# weekly brilliant problem#3, solving integral by symmetry, IIT JEE mains 2015 - Videos

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So, I finally have a chance to show you how to solve a definite integral by taking the advantage of its symmetry. Don’t just use brute force and integrate all the time (although it might be more fun).
I got this problem from: https://brilliant.org/problems/jee-mains-2015-1530/ , and there are plenty more!!!

integral of log(x)/(log(x)+log(6-x)) from 2 to 4,
integral of sqrt(cot(x))/(sqrt(cot(x))+sqrt(tan(x))),
JEE main 2015 integral,
integral by symmetry,
integral properties,

blackpenredpen,
math for fun

source

1. It is worthwhile to teach students to first LOOK at a plot of the integrand of a definite integral. See what you're up against. Things like symmetry stand out visually. And so does the fact that this function is quite close to linear over the interval. We happen to need the former and not the latter in this particular case, but irl it may well be the other way around.

2. I remember a video of you when you used the complex integral of x^i in order to evaluate the real integral of cos(ln(x))
You introduced complexes in a non complex problem to make your life easier.
I'm looking forward other videos where you'll use complex to easily solve real equations!!

3. @blackpenredpen I wonder if you can find the Integral of x^x from 0 to pi.

Edit: Bonus: Integral of same function from 0 to e, and more hilarious Integral of same function from 0 to i?
Comedic Edit: I'm a fool, I should have seen you video on Integral of x^x before saying this. 🙁

4. This uses a basic symmetry property:

int_a^b f(x) dx = int_a^b f(a + b – x) dx

Thus, our integral

I = int_2^4 (log x)/(logx + log(6 – x)) dx

becomes

I = int_2^4 (log(6 – x))/(log(6 – x) + log(x)) dx

Then,

2I = int_2^4 dx = 4 – 2 = 2
I = 1

Spoiler: The next video also uses this property 🙂

5. For the integral at the end of the video, I think it's pi/4 (???)

I simplified it to just cos and sin and did a weierstrass substitution followed by partial fraction decomposition and was able to integrate, and at the end it was just arctan(1).

Probably did it more complicated than it had to be but it seemed to work lol

If anyone could check me, I'd be grateful

6. As a student of Calculus, I m thankful to you sir for solving this problem. It would be really nice of you if you do more of these videos based on IIT-JEE Questions because these are quite challenging plus you will also get a boost in your views becuase IIT JEE is a big thing in India…!

7. If you let u=3-x, I think things cancel out easier.

integral argument becomes:
u=1 to u=-1
log(3-u) / [ log(3-u) + log(6-3-u) ] * (-du)

this can factor to
u=-1 to u=1
[ log(3-u) / log(3-u) ] * [ 1/ (1 + 1) ] du
=>
1 * (1/2) du

then
u/2, from -1 to 1
=>
(1)/2 – (-1)/2 = 1

8. How can we let u=6-x, then rewrite the formula, and then just swap u for x, without paying attention to the definition we have given u? Can we just let go of u=6-x later in the equation do that 6-u==6-x? Seems odd to me

9. Hey man, I LOVE your videos, especially this one haha. One thing I dont quite get though is why you are able to just substitute x for u at 7:20 . I mean u = 6-x isnt it? So why can you say that u = x and just get the integral into the x world. It'd be great if someone helped me out here, its a beautiful integral and it'd be even more beautiful if i understood that part 🙂

10. this notation something I dont like about my maths book (ncert and other indian class 12) book that they use log.. for natural logs. It can quite confuse anybody unknown. and may lead to wrong answers due to misconception.
why not use standard ln for log base e and log for log base 10.