HSC 2016 MX2 Integration Marathon (archive) (1 Viewer)

Paradoxica · Aug 8, 2016

MX2 2016 Integration Marathon

juantheron said:
$Evaluation of $\int_{0}^{1}\frac{\sin \theta \left(\cos^2 -\cos^2 \frac{\pi}{5}\right)\cdot \left(\cos^2 -\cos^2 \frac{2\pi}{5}\right)}{\sin 5\theta}d\theta$

The integrand simplifies to 1/16 by the expanding the denominator (Use the fact that cos(π/5) = φ/2 and cos(2π/5) = 1/2φ, then use the quintuple angle identity)

So the answer is 1/16.

Paradoxica · Aug 8, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
Have fun

$\noindent$\int \frac{\sin^2x+\sin x}{\sin x + \cos x +1}$d$x \\ \int \frac{\cos^2x+\cos x}{\sin x + \cos x +1}$d$x \\ \int \frac{\cos 7x- \cos 8x}{1+ 2\cos 5x}$d$x \\ \int_0^\pi \ln(1 - 2 \alpha \cos x + \alpha^{2}) $d$x \quad \forall |\alpha| < 1\\ \int_0^\pi \ln(1 - 2 \alpha \cos x + \alpha^{2}) $d$x \quad \forall |\alpha| = 1\\ \int_0^\pi \ln(1 - 2 \alpha \cos x + \alpha^{2}) $d$x \quad \forall |\alpha| > 1 \\ \int_0^\pi \frac{\text{d}x}{(5 - 3 \cos x)^{3}} \\ \int^b_a \dfrac{dx}{\sqrt{(e^x-e^a)(e^b-e^x)}} \quad 0<a<b \\ \int\frac{1 + a x^{t}}{x+bx^{2t+1}} $d$x \quad \forall \, t \in \mathbb{N} \\ \int \frac{\cos{2x}}{\cot^4{x}}$d$x\\ \int \frac{x^4+1}{x^{2}\sqrt{x^4-1}}$d$x \\ \int \frac{\sqrt{1-x^2}}{2x^4-x^2} $d$x \\ \int \frac{\sqrt{1+x^2}}{x} $d$x \\ \int_0^a \frac{\log(1+ax)}{1+x^2} $d$x$

$$\noindent$\int \frac{\csc^2{x} - 2005}{\cos^{2005}{x}} $d$x \\ \int \left(\frac{\tan^{-1}{x}}{x-\tan^{-1}{x}}\right)^2 $d$x \\ \int \frac{(\sin^3 \theta-\cos^3 \theta-\cos^2 \theta)(\sin \theta+\cos \theta+\cos^2 \theta)^{2010}}{(\sin \theta \cdot \cos \theta)^{2012}}$d$\theta$

a few of these are untouched.

juantheron · Aug 12, 2016

Re: MX2 2016 Integration Marathon

$I = \int\frac{(\sin^3 \theta-\cos^3 \theta-\cos^2 \theta)(\sin \theta+\cos \theta+\cos^2 \theta)^{2010}}{(\sin \theta \cdot \cos \theta)^{2012}}d\theta$

$I = \int \left(\tan \theta \cdot \sec \theta -\cot \theta \cdot \csc \theta -\csc^2 \theta\right)\cdot (\sec \theta+\csc \theta+\cot \theta)^{2010}d\theta$

$Now put $(\sec \theta+\csc \theta+\cot \theta) = t\;,$

$Then $(\sec \theta \tan \theta-\cot \theta \csc \theta-\csc^2 \theta)d\theta = dt$

$So $I = \int t^{2010}dt = \frac{t^{2011}}{2011}+\mathcal{C} = \frac{(\sec \theta+\csc \theta+\cot \theta)^{2011}}{2011}+\mathcal{C}$

juantheron · Aug 12, 2016

Re: MX2 2016 Integration Marathon

$Evaluation of $\int_{-\infty}^{\infty}\frac{1}{(x^2+ax+a^2)(x^2+bx+b^2)}dx\;,$ Given $a,b>0$

Paradoxica · Aug 12, 2016

Re: MX2 2016 Integration Marathon

juantheron said:
$Evaluation of $\int_{-\infty}^{\infty}\frac{1}{(x^2+ax+a^2)(x^2+bx+b^2)}dx\;,$ Given $a,b>0$

It can be shown the partial fractional decomposition of the integrand is

$\frac{x-b}{(a-b)(a^2+ab+b^2)(x^2+ax+a^2)} - \frac{x-a}{(a-b)(a^2+ab+b^2)(x^2+bx+b^2)}$

The rest is trivial.

Paradoxica · Aug 17, 2016

Re: MX2 2016 Integration Marathon

$$\noindent$ \frac{\sin{(\alpha-\beta)}\sin{(\beta-\gamma)}\sin{(\gamma-\alpha)}}{\sin{(x-\alpha)}\sin{(x-\beta)}\sin{(x-\gamma)}} \equiv \frac{A}{\sin{(x-\alpha)}}+\frac{B}{\sin{(x-\beta)}}+\frac{C}{\sin{(x-\gamma)}} \\ $Given the above is an identity, solve for $A,B,C. \\ $Hence, evaluate:$ \int \frac{\sin{(\alpha-\beta)}\sin{(\beta-\gamma)}\sin{(\gamma-\alpha)}}{\sin{(x-\alpha)}\sin{(x-\beta)}\sin{(x-\gamma)}} \text{d}x$

InteGrand · Sep 2, 2016

Re: MX2 2016 Integration Marathon

$\noindent Let $a,b\in \mathbb{R}$ such that $a+b\sin x$ is always non-zero for $x\in \mathbb{R}$. Find $\int _{0}^{2\pi} \frac{1}{a+b\sin x}\, \mathrm{d}x$.$

Paradoxica · Sep 2, 2016

Re: MX2 2016 Integration Marathon

InteGrand said:
$\noindent Let $a,b\in \mathbb{R}$ such that $a+b\sin x$ is always non-zero for $x\in \mathbb{R}$. Find $\int _{0}^{2\pi} \frac{1}{a+b\sin x}\, \mathrm{d}x$.$

The integral covers a full period so it is equivalently

$\int_0^{2\pi} \frac{\text{d}x}{a-b\sin{x}} = \int_0^{2\pi} \frac{a\, \text{d}x}{a^2 - b^2 \sin^2{x}}$

The improvised integrand has period π, and symmetry around π/2, so that reduces it into

$4a \int_0^{\frac{\pi}{2}} \frac{\text{d}x}{a^2 - b^2\sin^2{x}}$

By inspection, the substitution t = tanx is sufficient to complete the integral, and the rest is entirely trivial.

juantheron · Sep 21, 2016

Re: MX2 2016 Integration Marathon

$(1) Evaluation of $\int\frac{1}{\sec x+\csc x}dx$

$(2) Evaluation of $\int^{\frac{\pi}{2}}_{0}\frac{\cos^6 x}{\sin x+\cos x}dx$

$(3) Evaluation of $\int^{\frac{\pi}{4}}_{0}\frac{\sin^2 x\cos^2 x}{\sin^3 x+\cos^3 x}dx$

KingOfActing · Sep 24, 2016

Re: MX2 2016 Integration Marathon

juantheron said:
$(1) Evaluation of $\int\frac{1}{\sec x+\csc x}dx$

$\begin{align*}\int \frac{1}{\sec{x} + \csc{x}}\,dx &= \int\frac{\cos{x}\sin{x}}{\cos{x} + \sin{x}}\,dx\\&=\frac12\int \cos{x} + \sin{x} - \frac{1}{\cos{x} + \sin{x}}\,dx\\&=\frac12\sin{x}-\frac12\cos{x} -\frac12\int \frac{2}{1+t^2}\times\frac{1}{\frac{2t}{1+t^2} + \frac{1-t^2}{1+t^2}}\,dt \;\;\;\;\;\;\;\fbox{t = tan(x/2)}\\&= \frac12\sin{x}-\frac12\cos{x} +\int \frac{1}{(t-1)^2-2}\,dt \\&=\frac12\sin{x}-\frac12\cos{x}+\frac{1}{4}\ln \left\vert \frac{t-1-\sqrt{2}}{t-1+\sqrt{2}}\right\vert + C \\&=\frac12\sin{x}-\frac12\cos{x}+\frac{1}{4}\ln \left\vert \frac{\tan{\frac{x}{2}}-1-\sqrt{2}}{\tan{\frac{x}{2}}-1+\sqrt{2}}\right\vert + C \end{align*}$

Paradoxica · Sep 24, 2016

Re: MX2 2016 Integration Marathon

juantheron said:
$(1) Evaluation of $\int\frac{1}{\sec x+\csc x} \text{d}x$

$(2) Evaluation of $\int^{\frac{\pi}{2}}_{0}\frac{\cos^6 x}{\sin x+\cos x} \text{d}x$

$(3) Evaluation of $\int^{\frac{\pi}{4}}_{0}\frac{\sin^2 x\cos^2 x}{\sin^3 x+\cos^3 x} \text{d}x$

$$\noindent$ \int^{\frac{\pi}{2}}_{0}\frac{\cos^6 x}{\sin x+\cos x} \text{d}x = \int^{\frac{\pi}{2}}_{0}\frac{\sin^6 x}{\sin x+\cos x} \text{d}x \\ = \frac{1}{2} \int_0^\frac{\pi}{2} \frac{\cos^6{x} + \sin^6{x}}{\cos{x}+\sin{x}} \text{d}x = \frac{1}{2} \int_0^\frac{\pi}{2} \frac{1-3\cos^2{x}\sin^2{x}}{\cos{x}+\sin{x}} \text{d}x \\ = \frac{1}{8} \int_0^\frac{\pi}{2} \frac{4-3((\cos{x}+\sin{x})^2-1)^2}{\cos{x}+\sin{x}} \text{d}x \\ = \frac{1}{8} \int_0^\frac{\pi}{2} \frac{1}{\cos{x}+\sin{x}} -3(\cos{x}+\sin{x})^3 + 6(\cos{x}+\sin{x}) \text{d}x \\ = \frac{1}{8} \int_0^\frac{\pi}{2} \frac{1}{\cos{x}+\sin{x}} -3(\cos{x}+\sin{x})(1+2\sin{x}\cos{x}) + 6(\cos{x}+\sin{x}) \text{d}x$

Note that the integral posesses symmetry around x=π/4. After halving the domain of integration we flip the border to simplify.

$$\noindent$ = \frac{1}{4} \int_0^\frac{\pi}{4} \frac{1}{\cos{x}+\sin{x}} +3(\cos{x}+\sin{x})(1-2\sin{x}\cos{x}) \text{d}x \\ = \frac{1}{4} \int_0^\frac{\pi}{4} \frac{1}{\sqrt{2}\cos{x}} + 6\sqrt{2}\cos{x}\sin^2{x} \text{d}x \\ = \frac{1}{4} \left[ \frac{\log{(\frac{1+\sin{x}}{\cos{x}})}}{\sqrt{2}} + 2\sqrt{2}\sin^3{x} \right]_0^\frac{\pi}{4} = \frac{1}{8} \left(2+\sqrt{2}\log{(\sqrt{2}+1)} \right)$

Paradoxica · Sep 24, 2016

Re: MX2 2016 Integration Marathon

juantheron said:
$(1) Evaluation of $\int\frac{1}{\sec x+\csc x}\text{d}x$

$(2) Evaluation of $\int^{\frac{\pi}{2}}_{0}\frac{\cos^6 x}{\sin x+\cos x}\text{d}x$

$(3) Evaluation of $\int^{\frac{\pi}{4}}_{0}\frac{\sin^2 x\cos^2 x}{\sin^3 x+\cos^3 x}\text{d}x$

$$\noindent$ \int^\frac{\pi}{4}_0 \frac{\sin^2 x\cos^2 x \text{d}x}{\sin^3 x+\cos^3 x} \\ = \int_0^\frac{\pi}{4} \frac{(2\sin{x}\cos{x})^2(\cos{x}+\sin{x}) \text{d}x}{2(1+2\sin{x}\cos{x})(2-2\sin{x}\cos{x})} \\ u = \cos{x}-\sin{x},\,\text{d}u = -(\cos{x}+\sin{x})\text{d}x \\ 1-u^2 = 2\sin{x}\cos{x} \\ \int_0^1 \frac{(1-u^2)^2 \text{d}u}{2(2-u^2)(1+u^2)} \\ = \frac{1}{2} \int_0^1 \frac{1}{3(2-u^2)} + \frac{4}{3(1+t^2)} - 1 \, \text{d}u \\ = \frac{\pi}{6} + \frac{\log{(\sqrt{2}+1)}}{6\sqrt{2}} - \frac{1}{2}$

InteGrand · Oct 7, 2016

Re: MX2 2016 Integration Marathon

$\noindent Find $\int _{0}^{\pi} \frac{1 + 2\cos x}{5 + 4 \cos x}\, \mathrm{d}x$.$

Paradoxica · Oct 7, 2016

Re: MX2 2016 Integration Marathon

InteGrand said:
$\noindent Find $\int _{0}^{\pi} \frac{1 + 2\cos x}{5 + 4 \cos x}\, \mathrm{d}x$.$

Using the Weierstraß Substitution, the integral is transformed into a rational integral.

$$\noindent$ 2 \int_0^\infty \frac{3-t^2}{(9+t^2)(1+t^2)} \text{d}t$

Apply the inverse mapping t → 3/t to obtain the negative of the integral.

$$\noindent$ 2 \int_\infty^0 \frac{3-\frac{9}{t^2}}{(9+\frac{9}{t^2})(1+\frac{9}{t^2})} \frac{\text{-3d}t}{t^2} = 2\int_0^\infty \frac{t^2-3}{(9+t^2)(1+t^2)} \text{d}t$

Thus, the integral is 0.

InteGrand · Oct 7, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
Using the Weierstraß Substitution, the integral is transformed into a rational integral.

$$\noindent$ 2 \int_0^\infty \frac{3-t^2}{(9+t^2)(1+t^2)} \text{d}t$

Apply the inverse mapping t → 3/t to obtain the negative of the integral.

$$\noindent$ 2 \int_\infty^0 \frac{3-\frac{9}{t^2}}{(9+\frac{9}{t^2})(1+\frac{9}{t^2})} \frac{\text{-3d}t}{t^2} = 2\int_0^\infty \frac{t^2-3}{(9+t^2)(1+t^2)} \text{d}t$

Thus, the integral is 0.

Correct!

likeneverbefore · Oct 17, 2016

Re: MX2 2016 Integration Marathon

Paradoxica · Oct 17, 2016

Re: MX2 2016 Integration Marathon

likeneverbefore said:
View attachment 33516

Divide, cancel and use the substitution t = x+1/x, then split the integral.

One side is the integral representation of the inverse secant and substitute u = 1/t² for the other side to obtain a simple irrational integral.

likeneverbefore · Oct 17, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
Divide, cancel and use the substitution t = x+1/x, then split the integral.

One side is the integral representation of the inverse secant and substitute u = 1/t² for the other side to obtain a simple irrational integral.

Hi Paradoxica! thanks for the reply, do you mind uploading your worked solutions haha im still a bit stuck!

thank you very much!!!

Paradoxica · Oct 17, 2016

Re: MX2 2016 Integration Marathon

$$\noindent$ \int \frac{(x^2-1)^2 (x^2+2x+1) \sqrt{x^4+x^2+1}}{(x^2+1)^2 (x^2-1)(x^4+x^2+1)} \text{d}x = \int \frac{(x^2-1) (x^2+2x+1)}{(x^2+1)^2 \sqrt{x^4+x^2+1}} \text{d}x \\ = \int \frac{(x+\frac{1}{x}+2) (1-\frac{1}{x^2}) \text{d}x}{(x+\frac{1}{x})^2 \sqrt{x^2+2+\frac{1}{x^2} -1}} \stackrel{t = x+ \frac{1}{x}}{=} \int \frac{t+2}{t^2 \sqrt{t^2 -1}} \text{d}t \\ = \int \frac{\text{d}t}{t \sqrt{t^2-1}} + \int \frac{2}{t^2 \sqrt{t^2-1}} \text{d}t = \sec^{-1}{t} + \int \frac{2 \text{d}t}{t^3} \frac{1}{\sqrt{1-\frac{1}{t^2}}} \\ \stackrel{u = \frac{1}{t^2}}{=} \sec^{-1}{t} + \int \frac{-\text{d}u}{\sqrt{1-u}} = \sec^{-1}{t} + 2\sqrt{1-u} + \mathcal{C} \\ = \sec^{-1}{\left(\frac{x^2+1}{x} \right)} + \frac{2\sqrt{x^4+x^2+1}}{x^2+1} + \mathcal{C}$

juantheron · Oct 18, 2016

Re: MX2 2016 Integration Marathon

$Evaluation of $\int^{\infty}_{0}\lfloor ne^{-x} \rfloor dx,$ where $n\in \mathbb{N},n>1$

$and $\lfloor x \rfloor$ represent floor of $x$

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