Question

Please help me get started on the following integral (click to see).

Answer 1

\[\int\limits_{}^{}\tan ^{3}(4x)dx\]

Answer 2

\[\int\limits_{}^{}\tan(4x)(\sec ^{2}(4x)-1)dx\]
if i use a relevant identity as shown here i would use u=secx but where would i got from there, the integrand does not reduce full if i do that u sub.

Answer 3

Distribute the tan(4x) then you will have an integral of the form sec^nf(x) dx -tan(fx)dx so you would have the \[\int\limits_{}u ^{n}du-\int\limits_{}^{}\tan(4x)dx\] with the substitution you came up with. If you let u=sec(4x), then du=4sex(4x)tan(4x)dx and so all you would need to do is divide du by 4. The second integral is a simple integral.

Answer 4

i'm not following you. the integrals would be \[\int\limits_{}^{}\sec ^{2}(4x)\tan(4x)dx-\int\limits_{}^{}\tan(4x)dx\]

Answer 5

if u=sec(x) then dx=(du)/4sec(4x)tan(4x). plugging that back into the left integral i would still have sec(x) remaining.

Answer 6

Yeah, so separate the sec^2(4x) into u=sec(4x) and du=sex(4x)tan(4x)dx

Answer 7

oops forgot the 4.

Answer 8

I would have two u's then. or can you have one sec(x) and one (u)?

Answer 9

the second sex(4x) is part of du

Answer 10

focusing on the left-integral...
\[\int\limits_{}^{}\sec(4x)\sec(4x)\tan(4x)dx\]
if i make u=sec(4x) all i will have left is this...
\[\int\limits_{}^{}u ^{2}\tan(4x)\frac{ du }{ 4\sec(4x)\tan(4x) }\]
I would still have 1/4sec(4x) remaining

Answer 11

or can I use u as a substitution for one of the sec(4x)'s like this...
\[\int\limits_{}^{}u \sec(4x)\tan(4x)\frac{ du }{ 4\sec(4x)\tan(4x) }\]

Answer 12

Whoa...hang on a sec. I had to step away.

Answer 13

I am going to write it out as I can do it faster than using the equation editor.