with these trig integrals.. 1st: 3tan3tsec3t and the integral of 2cos3t
1st one: rewrite it first, remember: tan(3t) = sin(3t)/cos(3t) sec(3t) = 1/cos(3t) now if we rewrite it, we get: \[\int\limits_{}^{} 3[(\sin3t)/(\cos3t)] (1/\cos3t) dt\] \[\int\limits_{}^{} 3 (\sin3t/\cos^23t) dt\] let u = cos3t du = -3sin3t dt \[\int\limits_{}^{}-(1/u^2) du\] which just becomes: = (1/u) back substituting we get: 1/cos3t
Second one: \[\int\limits_{}^{} 2\cos(3t) dt\] let's try a straight forward u-substitution: u = 3t du = 3dt dt = du/3 \[\int\limits_{}^{}2\cos(u)du/3\] \[\int\limits_{}^{}2/3\cos(u)du\] lets drag the 2/3 out of the integral: \[2/3 \int\limits_{}^{}\cos(u)du\] What's the integral of cos(u) ? it's just sin(u) right! 2/3 (sin(u)) back substituting, we get: 2/3 (sin3t)
Hermeezey's answer is great, unless you meant\[\int3\tan^3t\sec^3tdt\]which is a different problem did Hermeezy have it right?
\[\cos3t(\sec ^{2}3t)/3 = 2\cos3t??????\]
its right..just makin sure im integrating the right integral for the second one.. BTW this is part of a differential equation problem and thanks for your help!
well cos3t*sec^2(3t)=sec(3t) but I don't see where that appears above
my fault..supposed to be 9sec^2(3t)
sorry, did you have a question?
yeah \[\cos3t(\sec ^{2}3t)/3=2\cos3t\]
omg \[\cos3t(9\sec ^{2}3t)/3=2\cos3t\]
thats the real question
the question is solve for t ?
does the left simplify to get the right?
nope it would be like you said above.. = 3sec3t
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