Calculus  Question
posted by Robert
Am I allowed to do this?
for the integral of
∫ sec^4 (3x)/ tan^3 (3x) dx
I change it to
∫ sec^4 (3x) tan^3 (3x)
From here I use the rule for trigonometry functions.

Robert
Or do I use the rule of ∫ sec^n x dx divided by the rule of ∫ tan^n x dx?

Robert
If I go with my first assumption I get this:
∫ sec^4 (3x) tan^3 (3x) = ∫ sec^3 (3x) tan^4 (3x) sec(3x)tan(3x) dx
= ∫ sec^3 (3x) (sec^4 3x  1)sec(3x)tan(3x) dx
u = sec(3x) dx
du = 3sec(3x)tan(3x) dx > 1/3du
= 1/3 ∫ u^3 (u^4  1) du
= 1/3 ∫ (u^7  u^3) du
= 1/3 (u^6/6  u^4/4) + C
= 18/sec^6(3x)  sec^4 (3x)/12 + C
Would this be the right method and answer? Thank you
Respond to this Question
Similar Questions

Integration
Intergrate ¡ì sec^3(x) dx could anybody please check this answer. are the steps correct? 
calculus
find dy/dx y=ln (secx + tanx) Let u= secx + tan x dy/dx= 1/u * du/dx now, put the derivative of d secx/dx + dtanx/dx in. You may have some challenging algebra to simplify it. Use the chain rule. Let y(u) = ln u u(x) = sec x + tan x … 
calculus
Use integration by parts to evaluate the integral of x*sec^2(3x). My answer is ([x*tan(3x)]/3)[ln(sec(3x))/9] but it's incorrect. u=x dv=sec^2(3x)dx du=dx v=(1/3)tan(3x) [xtan(3x)]/3  integral of(1/3)tan(3x)dx  (1/3)[ln(sec(3x))/3] … 
calculus (check my work please)
Not sure if it is right, I have check with the answer in the book and a few integral calculators but they seem to get a different answer ∫ sec^3(x)tan^3(x) dx ∫ sec^3(x)tan(x)(sec^2(x)1) dx ∫ tan(x)sec(x)[sec^4(x)sec^2(x)] … 
Calculus
Calculate the following integral: ∫ sec^4 (3x)/ tan^3 (3x) dx For this one, can I bring up the tan to tan^3? 
Calculus AP
I'm doing trigonometric integrals i wanted to know im doing step is my answer right? 
calculus II
∫ tan^2 x sec^3 x dx If the power of the secant n is odd, and the power of the tangent m is even, then the tangent is expressed as the secant using the identity 1 + tan^2 x = sec^2 x I thought that since tan is even and sec is … 
Calculus 2
∫ tan^2 (x) sec^4 (x) dx ∫ [tan^2 (t) + tan^4 (t)] dt ∫ [1tan^2 (x)] / [sec^2 (x)] dx Trigonometric integral Please show steps so I can understand! 
Calculus 2 Trigonometric Substitution
I'm working this problem: ∫ [1tan^2 (x)] / [sec^2 (x)] dx ∫(1/secx)[(sin^2x/cos^2x)/(1/cosx) ∫cosxsinx(sinx/cosx) ∫cosx∫sin^2(x)/cosx sinx∫(1cos^2(x))/cosx sinx∫(1/cosx)cosx sinx∫secx∫cosx … 
calculus trigonometric substitution
∫ dx/ (x^2+9)^2 dx set x = 3tan u dx = 3 sec^2 u du I = 3 sec^2 u du / ( 9 tan^2 u + 9)^2 = 3 sec^2 u du / ( 81 ( tan^2 u + 1)^2 = sec^2 u du / ( 27 ( sec^2 u )^2 = du / ( 27 sec^2 u = 2 cos^2 u du / 54 = ( 1 + cos 2u) du / 54 …