Question

help with these trig integrals.. 1st: 3tan3tsec3t and the integral of 2cos3t

Answer 1

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

Answer 2

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)

Answer 3

Hermeezey's answer is great, unless you meant\[\int3\tan^3t\sec^3tdt\]which is a different problem
did Hermeezy have it right?

Answer 4

\[\cos3t(\sec ^{2}3t)/3 = 2\cos3t??????\]

Answer 5

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!

Answer 6

well cos3t*sec^2(3t)=sec(3t)
but I don't see where that appears above

Answer 7

my fault..supposed to be 9sec^2(3t)

Answer 8

sorry, did you have a question?

Answer 9

yeah \[\cos3t(\sec ^{2}3t)/3=2\cos3t\]

Answer 10

omg \[\cos3t(9\sec ^{2}3t)/3=2\cos3t\]

Answer 11

thats the real question

Answer 12

the question is solve for t ?

Answer 13

does the left simplify to get the right?

Answer 14

nope it would be like you said above.. = 3sec3t