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I don't think this property is abused. For questions like #184, this property can help you avoid IBP. Of course, there are also questions like #183 and #185, where you actually need to get the primitive function by IBP.

Change the limit to force you to find the primitive function😈 \int_{3\pi/2}^{5\pi/4} \frac{\sin x}{\sqrt{1+\sin 2x}}\,dx = \frac \pi 8 + \frac {\ln2} 4

Maybe it wasn't too difficult for mathematical olympiad...but should be enough of a headache for MX2. If it appears in MX2 paper (final chance in 2019!!!), the performance will likely be worse than the geometry question last year.

Geometry question could be a killer :devil: but I think it will be gone from MX2 soon. ABCD is a quadrilateral with three equal sides AB,BC and CD. Show that the mid-point of AD lies on a circle with diameter BC if and only if the area of ABCD is a quarter of the product of its diagonals.

\int_{4}^{9}\left(1+\csc\left(\ln x\right)\right)\sqrt{\frac{\tan\left(\ln\sqrt{x}\right)}{x}}dx=6\sqrt{\tan\left(\ln3\right)}-4\sqrt{\tan\left(\ln2\right)}

A couple more \int_0^{\pi^2}\frac{1}{\csc\sqrt{x}+\sin\sqrt{x}}dx=\pi\sqrt{2}\ln(\sqrt{2}+1) \int_{\frac{\pi^2}{16}}^{\frac{9\pi^2}{16}}\frac{\sin\sqrt{x}}{1+\cos^4\sqrt{x}}dx=\frac{\pi\sqrt{2}}{4}(\ln5+2\tan^{-1}2)

This one closely resembles an earlier integral...shouldn't take too long to solve. \int_0^{\frac{\pi^2}{4}}\frac{\cos\sqrt{x}+6\sin\sqrt{x}}{\left(5\sin\sqrt{x}-30\cos\sqrt{x}+31\right)^2}dx

I regret I have picked these numbers as upper and lower limits. It is quite messy to rationalize the answer.o_O Nevertheless, it is a good practice for rationalization skills. Isn't it? \int_{\frac{3}{8}}^{\frac{5}{12}}\frac{\pi\left(4x^2+1\right)-2}{\left(2x\sin\pi x+\cos\pi...

This is not challenging. \int_{\frac{\pi}{6}}^{\frac{2019\pi}{2}}\frac{\sin x+\ln\left(x^{x\cos x}\right)}{x^{1+\sin x}}dx=\sqrt{\frac{6}{\pi}}-\frac{2019\pi}{2}

Well done!!! The integral of square root tangent is notorious.😈 \int_0^{\frac{\pi}{4}}\frac{x\sqrt{2\tan x}}{\sin2x}dx =\int_0^{\frac{\pi}{4}}\frac{x\sec^2x}{\sqrt{2\tan x}}dx =\left[x\sqrt{2\tan x}\right]_0^{\frac{\pi}{4}}-\int_0^{\frac{\pi}{4}}\sqrt{2\tan x}dx...

This one should lead you to a well-known tedious integral. \int_0^{\frac{\pi}{4}}\frac{x\sqrt{2\tan x}}{\sin2x}dx=\ln(\sqrt{2}+1)+\frac{\pi}{2\sqrt{2}}-\frac{\pi}{2}

Interesting. Do you still remember that question?

Was this one solved or not? The trick is quite similar. \int_2^4\left(\sqrt{x} + \sqrt[5]{\frac{x^{\frac{76x+75}{4x}} - \left(\log_3 x\right)^{19}}{x^{\frac{72x-73}{3x}} + \left(\log_5 x\right)^{24}}} + \frac{1}{\sqrt{x}}\right)\left(\sqrt{x} - \sqrt[5]{\frac{x^{\frac{76x+75}{4x}} - \left(\log_3...

It's correct. Tabular integration also works quite well. D \qquad\qquad I u^2\qquad e^u\sin u 2u\qquad\frac{e^u}{2}\left(\sin\ u-\cos u\right) 2\qquad-\frac{e^u}{2}\cos u 0\qquad-\frac{e^u}{4}\left(\sin\ u+\cos u\right)

If you have read #162, this one should not be challenging. \int\left(\ln^2x\right)\sin\left(\ln x\right)dx=\frac{x\left(\ln^2x\right)\left(\sin\left(\ln x\right)-\cos\left(\ln x\right)\right)}{2}+x\left(\ln x\right)\cos\left(\ln x\right)-\frac{x\left(\sin\left(\ln x\right)+\cos\left(\ln...

Using similar technique, it should be straight-forward to show that \int_0^{\pi}\frac{\sqrt[2019]{\sin x}\cdot\sin2x}{e^{\sin x}\ln\left(1+\sin^{2019}x\right)}dx=0:cool:

Let\ v=x^x. \ln v=x\ln x \frac{dv}{v}=\left(1+\ln x\right)dx When\ x=0.25,\ v=0.25^{0.25}=\frac{1}{\sqrt{2}}. When\ x=0.5,\ v=0.5^{0.5}=\frac{1}{\sqrt{2}}. \int_{0.25}^{0.5}\left(1+\ln x\right)\sin\left(2x^x\right)dx =\int_{\frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}}\frac{\sin\left(2v\right)}{v}dv =0

Just spent some time to type my solution. If you see any typo, please let me know. It is assumed that readers can derive: \int e^x\sin x=\frac{e^x}{2}\left(\sin\ x-\cos x\right)+c \int e^x\cos x=\frac{e^x}{2}\left(\sin\ x+\cos x\right)+c Let's look at this integral first. I've added DI table...

There's no closed form. Take a look at the upper and lower limits.:cool:

It should be well known enough that \int e^x\sin x=\frac{e^x}{2}\left(\sin\ x-\cos x\right)+c and \int e^x\cos x=\frac{e^x}{2}\left(\sin\ x+\cos x\right)+c (can be derived by integration by parts twice) This is how you may make the life easier. \int_{ }^{ }\left(x+1\right)e^x\sin xdx...