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Some more interesting expressions but no success \int_{0}^{\frac{\pi}{4}}\tan^{-1}\left(\frac{1-\sin x\cos x}{\sin x+\cos x}\right)dx \int_{0}^{\frac{\pi}{4}}\tan^{-1}\left(\frac{2-\cos2x}{2\sqrt{2}\cos x}\right)dx \int_{0}^{\frac{\pi}{4}}\tan^{-1}\left(\frac{1+2\sin^{2}x}{2\sqrt{2}\cos...

A few more alternative expressions that look interesting but not leading to a solution:frown2: \frac{1}{2}\int_{0}^{\frac{\pi}{2}}\tan^{-1}\left(\frac{\sec x-\cos x}{2}\right)dx \frac{1}{2}\int_{0}^{\frac{\pi}{2}}\tan^{-1}\left(\frac{\sin x\tan x}{2}\right)dx...

Do you have more hints?

Please correct me if there is any typo. \frac{1}{1+r}=\sum_{n=0}^{\infty}\left(-r\right)^{n} \int_{0}^{x}\frac{1}{1+r}dr=\int_{0}^{x}\sum_{n=0}^{\infty}\left(-r\right)^{n}dr \ln\left(1+x\right)=\sum_{n=0}^{\infty}\frac{\left(-1\right)^{n}x^{n+1}}{n+1}...

Did you prove or disprove the hypothesis?:p

Was this one solved? $\noindent Let $T_n$ be a sequence such that $T_1=1$, $T_2=2$ and $T_{n+2}=T_{n+1}+T_n$. Show by mathematical induction that $T_nT_{n+2}-(T_{n+1})^2=\left(-1\right)^n$. If you like to play with induction, you may also try $\noindent Let $a$ and $b$ be two relatively prime...

Many alternative forms that lead me to nowhereo_O Why is pi^2/8 appearing so often? \int_{0}^{\infty}\frac{\tan^{-1}x}{\sqrt{x^{2}+1}\left(2x^{2}+1\right)}dx \frac{\pi^{2}}{8}-\int_{0}^{\infty}\frac{x\tan^{-1}x}{\sqrt{x^{2}+1}\left(2+x^{2}\right)}dx...

A solution based on elementary geometry will just be more convoluted. If you have attempted the hardest easy geometry problem (google it if you haven't), then you should see that sine rule can save you from constructing many additional lines.

haha...:awesome: There are other possibilities but it's good to enrich the toolbox. When reverse quotient rule works, it is really amazing. This is an example with denominator of the form (something)^2...but reverse quotient rule may not be a good idea...

geometry again😈 In the figure, ABC, ADE and AFG are equilateral triangles. H, I and J are the mid-points of CD, EF and GB respectively. Find HI:IJ.

I'm really lost. I tried to express the integral in various alternative ways but I can't solve any of these.:mad: \frac{\pi^{2}}{8}-\int_{0}^{\frac{\pi}{2}}\frac{x\sin x}{1+\cos^{2}x}dx \frac{\pi^{2}}{8}-\int_{0}^{\frac{\pi}{2}}\tan^{-1}\left(\sin x\right)dx...

Why not have a crossover of trig and exponential? :p Here, I'm using re-arranged version that I finds more convenient. \int_{b}^{a}f\left(x\right)dx=\int_{b}^{\frac{a+b}{2}}f\left(x\right)+f\left(a+b-x\right)dx=\int_{\frac{a+b}{2}}^{a}f\left(x\right)+f\left(a+b-x\right)dx...

Yes

Shall we have geometry again? It would be interesting if geometry can be given more weight in its final MX2 year. AB is a chord of a circle with centre O. Extend AB to C and extend OB to D such that AC=OC, AD||OC and ∠ACD=15°. Find BC:BD.

Yes.

Let's have another geometry problem. This should be simpler than the one yesterday. O is a point inside equilateral triangle ABC such that 6∠OAB=3∠OBC=2∠OCA. Find OA:OB.

Any hint?🤔

h is not uniformally distributed. You need to take into account the probability density function of h.🙄

The version that I find most useful is \int_{b}^{a}f\left(x\right)dx=\int_{b}^{\frac{a+b}{2}}f\left(x\right)+f\left(a+b-x\right)dx=\int_{\frac{a+b}{2}}^{a}f\left(x\right)+f\left(a+b-x\right)dx Two common cases are b=-a and b=0...

Let's continue to have fun with geometry. If a similar question appears in MX2 paper, I guess half of the candidates may give up.:p A and B are two points on a circle with centre O. Extend OA to C and OB to D such that ∠ADC=∠BCD. Prove that AB||CD. Hint: Consider two cases - AB is the diameter...