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\int_{0}^{a}f\left(x\right)dx=\int_{0}^{\frac{a}{2}}\left(f\left(x\right)+f\left(a-x\right)\right)dx \int_{0}^{\frac{\pi}{2}}\sin^{2}\left(x\right)\ln\left(\tan\ x\right)dx =\int_{0}^{\frac{\pi}{4}}\left(\sin^{2}x\ln\left(\tan\ x\right)+\cos^{2}x\ln\left(\cot\ x\right)\right)dx...

With the usual trick, f(x)+f(pi/2-x)=-cos(2x) ln(tan x), which can be evaluated by IBP.

I guess you must have got an answer close to zero. \int_{0}^{\frac{\pi}{4}}\left(\tan x-\tan^{2}x\right)^{4}dx =\int_{0}^{1}\frac{\left(t-t^{2}\right)^{4}}{1+t^{2}}dt =\int_{0}^{1}\left(t^{6}-4t^{5}+5t^{4}-4t^{2}+4-\frac{4}{1+t^{2}}\right)dt...

The algebra is quite annoying.o_O \int_{0}^{1}\frac{1}{2^{x+1}+\sqrt{4^{x}+2^{x}}+\sqrt{4^{x}-2^{x}}}dx=\frac{6+9\sqrt{2}+6\sqrt{3}-9\sqrt{6}-2\pi}{6\ln2}

All the mentioned definitions are equivalent. My prof preferred to take this formula as the definition because it avoids introducing complex power series at the very beginning of the course.

You need to define e^x in the complex plane before you can exponentiate a complex number. There are different ways to do it. One approach is to define e^x as a complex power series, which would lead to a question why it converges in the same way as the real counterpart. Another approach is to...

I'm not saying anyone is not credible. However, any attempt to "prove" this formula will require some "hidden" assumptions that are definitely out of reach for 4U students. I have seen various versions of sloppy "proof" that omits those assumptions.

This approach would have to establish that the differential operator applies to a complex function e^(ix) in the same way as a real function e^(kx)...which is not trivial.

Well, I guess your teacher assumes that there exists a Taylor series for e^(ix) in the complex plane first...which is not trivial.

I would rather call it a definition. I don't think there exists a rigirous "proof" unless you make various assumptions that complex analysis behaves similarly to real analysis.

Yes, it is a good idea (unless you want to use trig identities many times or prove a reduction formula first)

I'll try @sharky564 question later.o_O This question should be more accessible to 4U students.:cool: \int_{0}^{\frac{\pi}{4}}\left(\tan x-\tan^{2}x\right)^{4}dx

\int\frac{x\cos x}{1+\sin^{2}x}dx =x\tan^{-1}\left(\sin x\right)-\int\tan^{-1}\left(\sin x\right)dx Let's try Feynman trick. I\left(a\right)=\int_{0}^{\frac{\pi}{2}}\tan^{-1}\left(a\sin x\right)dx I'\left(a\right)=\int_{0}^{\frac{\pi}{2}}\frac{\sin x}{1+\left(a\sin x\right)^{2}}dx...

Thanks God! I finally have a way out for @sharky564 integral. I will try to type the solution over the weekend.

Continuing from above \int_{0}^{\pi^{2}}\frac{\left(1+\sin\sqrt{x}\right)^{n}+\left(1-\sin\sqrt{x}\right)^{n}}{\left(1+\sin\sqrt{x}\right)^{n+1}+\left(1-\sin\sqrt{x}\right)^{n+1}}dx =2\pi\int_{0}^{1}\frac{1+t^{2n}}{1+t^{2n+2}}dt...

\int_{0}^{\pi^{2}}\frac{\left(1+\sin\sqrt{x}\right)^{n}+\left(1-\sin\sqrt{x}\right)^{n}}{\left(1+\sin\sqrt{x}\right)^{n+1}+\left(1-\sin\sqrt{x}\right)^{n+1}}dx =2\int_{0}^{\pi}u\frac{\left(1+\sin u\right)^{n}+\left(1-\sin u\right)^{n}}{\left(1+\sin u\right)^{n+1}+\left(1-\sin u\right)^{n+1}}du\...

It's difficult to judge the difficulty from appearance. sqrt(tan x) may look simple but it's more tedious to integrate than some horrible looking functions.

This question can take ages if you're not on the right track.:mad: \frac{\int_{0}^{1}\left(1+a+b+c\right)^{m}x^{n}dx}{\int_{0}^{1}\log_{2}\left(\sqrt[n+1]{\frac{\left(16^{\sqrt{x}}+k\cdot4^{\sqrt{x}}+4\right)\left(16^{1-\sqrt{x}}+k\cdot4^{1-\sqrt{x}}+4\right)\left(\sec^{2}\frac{\pi...

One more question \lim_{n\to\infty} \int_{0}^{\pi^{2}}\frac{\left(1+\sin\sqrt{x}\right)^{n}+\left(1-\sin\sqrt{x}\right)^{n}}{\left(1+\sin\sqrt{x}\right)^{n+1}+\left(1-\sin\sqrt{x}\right)^{n+1}}dx (It can be solved without knowing the result in #217.🤓)

I have decided to give up @sharky564 question. This is a new question that should be solvable in 4U. $Given that $\int_{0}^{\infty}\frac{1}{1+x^{k}}dx=\frac{\pi}{k}\csc\frac{\pi}{k}$ for $k>1$. Find...