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I know @vernburn has correctly split the interval. However, periodicity actually made the calculation easier rather than causing trouble. Depending on how you express the antiderivative, the constant of integration may be different.

Periodicity is not the cause of this trouble. If you integrate cos x, then you get sin x which is periodic. However, the substitution still works as usual. The cause of this trouble is a different constant of integration at different intervals. Instead of +c for all real numbers, you actually...

It's not too hard to get the antiderivative in terms of elementary function...but the evil is in the substitution of upper and lower limits.😈 \int_{\frac{\pi}{3}}^{\frac{2021\pi}{2}}\frac{1}{2-\cos x}dx

This is the product of two functions. You cannot substitute and multiply this way.

another one Feel free to share your attempt. \int_{\frac{1}{4}}^{\frac{1}{2}}\sin\left(x^{x}+\ln x\right)\cos\left(x^{x}-\ln...

Your approach is correct but unfortunately a factor of 1/2 is missing somewhere. \int_{-\frac{1}{2\sqrt{2}}}^{\frac{1}{2\sqrt{2}}}\frac{\sqrt{64x^{6}-16x^{4}+x^{2}}}{\left(\sqrt{1-16x^{4}}+\sqrt{12x^{2}-48x^{4}}\right)\left(1+\pi^{x}\right)}dx=\frac{\pi}{48}-\frac{\sqrt{3}}{32}\pi+\frac{1}{8}

It seems harder 3U and graphs are covered in Part IA and IB respectively. http://www.angelfire.com/ab7/fourunit/4usyllabus2.pdf

a new one...shouldn't be too hard:p Feel free to share your attempt. Find the area bounded by x-axis and the curve y=\frac{\sqrt{64x^{6}-16x^{4}+x^{2}}}{\left(\sqrt{1-16x^{4}}+\sqrt{12x^{2}-48x^{4}}\right)\left(1+\pi^{x}\right)}.

This one should be easier. \int_{\frac{\pi}{10}}^{\frac{3\pi}{20}}\ln\left(\frac{1}{4}+\frac{3\tan x-1}{1-3\tan^{2}x}\right)dx

Once the substitution is uncovered, it is not much fun.

I think you ignored the base of the log and got (pi/12) ln 6.

Anyone has guessed the substitution?

This one should be harder. I think I've masked the underlying substitution quite well.:cool: \int_{-\frac{\pi}{12}}^{\frac{\pi}{6}}\frac{\log_{6}\left(\frac{6+2\sqrt{3}\tan x}{5-\sqrt{3}-\left(1+\sqrt{3}\right)\tan x}\right)}{\sin^{2}\left(\frac{\pi}{3}-x\right)+\cos^{2}x}dx

It is very quiet recently.:( Feel free to share your attempt. \int_{0}^{\pi^{2}}\frac{1+\cos\sqrt{x}}{\sin\sqrt{x}}dx=4\pi\ln2

I guess this is a typo. I think the whole page is poorly worded.

I agree induction is fine. By the way, we have common "difference" in AS and common "ratio" in GS.

It is just the derivative of a finite GS. As there is a finite number of terms, why do you have the concern of convergence?

f is continuous and differentiable everywhere in its domain.

If the decomposition step is only valid for positive integer values of x, then it's simply not continuous and therefore not differentiatable.