Integrate [sin( x)/ (sin x + cos(x))]

This strategy came by keen observation. It is the first time I ever used it abd I know it works.

[Fragged]

what i like to do for this integral is i multiply and divide by 2 so its

integral of1/2( 2sinx/(sinx+cosx))dx

then i would add and substract cosx from the numerator so it becomes

integral dx

then i seperate the fraction so it becomes

integral dx
and from there its a straight forward u-substitution keeping u = sinx+cosx

[sodabutnofizz]

This question without limits: 💀
This question with limits: 🥰

[arnavthescientist]

Hi third year mechanical engineering student here from India and thanks to your videos I can still be in touch with my calculus skills even though I barely need them in engineering.

[sandem]

For this I used the fact that then let u = x+pi/4 and then used the sine sum formula to expand the top. From there it was trivial.

[SQT]

This is a way I saw in a book. It re-wrote the sinx in the numerator as ½(sinx+cosx) -½(cosx-sinx). Put that over the given denominator, split the fractions you get integral of ½ -½(cosx-sinx)/(sinx+cosx), which is ½x -½In(sinx+cosx). Yours was a nice solution as well.

[Philippe-coder]

this is a tricky and beautiful way of solving this integral.

[anestismoutafidis]

∫ [sinx/sinx+(1-sinx)+c] <=>
-cosx/-cosx-(x+cosx)+c
-cosx/-cosx-x-cosx +c <=>
-cosx/-2cosx-x +c <=>
[-cosx/-2cosx + -cosx/-x] =
[cosx/x=1 because there is a solution cos(x)=x], for this, we continue to the final solution
∫ [1/2 + 1 +c] = [1/2+2/2+c] =
[1, 5+c] ore [3π/2+c]
Thank you very much for your brilliant support of teaching- material!

[anoopkumar-dtwp]

I liked your method. However, I wrote the numerator as So I put

sin x = A(sin x + cos x) + B(d/dx(sin x + cos x)

sin x = A(sin x + cos x) + B(cos x - sin x)

sin x = Asin x + Acosx + Bcos x - Bsin x

sin x = sin x(A - B) + cos x(A+B)

A - B = 1
A + B = 0

A= 1/2
B=-1/2

Substitute it back, solve and we will get

1/2 x - 1/2 ln (sin x + cos x)


For any complicated fraction, I will always try whether this method works or not. Luckily for many it works as it worked here as well.. Thank you😊

[holyshit]

- cos(x)))
then seperate fractions
Other option is divide numerator and denominator by cos(x)
then factor out cos^2(x) from denominator then substitute t = tan(x)

[Aek]

Integrate:
I= 1/1+cotx=
cosec^2 x - cot^2 x/1+cotx=
-ln|1+cotx| - (cot^2 x-1+1)/1+cotx=
-ln|1+cotx| - lnt.(cotx-1)-I
2I=-ln|1+cotx|-ln|sinx|+x
Άλλη όμορφη λύση του ίδιου ολοκληρώματος, δική μου σκέψη!!❤

[pankhapinaxx]

Similar to your approach, you can generalize the integral of (p.sinx + q.cosx + r)/(a.sinx + b.cosx + c). The numerator can be written as:
p.sinx + q.cosx + r = l.(a.sinx + b.cosx + c) + m.(a.cosx -b.sinx) + n, you will get system of 3 variables, with 3 equations,
a.l - b.m = p
b.l + a.m = q
c.l + 1.n = r
solving these three equations for l, m and n you will get a simplified version of original integral.

[kamalmukhtar]

Wow!! Amazing!!!! I love the simplicity!!!, 👍

[wayneosaur]

Another formula the reveals a relationship between e and trig functions. I like it.

[maxborn]

You can also divide numerator/denominator by cos(x), get tan(x)/[tan(x)+1]. Now tan(x-pi/4) = [tan(x)-1]/[tan(x)+1], so completing that part gives back the result.

[cristianvita]

Interesting method, thank you sir for putting it that way!
I was thinking of factorizing sin(x) in the denominator, then cancel it out with the sin(x) of the above, and the integral would remain A = integral of (1/(1+cot(x))

[youngmustang]

Wow I have never seen anyone solve an integral this way before

[CalculusIsFun]

I just multiplied by cos(x) - sin(x) on the top and bottom. Bottom becomes cos(2x), top becomes (1/2)sin(2x) - sin^2(x)

split apart the integral. The first one, the (1/2)(sin(2x)/cos(2x)) just works out to (-1/4)ln|cos(2x)|. For the remaining integral, sin^2(x) = (1/2)[1 - cos(2x)| so this integral becomes the integral of (
-(1/2)sec(2x) + (1/2) with respect to x which works out to -(1/4)ln|sec(2x) + tan(2x)| + (1/2)(x).

that all together is (1/4)ln|cos(2x)| - (1/4)ln|sec(2x) + tan(2x)| + (x/2)


I used property of logarithm ln(A) + ln(B) = ln(AB) to simplify a bit.

My final answer was -(1/4)ln|1+sin(2x)| + (x/2) + C


EDIT: just for fun, I decided to prove that your solution is equivalent to mine. Mines was (-1/4)ln|1 + sin(2x)| + (x/2) + C

notice how sin(2x) = 2sin(x)cos(x) and 1 = cos^2(x) + sin^2(x)

so my solution equals (-1/4)ln|cos^2(x) + 2sin(x)cos(x) + sin^2(x)| + (x/2) + C

= -(1/4)ln|(cos(x) + sin(x))^2) + (x/2) + C

Now let’s just use properties of logarithms. ln(a^b) = bln(a) therefore my solution equals -(1/2)ln|sin(x) + cos(x)| + (x/2) + C

[mdasifeqbal]

The method is simple but is lengthy also.
Numerator= ½(2sinx) = ½[(sinx + cosx) + (sinx - cosx)]
Separate them to get 2 integrals.
1st Integral is x
2nd integral is ln|sinx + cosx|

[robertveith]

@ Prime Newtons -- You have the correct answer in a nox at the bottomleft, bot you are missing a denominator of 2 at the far right for the natural log term.

[SSRSSR]

Yes whenever i used to see linear sine and cosine in fraction i use to substitute everyhing in terms of tan (theta/2)