Integral with Logarithms

∫π/20log(cos(x))log(sin(x)) dx One solution: Consider : F(m,n)=∫π/20sin2m−1xcos2n−1xdx To solve this put sin2x=t to get our integral as : F(m,n)=12∫10tm−1(1−t)n−1dt=β(m,n)2 Where β(m,n) is the beta function. F(m,n)=Γ(m)Γ(n)2Γ(m+n) Hence we have : Γ(m)Γ(n)Γ(m+n)=2∫π/20sin2m−1xcos2n−1xdx First differentiating both sides with respect to m we have : Γ(n)(Γ(m+n))2(Γ′(m)Γ(m+n)−Γ(m)Γ′(m+n))=4∫π/20log(sin(x))sin2m−1xcos2n−1xdx Better written as : Γ(m)Γ(n)Γ(m+n)(ψ(m)−ψ(m+n))=4∫π/20log(sin(x))sin2m−1xcos2n−1xdx where ψ(x) is the digamma function. Now differentiate … Read more

Help with an Inverse Trigonometry Integral 2

Evaluate $$\int^{1/{\sqrt{3}}}_{-1/{\sqrt{3}}} \frac{x^4}{1-x^4}\cos^{-1}\frac{2x}{1+x^2} \mathrm{d}x$$ The Solution: We try to eliminate $\cos^{-1}\frac{2x}{1+x^2}$ by using the relation $$\pi – cos^{-1}(a) = cos^{-1}(a)$$ Consider the first 3 steps: $$I=\int^{1/{\sqrt{3}}}_{-1/{\sqrt{3}}} \frac{x^4}{1-x^4}\cos^{-1}\frac{2x}{1+x^2} \mathrm{d}x$$ Putting $x=-t$ $$I=\int^{1/{\sqrt{3}}}_{-1/{\sqrt{3}}} \frac{(-t)^4}{1-(-t)^4}\cos^{-1}\frac{2(-t)}{1+(-t)^2} (-dt)$$ $$I=\int^{1/{\sqrt{3}}}_{-1/{\sqrt{3}}} \frac{t^4}{1-t^4}(\pi -cos^{-1}\frac{2t}{1+t^2}) (-\mathrm{d}t)$$  Now, how do we add both of these: $$\int^{1/{\sqrt{3}}}_{-1/{\sqrt{3}}} \frac{x^4}{1-x^4}\cos^{-1}\frac{2x}{1+x^2} \mathrm{d}x +\int^{1/{\sqrt{3}}}_{-1/{\sqrt{3}}} \frac{t^4}{1-t^4}(\pi -cos^{-1}\frac{2t}{1+t^2}) (-\mathrm{d}t)$$ Could … Read more

Convert into Beta Function

Could somebody show me how to convert $$\int _{ 0 }^{ 1 }{ { \bigg [ \left( 1-{ x }^{ 1/17} \right) }^{ 23 } } +{ \left( 1-{ x }^{ 1/23 } \right) }^{ 17 } \bigg ] dx$$ into the Beta Function? I could recognize the similarity of the Integrand and the limits … Read more

A function f(x) satisfies f(x)=sinx+∫x0f′(t)(2sint−sin2t)dtf(x) = \sin x +\int^x_0 f'(t)(2\sin t-\sin^2t)dt, then find f(x)

Problem : A function f satisfies f(x)=sinx+∫x0f′(t)(2sint−sin2t)dt, then find f(x). My approach : Differentiating both sides we get : f′(x)=cosx+f′(x)(2sinx−sin2x) Using Leibnitz’s rule : ddx(∫ψ(x)ϕ(x)f(t)dt)=ddx{ψ(x)}f(ψ(x))−ddx{ϕ(x)}f(ϕ(x)) Could you please guide me further as it seems application of Leibnitz’s rule is incorrect. Please guide will be of great help thanks. Answer You applied Newton-Leibniz rule correctly. f′(x)=cosx1−2sinx+sin2x=cosx(sinx−1)2 … Read more

Why does this sum equal zero?

Let $\gamma$ be a piece-wise, smooth, closed curve. Let $[t_{j+1}, t_{j}]$ be an interval on the curve. Prove, $$\int_{\gamma} z^m dz=0$$ In the proof it states $$\int_{t_{j}}^{t_{j+1}} \gamma^m(t) \gamma'(t)=\frac{1}{m+1} [\gamma^{m+1}(t_{j+1})-\gamma^{m+1}(t_{j})]$$ Which I can see why. Next it claims, $$\int_{\gamma} z^m dz=\sum_{j=0}^{n-1} [\frac{1}{m+1} [\gamma^{m+1}(t_{j+1})-\gamma^{m}(t_j)]]=\frac{1}{m+1} [\gamma^{m+1}(b)-\gamma^{m+1}(a)]=0$$ I understand why it is $0$ but how did they … Read more

Definite Integration with Trigonometric Substitution

I’m working on a question that involves using trigonometric substitution on a definite integral that will later use u substitution but I am not sure how to go ahead with this. ∫211×2√4×2+9dx My first step was to use √a2+x2 as x=atanθ to get… 2x=3tanθ:x=32tanθ dx=32sec2θ Substituting: ∫32sec2θ94tan2θ√9tan2θ+9 The problem here is how do I change … Read more

How to prove ∫1001⌊arctanx⌋dx=100−tan1\int_{1}^{100} \lfloor \arctan x \rfloor dx = 100 – \tan 1?

The question was to find ∫1001⌊arctanx⌋dx. I hesitated because I learnt from illustrations in my book that when there is step up function, it is compulsory to break it at integral limits. Did that mean I’ve to break the limits ∫21⋯∫10099? I always had problem what that means-arctan100∘orarctan100. If it was later, I felt very … Read more