/sci/ - Science & Math

>>10794146
Yeah, R E S I D U E S

>>10794146
Trig substitution [sqrt(3)/sqrt(2)]*sectheta.

If you get creative you will find you can solve almost anything using trig substitution.

>>10794146
You can't solve it because you were clamped and irridated, move to the woods to escape the flouride and the solution will come in time.

>>10794146
Move the denominator to the numerator algebraically making the power negative then use the general power rule for integrals. Done!

>>10794163
how I end up with 1/8*integral of secx/tan^5x

>>10794173
Additionally, the final answer should be
-1 ÷ 2(2x^2 - 3)^2

>>10794180
Zero clue.

>>10794146
Did I help with it? You good now OP?

>>10794146
Its pretty obvious that you can do it with partial fractions
If you want to do it with 6 denominators, namely (sqrt(2)*x + sqrt(3)), (sqrt(2)*x + sqrt(3))^2, (sqrt(2)*x + sqrt(3))^3, (sqrt(2)*x - sqrt(3)), (sqrt(2)*x - sqrt(3))^2, (sqrt(2)*x - sqrt(3))^3, then you can compute the numerators with polynominal division. Use the Extended Eucledian algorithm on (sort(2)*x + sqrt(1.5))^3 and (sqrt(2)*x - sqrt(1.5))^3 and get the coefficients (the gcd should be 1, so you screwed up if its not). Take the coefficients, and divide them by the opposing polynominals, so, if p(x) is your coefficient for (sqrt(2)*x + sqrt(1.5))^3, then divide it by (sqrt(2)*x - sqrt(1.5))^k, where k is 0, 1, or 2, and if q(x) is your coefficient for (sqrt(2)*x - sqrt(1.5))^3, then divide it by (sqrt(2)*x + sqrt(1.5))^k, where k is 0, 1 or 2. The quotients after each division step should always be constants (the remainders should be of lower degree and get sent to the k-1 power division step) and they should be the numberators of each fraction.

>>10794180
So first t=theta for this thread and I will forget about the constant multiple for now.

So then expand sect*(1/tan^5t) into (1/cost)*(cos^5t/sin^5t) and simplify to cos^4t/sin^5t. Then expand the top into (1-sin^2t)^2/sin^5t. Then expand the top part and split up the integral. You should have sec^5t, -2sec^3t, and sect.

>>10794183
thats wrong

>>10794203
Partial fractions is retarded and autistic. Literally just use trig sub. It's also a lot harder to make a mistake using trig subs.

>>10794203
>>10794209
I came to the conclusion that if I get something like that in test I wont waste my time

>>10794173
kek, that's a classic mistake my friend

>>10794146
just do trig sub, it might look ugly and long but it will work, maybe you can also do hyperbolic trig or exponential sub but idk

>>10794709
>just do trig sub
didnt work I am looking into recursive

>>10794146
trig sub retard and don’t post stupid nonsense like this outside brainlet containment thread
>>10794203
use Latex for fuck’s sake this hurts my eyes

>>10795069
I tried it

>>10795140
>2^3

>>10794146
I have wolfram alpha pro but I am scared they will suspend my account if I type here the full solution.

>>10795140
What is the issue? That trig integral is a bitch but it is definitely solvable.

>>10795991
Its not as good as the partial fractions method

>>10794146
>>10795692

>>10794146
Pretty sure using partial fractions would be fine although not very enjoyable. Did you try u-v sub maybe?

>>10796054
Trivial

>>10796054
kek