Integrating A Rational Function of Sine and Cosine

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[@rhysjarvis6629]

Can't you multipy top and bottom by (sinx + cosx - 1)? bottom simplifies to 2 sinx cosx, and top is sinx + cosx - 1.

[@phill3986]

You'd be okay with the (sinx + cosx)/2sinxcosx but what about the -1/2sinxcosx?

[@yamikira6512]

​@@phill3986 it becomes 1/sin(2x)

[@phill3986]

​​@@yamikira6512oh yeah, it was right there. Thanks!

[@joyneelrocks]

Instead of using Weierstrass’ Substitution, you can change sin(x) + cos(x) = cos(x - π/4). Then the denominator becomes:
sin(x) + cos(x) + 1 = 1 + cos(x - π/4)
Then use the product of two cos() identity:
cos^2(a) = [1 + cos(2a)] / 2
2cos^2(a) = 1 + cos(2a)
Input 2a = x - π/4 ==> a = x/2 - π/8 ==> da/dx = 1/2 ==> da = 1/2 * dx to now have:
∫ 1/[sin(x) + cos(x) + 1] dx = ∫ sec^2(a) da = tan(a) + C = tan(x/2 - π/8) + C
Simple and its faster…

[@DonEnsley-yi2ql]

Let u = tan x/2 + 1
The integral then becomes
∫du/u = ln |u|+c
= ln | tan(x/2+1)|+ c

[@scottleung9587]

Cool!

[@pauljackson3491]

I'm trying to solve a rational quintic but the quintic is the denominator:
INT 1/(x^5+x+1) dx
Since the trinomial is not factorable then partial fractions don't work.
Using weird trig substitutions it can change into something like:
INT sin(x)/sin(8*x) dx
I don't know that one either.
Are they even solvable; do they need Bring radicals; can it be done using Feynman's technique?

[@DupekGownienko]

wtf do you mean it's not factorable?
x^5+x+1=(x^3-x^2+1)(x^2+x+1)

[@giuseppemalaguti435]

Posto t=tg(x/2),uso le formule parametriche...I=ln(t+1)=ln(1+tgx/2)+c

[@phill3986]

😊😊😊🎉🎉🎉👍👍👍

[@MortezaSabzian-db1sl]

It was so simple, it didn't matter at all
Integral(1/(sin(x)+cos(x)+1))
Integral(1/[2*cos^2(x/2)(tan(x/2)+1])
ln(1+tan(x/2))+C
Like fans

[@honestadministrator]

(dx /( cos ( x) + 1))
/( 1 + sin ( x) /( 1 + cos ( x)))
[ again d ( sin ( x) / ( 1 + cos ( x))
= cos ( x) (1 + cos ( x) +sin^2 ( x)) dx
/( 1 + cos ( x)) ^2
=( 1 + cos ( x)) dx / /( 1 + cos ( x)) ^2
= dx /( 1 + cos ( x))
Hereby
dx /( sin ( x) + cos ( x) + 1)
= d ( sin( x) /( cos ( x) + 1))
/( sin ( x) + cos ( x) + 1)
= d ( ln (sin (x) / (cos ( x) + 1) + 1))

[@bernarddoherty4014]

U du dx/dz ez