how do u evaluate this?
integral [3x (sinx/cos^4x)] dx
Use Integration by Parts,
Let,
u=3x and v'=sin x/cos^4 x
Then,
u'=3 and v=INTEGRAL sin x/cos ^4 x dx
But how do we find that integral?
We have, (multiply by negative)
-INTEGRAL -sin x/cos^4 x dx
Let, t=cos x then t'=-sin x
Apply substitution rule,
-INTEGRAL t'/t^4 dx=INTEGRAL 1/t^4 dt
Thus, by using t^{-4} and exponent rule we have,
-1/3*t^{-3}
Substitute back,
(-1/3)cos^{-3} x
Thus by parts,
-xcos^3 t+(1/3)INTEGRAL 1/cos^3 x dx
The problem is how to deal with that integral.
I believe that if you can express it then you should write,
sec x*sec^x
Apply integration by parts.
the method that you're using looks a little complicated, i got some hints on this question, it is to first find ∫ sec^3(x) dx
∫ sec^3(x) dx
= ∫ sec x d tan x
= sec x tan x - ∫ tan x d sec x
= sec x tan x - ∫ sec x tan^2(x) dx
then i kinda got stuck here, how do u further simplify this?