Integral of the Day 5.2.24 | Trig Integral...Can you make it Boomerang?? | Math with Professor V

Here's your latest Integral of the Day! I solved it by adding and subtracting terms from the numerator to make it "boomerang"! Did you solve it differently? Let me know!

Here's links to some other integrals that "boomerang":

Here's the video on the Weierstrass substitution I mentioned as well:

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xoxo,
Professor V

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[jole]

You're the best maths teacher every

[siyabongashoba]

Dear Prof V thank you so much for the video it is on another level! I have never saw this trick before. Prof I got 1/2 *x - 1/4 *ln| sec(2x)+tan(2x) | - 1/4 * ln| 2(sinx)^2-1 | + C I am not sure if it is right or wrong. Firstly I time everything by 1 (sinx - cosx)/ (sinx - cosx) and broke it up into two part, part 1 integral of (sinx)^2/((sinx)^2-(cosx)^2) and part 2 was integral of (sinx*cosx)/ ((sinx)^2 - (cosx)^2) part 2 was not that hard I used identity and output - 1/4 *ln| 2(sinx)^2 - 1 | part 1 I used the double angle of cos to change every square in terms of cos(2x) and output 1/2x - 1/4 *ln| sec(2x) + tan(2x) | ya ne almost 1 hour

[Dissapointed]

Solution: multiply the top and bottom by sin(x)-cos(x). So we have: Int{ (sin^2(x)- cos(x)sin(x))/ sin^2(x)-cos^2(x)}
The identity cos2x= cos^2x - sin^2(x) can be used such that the denominator becomes -cos(2x).
Also sin(x)cos(x)= 1/2sin(2x).
So know we have
Int{ (sin^2(x) -sin2x)/(cos2x)}
This can be broken up into pieces know.
Int{sin^2(x)/cos(2x)}- Int{sin2x/cos2x}
sin^2(x)= (cos2x-1/(-2) via identity. Therefore;
Int{cos2x-1/-2cos2x} -Int{sin2x/cos2x}
This works out be
x/2

[kninh]

For the integral, can you rewrite it as sinx(sinx+cosx)/(sin^2x + cos^2x)

[Algebrainiac]

This one was too hard I couldn’t do it lol 😭😭