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 1/2(sinx+cosx+sinx-cosx)/(sinx+cosx) dx

then i seperate the fraction so it becomes

integral 1/2+1/2(sinx-cosx)/(sinx+cosx) dx
and from  there its a straight forward u-substitution keeping u = sinx+cosx

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

For this I used the fact that sin(x)+cos(x)=sqrt(2)*sin(x+pi/4) then let u = x+pi/4 and then used the sine sum formula to expand the top. From there it was trivial.

This guy has the best chill approach.

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.

This was absolutely beautiful!!!!!!!

Wow. This solution exudes elegance!

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

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

What a brilliancy!!

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.

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

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.

genius!!!

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

That was brilliant

OMG I love this channel

I watch at least one Prime Newtons video a day. It brings a smile to my face

∫ [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!

My approach is to rewrite sinx + cosx as ksin(x+α) by using the reduction formula
k = √2 , α = π/4
So, it will be
∫ sinx / √2 sin(x+π/4) dx

Now, take the √2 from the denominator outside the integral: it will be √2/2 ∫ sin(x) / sin(x+π/4) dx

Since the denominator is sin (x+π/4)we can write it using its reciprocal:

now it will be
√2/2 ∫ sinx•csc(x+π/4) dx

now do quick u-sub
u = x+π/4 —> x = u-π/4 , du = dx

it will be

√2/2 ∫ sin(u-π/4)• csc(u) du
now apply the formula sin(a-b) = sina•cosb-cosa•sinb

you’ll get √2/2 ∫ (√2/2 sinu -√2/2 cosu ) cscu du

Now distribute cscu

you’ll get
√2/2 ∫ √2/2 sinu • cscu - √2/2 cosu• cscu du

now as you know that , sinu • cscu = 1 and cosu•cscu = cotu

you’ll get
√2/2 ∫ √2/2-√2/2 cotu du

take √2/2 as common factor and put it out the integral

you’ll get
1/2∫ 1-cotu du

—> 1/2 [ ∫ 1 du - ∫ cotu du ]

—> 1/2[ u - ∫ cotu du ]

now , cotu = cosu/sinu
So , ∫ cosu / sinu du
Now, you can notice that the ‏derivative of sinu is cosu and it is there in the numerator
so the integral will be the absolute value of the thing that in the denominator,so it will be
∫ cosu / sinu du = Ln| sinu | +c

So , 1/2 [ u - Ln| sinu | + c

1/2u -1/2Ln| sinu | + c
1/2 (x+π/4) -1/2Ln | sin(x+π/4) | + c

1/2 x + π/8 -1/2 Ln | sin(x+π/4 | +c
now you notice that you have π/8 as a constant and c , so you can write them as a new constant c₂

So the final answer is
1/2 x - 1/2Ln | sin(x+π/4) | + c₂
which is equivalent to the answer that in the video