Question

Calculus II "Integrals of Trig Products" homework question **Due by midnight
If the integral tan^2(x) sec(x) dx from 0 to pi/4 = I,
express the value of tan^4(x) sec(x) dx in terms of I.

Answer 1

hint \[\tan^2x=\sec^2x-1\]

Answer 2

I've got it down to \[\int\limits_{0}^{\pi/4} \tan^4 (x) \sec (x) dx = \tan^3(x)\sec(x) - 3\int\limits_{0}^{\pi/4} \sec^2 (x) \tan(x)dx\] using \[\int\limits_{}^{} udv = uv - \int\limits_{}^{} vdu\]
but I don't know where to go from there.

Answer 3

hmm let see
try \[\tan^4x\sec x=(\sec^2x-1)^2\sec x\]

Answer 4

give it a try and if you need further help tag me ^_-

Answer 5

it would work with IBP too but you did something wrong along the way

Answer 6

just be persistent to find the solution hehe
don't give up easily :)

Answer 7

I'm not sure that I can figure it out within the time limit.

Answer 8

\[\int \tan^4x\sec xdx=\int (\sec^2x-1)\tan^2x\sec xdx\]
\[=\int \tan ^2x \sec^3 xdx-I\]

Answer 9

take it from there

Answer 10

i gotta leave to sleep

Answer 11

Oh, okay. Thanks for the help. :)
I'll work on it.

Answer 12

if you really need to have this answered, i believe i have it

Answer 13

But its been a while since calc II so i dont really know if its correct

Answer 14

I'm out of time now, so figuring out how to get the answer is more important now.

Answer 15

ok, well here is my answer for a similar problem i did in calc ii

Answer 16

If I= integral from 0 to pi/4 of tan^6(x)sec(x)dx, then express the value of the integral from 0 to pi/4 of tan^8(x)sec(x)dx in terms of I.

Answer 17

my answer for this (very similar) problem, was as follows:
∫ [0,π/4] tan⁸ x sec x dx

= ∫ [0,π/4] tan⁶ x tan² x sec x dx

= ∫ [0,π/4] tan⁶ x ( sec² x - 1 ) sec x dx

= ∫ [0,π/4] tan⁶ x sec³ x dx - ∫ [0,π/4] tan⁶ x sec x dx

= ∫ [0,π/4] tan⁶ x sec³ x dx - I ............ from (1) ................... Ans.

Answer 18

this is NOT your problem, but it is similar

Answer 19

The answer is supposed to be \[\frac{ \sqrt{2} }{ 4 } - \frac{ 3 }{ 4 } I\]
but I haven't figured out how to get there.

Answer 20

no frills solution

Answer 21

Thanks! :)