integral of 1/(sin(x)+cos(x))^2, calculus 2 integrals

angle additional formula: a*sin(x)+b*cos(x)=Acos(x+𝛼)

blackpenredpen,
math for fun

[qilinxue]

Giving a shot at this.

Using the method shown in this video, sinx+cosx=sqrt(2)*cos(x-pi/4)

This simplifies the integral to 1/sqrt(x) * int[ 1/cos(x-pi/4) dx]
Looking just inside this integral, it's

sec(x-pi/4)

The integral of sec(x) is ln(|tan(x)+sec(x)|)

Therefore, the integral of sec(x-pi/4) is ln(|tan(x-pi/4)+sec(x-pi/4)|)

The final answer is:

[FunctionalIntegral]

This integral is equivalent to 1/2 integral 1/(1+sin(u)) du with u = 2x which gives - 1/(tan(x)+1) + C.
This is a much faster way.

[gnikola]

I didn't know about the a*sin(x) + b*cos(x) = A*cos(x), will you do a video about that? It's awesome!

[gandesto]

FINALLYYYY i have two hours trying resolve this integrals, thank u :''u

[joeli]

(divided by root(2)) Got to use Weierstrass Sub and partial fractions decomposition and a U-Sub.

[gopalguchhait]

We can put 1=(sin^2 x+cos^2 x) then we written as (a^2+b^2)= (a+b)^2 -2ab

[link_z]

Using the same formula as you for the second integral,
Therefore we want to calculate the ∫1/(sqrt(2)*cos(x-(pi/4)))dx.
First of all I'll use the substitution: u=x-(pi/4), du=dx and take the sqrt(2) over the integral: (1/sqrt(2))*∫(1/cos(u)du) or (1/sqrt(2))*∫(sec(u)du).

∫(sec(u)du)=∫(sec(u)* du= du

w=sec(u)+tan(u)
dw=sec(u)^2+sec(u)*tan(u)du

Plug it in:

ln|w|
Revert the w-->u-->x and add a constant of integration C.

* + C

[andresvivas]

Hi, for your question I used Weierstrass Substitution.. so taking
U=tan (x/2)
Sinx=2u/(1+u^2)
Cosx=(1-u^2)/(1+u^2)
dx=2/(1+u^2)du

Doing the replacement and all the algebra cancelations we will have integral of:
2/(-u^2+2u+1)du

Factoring out -1 and completing the square in the denominator we will have integral of:

-2/((u-1)^2 - 2) ... Direct integral of inverse hyperbolic tangent... so the anwer in terms of u is equal to:
square (2) arctanh((u--1)/square (2))

The final anwer is:
square (2) arctanh((tan (x/2)--1)/square (2)) + C..

I hope is fine... I don't know if I can do more cancelations haha...

Great lesson BTW... Keep your awesome videos...

[RutvikPhatak]

The answer to 7:41 is


1/√2{ ln } + C

Edit- forgot the c part

[holyshit]

For those who like integration by parts




First integral can be calculated by parts and in second integral integrand can by simplified by cancelling numerator and denominator

[ZeroByte__Hacks]

1/1+sin2x = 1/{1+2tanx÷(1+tan²x)} = sec²x/(1+tanx)², then Let t = 1+tanx then sec²x and finally Ans. -1/1+tanx + C

[alisheikh]

You can write sinx+cosx = root(2)sin(x+π/4)

Then I=1/2cosex^2(x+π/4)

[mustaphasahnoune]

Integrating 1/(a+cos x)^2 from 0 to 2pi

[Theraot]

I think I found a new favorite integral

[danniamejia]

Would there be a possibility that there are Spanish subtitles in your videos? Well, it's for a better understanding of it. Please! :c

[Zonnymaka]

...and now the ghetto way :)
We have a squared function on the denominator, so we can make an educated guess and claim that might be the outcome of derivating something like Y/(sin(x)+cos(x))
Hence
Nah, we aren't going to solve the ODE
We notice that [Y'*cos(x)+Y*sin(x)]+ [Y'*sin(x)-Y*cos(x)] = 1
RedPen would mumble "Mmmm, what can we do now?"
If Y=sin(x) the first bracket would be equal to 1 and the second bracket would be equal to 0! :)
We check and it works! Done!

[matthewstevens]

Can you prove the half angle identities? A bit perplexed on them

[girlchannel]

Sir can u help me please, i've got a homework and the question is
Integrate (sin x cos x)^5 dx =..

[kimrocha]

There is a third way to do this. If you handle trigonometric functions to exponential form, you'll get -1/(exp(2ix)+i).


Check at wolfram alpha :))

I love your channel, please, continue to do it!

[sarvagya-sharma]

Blackpenredpen...












BLUE PEN