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The same old trick also applies here. \int_{1-\sqrt{2}}^{\sqrt{2}-1}\frac{\left(1+x\tan^{-1}x\right)^{2}}{\left(1+x^{2}\right)\left(1+\pi^{\tan x}\right)}dx =\int_{0}^{\sqrt{2}-1}\frac{\left(1+x\tan^{-1}x\right)^{2}}{1+x^{2}}\left(\frac{1}{1+\pi^{\tan...

Let's use the same old trick that has appeared in this thread many times. \int_{b}^{a}f\left(x\right)dx=\int_{\frac{a+b}{2}}^{a}\left(f\left(x\right)+f\left(a+b-x\right)\right)dx \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\cos x}{e^{\frac{1}{x}}+1}dx =\int_{0}^{\frac{\pi}{2}}\left(\frac{\cos...

First question for 2020 \int_{1-\sqrt{2}}^{\sqrt{2}-1}\frac{\left(1+x\tan^{-1}x\right)^{2}}{\left(1+x^{2}\right)\left(1+\pi^{\tan x}\right)}dx=\frac{192\pi+24\sqrt{2}\pi^{2}-24\pi^{2}-\pi^{3}}{1536}

Last question of the year Show by mathematical induction that \int_{0}^{1}x^{n}e^{ax}dx=\frac{n!}{\left(-a\right)^{n+1}}+\frac{e^{a}}{a}-\sum_{r=1}^{n}\frac{n!e^{a}}{\left(n-r\right)!\left(-a\right)^{r+1}} for all positive integers n and non-zero real number a.

As mentioned by DrDusk, with a lot of practice, you will develop an intuition for certain integrals. For u-substitution, quite often you can factor out the derivative of a familiar function. \int\frac{\cos x\ dx}{\sqrt{\sin x}}=\int\frac{d\left(\sin x\right)}{\sqrt{\sin x}} \int\frac{2x...

I also find the given substitution unexpected. I would straight away think of the trig substitution x=sec t.

It seems no one has attempted this interesting question yet. (a)(i) Yes. When x=0, dy/dx=0. (ii) No. For any interval -𝛿<x<𝛿, there exists x such that y<0. (iii) No. For any interval -𝛿<x<𝛿, there exists x such that y>0. (iv) No. The graph is neither concave nor convex near x=0. (b) Yes. For...

It seems no one has attempted yet. It's actually a combination of several results. The last one is the most interesting.😈 \int_{0}^{1}\left(1+a+b+c\right)^{m}x^{n}dx=\frac{\left(1+a+b+c\right)^{m}}{n+1} \int_{0}^{1}\log_{2}\left(\sec^{2}\frac{\pi x}{4}+2\tan\frac{\pi x}{4}\right)^{a}dx...

This one is similar but the algebra is slightly longer. \frac{d}{dk}\int_{0}^{2\tan^{-1}k}\frac{x}{1+\cos x+k\sin x}dx=\frac{2k^{2}\tan^{-1}k+k\ln\left(k^{2}+1\right)-\left(k^{2}+1\right)\left(\tan^{-1}k\right)\ln\left(k^{2}+1\right)}{k^{2}\left(k^{2}+1\right)}

\int_{0}^{\tan^{-1}k}\ln\left(1+k\tan x\right)dx=\frac{\tan^{-1}k}{2}\ln\left(k^{2}+1\right) (where k is a real constant)

(a) Consider the graph of y=\sqrt[3]{x^{5}}\sin\frac{1}{x}. (i) Is (0,0) a stationary point? (ii) Is (0,0) a minimum point? (iii) Is (0,0) a maximum point? (iv) Is (0,0) a point of inflection? (b) Consider the graph of y=\left|\sqrt[3]{x^{5}}\sin\frac{1}{x}\right|. Is (0,0) a minimum point...

Yes. It's very difficult (or nearly impossible) to choose a delta without working backwards on a rough work paper.o_O

I guess the question is not expecting you to apply trigonometric identities for product to sum or sum to product. Otherwise, there's no point to bring complex number into the question. Having said that, I think a trigonometric approach is more elegant, especially for the second part of the...

I think "Reverse the steps to get the proof required" actually means you have to write the whole stuff again in reversed order. The technique of working backwards is extremely useful when you learn how to prove limit in university. (It would be very difficult to think of a correct bound...

This one isn't too hard, right? The same old trick applies. \int_{0}^{\pi^{2}}\frac{1}{1-\sin\sqrt{x}\cos\sqrt{x}}dx =\int_{0}^{\pi}\frac{2x}{1-\sin x\cos x}dx =\int_{0}^{\frac{\pi}{2}}\frac{2x}{1-\sin x\cos x}dx+\int_{\frac{\pi}{2}}^{\pi}\frac{2x}{1-\sin x\cos x}dx...

It is too abstract. I'll leave it for your university lecturer and tutor.:cool:

It requires uniform convergence. Does the syllabus cover techniques to prove uniform convergence?o_O

A new one \int_{0}^{\pi^{2}}\frac{1}{1-\sin\sqrt{x}\cos\sqrt{x}}dx=\frac{5\sqrt{3}\pi^{2}}{9}

I have skipped the steps of rationalization. \int_{0}^{1}\frac{1}{2^{x+1}+\sqrt{4^{x}+2^{x}}+\sqrt{4^{x}-2^{x}}}dx =\int_{0}^{1}\frac{2^{-x}}{2+\sqrt{1+2^{-x}}+\sqrt{1-2^{-x}}}dx =-\frac{1}{\ln2}\int_{1}^{\frac{1}{2}}\frac{1}{2+\sqrt{1+u}+\sqrt{1-u}}du...

This one is absolutely painful.🤬 \int_{0}^{1}\left(\frac{3e^{x}+1}{e^{x}+\sqrt{x+e^{x}}}+\frac{\left(e^{x}+1\right)^{2}}{x+e^{x}+\sqrt{e^{3x}+xe^{2x}}}\right)dx =\ln\left(e^{2}+e+2e\sqrt{e+1}+1\right)-2+2\sqrt{e+1}-\ln4