HSC 2016 MX2 Integration Marathon (archive) (8 Viewers)

omegadot · Nov 28, 2015

Re: MX2 2016 Integration Marathon

Drsoccerball said:
My final integral is : $\int{\frac{du}{u^2\sqrt{1-2u^2}}}du$

$$\noindent Let $u = \frac{1}{\sqrt{2}} \sin \theta, du = \frac{1}{\sqrt{2}} \cos \theta \, d\theta$. So\\\begin{align*}\int \frac{du}{u^2 \sqrt{1 - 2 u^2}} &= \int \frac{\frac{1}{\sqrt{2}} \cos \theta}{\frac{1}{2} \sin^2 \theta \sqrt{1 - \sin^2 \theta}} \, d\theta\\&=\frac{2}{\sqrt{2}} \int \frac{\cos \theta}{\sin^2 \theta \cos \theta} \, d\theta\\&= \frac{2}{\sqrt{2}} \int \mbox{cosec}^2 \theta \, d\theta\\&=-\frac{2}{\sqrt{2}} \cot \theta + \mathcal{C}\end{align*}$

$$\noindent As $u = \frac{1}{\sqrt{2}} \sin \theta$, from a simple right-angled triangle we have $\cot \theta = \frac{\sqrt{1 - 2u^2}}{\sqrt{2} u}$. Thus$\\\begin{align*} \int \frac{du}{u^2 \sqrt{1 - 2 u^2}}&= - \frac{\sqrt{1 - 2u^2}}{u} + \mathcal{C}.\end{align*}$

omegadot · Nov 28, 2015

Re: MX2 2016 Integration Marathon

InteGrand said:
$\textbf{ALTERNATE METHOD}$

$Let $ \mathcal{I} = \int x^2 \cot ^{-1} x \text{ d}x$. Using IBP by integrating the $x^2$ and differentiating the $\cot ^{-1} x$ yields$

$\begin{align*} \mathcal{I} &= \frac{x^3}{3}\cot ^{-1}x -\frac{1}{3} \int x^3 \cdot \frac{-1}{1+x^2}\text{ d}x \\ &= \frac{x^3}{3} \cot ^{-1}x + \frac{1}{3} \int \frac{x^3}{1+x^2} \text{ d}x \\ &= \frac{x^3}{3}\cot ^{-1}x +\frac{1}{3}\int \left( x -\frac{x}{1+x^2}\right) \text{ d}x \\ &= \frac{x^3}{3}\cot^{-1}x +\frac{1}{3}\left(\frac{x^2}{2}-\frac{1}{2}\ln \left(1 + x^2\right) + \mathcal{K} \right) \\ &= \frac{x^3}{3}\cot ^{-1}x + \frac{x^2}{6} -\frac{1}{6} \ln \left( 1+x^2 \right) + \mathcal{C}\quad \Big{(}\text{where }\mathcal{C} =\frac{1}{3}\mathcal{K}\Big{)}. \end{align*}$

Mmmm...seem to be having a particularly bad day at the office. I should have spotted this one a mile off -- 2016ers take note.

Drsoccerball · Nov 28, 2015

Re: MX2 2016 Integration Marathon

omegadot said:
$$\noindent Let $u = \frac{1}{\sqrt{2}} \sin \theta, du = \frac{1}{\sqrt{2}} \cos \theta \, d\theta$. So\\\begin{align*}\int \frac{du}{u^2 \sqrt{1 - 2 u^2}} &= \int \frac{\frac{1}{\sqrt{2}} \cos \theta}{\frac{1}{2} \sin^2 \theta \sqrt{1 - \sin^2 \theta}} \, d\theta\\&=\frac{2}{\sqrt{2}} \int \frac{\cos \theta}{\sin^2 \theta \cos \theta} \, d\theta\\&= \frac{2}{\sqrt{2}} \int \mbox{cosec}^2 \theta \, d\theta\\&=-\frac{2}{\sqrt{2}} \cot \theta + \mathcal{C}\end{align*}$

$$\noindent As $u = \frac{1}{\sqrt{2}} \sin \theta$, from a simple right-angled triangle we have $\cot \theta = \frac{\sqrt{1 - 2u^2}}{\sqrt{2} u}$. Thus$\\\begin{align*} \int \frac{du}{u^2 \sqrt{1 - 2 u^2}}&= - \frac{\sqrt{1 - 2u^2}}{u} + \mathcal{C}.\end{align*}$

I wasn't stuck I just couldn't be bothered doing it.

Paradoxica · Nov 28, 2015

Re: MX2 2016 Integration Marathon

omegadot said:
$$\noindent \textbf{Next Question}$

$$\noindent Find $\int \frac{dx}{\cos (2x) \sin (4x)}.$

$\int \frac{dx}{\cos (2x) \sin (4x)} = \int \frac{2\sin{2x}dx}{\sin^2{4x}} = -\int \frac{d(\cos{2x})}{1-\cos^2{4x}}$

$= -\int \frac{d(\cos{2x})}{1- (2\cos^2{2x}-1)^2} = \int \frac{d \psi}{4\psi^4 - 4\psi^2} = \dots$

$= \frac{1}{4\psi} + \frac{1}{4} \ln{\sqrt{\frac{1-\psi}{1+\psi}}} + C = \frac{1}{4\cos{2x}} + \frac{1}{4} \ln{\sqrt{\frac{1-\cos{2x}}{1+\cos{2x}}}} + C$

$= \frac{1}{4\cos{2x}} + \frac{1}{4} \ln{|\tan{x}|} + C$

$\noindent Hmmmm.... that absolute looks troublesome. I do not recommend doing it this way.$

Drsoccerball · Nov 28, 2015

Re: MX2 2016 Integration Marathon

$An alternative would be to change the integral to $ I=2\int{\frac{1}{sin(2x)+sin(6x)}}dx $ and work from there$

omegadot · Nov 28, 2015

Re: MX2 2016 Integration Marathon

$$\noindent \textbf{Next Question}$

$$\noindent Evaluate $\int^{\frac{\pi}{2}}_0 \frac{a \cos x + b \sin x}{\cos x + \sin x} \, dx.$

Drsoccerball · Nov 29, 2015

Re: MX2 2016 Integration Marathon

omegadot said:
$$\noindent \textbf{Next Question}$

$$\noindent Evaluate $\int^{\frac{\pi}{2}}_0 \frac{a \cos x + b \sin x}{\cos x + \sin x} \, dx.$

$$ Let $ I=\int_0^{\frac{\pi}{2}}{\frac{acosx+bsinx}{cosx+sinx}}dx$

$I=\int_0^{\frac{\pi}{2}}{\frac{asinx+bcosx}{cosx+sinx}}$

$2I=\int_0^{\frac{\pi}{2}}{(a+b)}dx$

$I= \frac{\pi}{4}(a+b)$

omegadot · Nov 29, 2015

Re: MX2 2016 Integration Marathon

$\noindent \textbf{Follow-up Question}$

$$\noindent Find $\int \frac{a \cos x + b \sin x}{\cos x + \sin x} \, dx.$

dan964 · Nov 29, 2015

Re: MX2 2016 Integration Marathon

Drsoccerball said:
$$ Let $ I=\int_0^{\frac{\pi}{2}}{\frac{acosx+bsinx}{cosx+sinx}}dx$

$I=\int_0^{\frac{\pi}{2}}{\frac{asinx+bcosx}{cosx+sinx}}$

$2I=\int_0^{\frac{\pi}{2}}{(a+b)}dx$

$I= \frac{\pi}{4}(a+b)$

what substitution did you use for the second line

leehuan · Nov 29, 2015

Re: MX2 2016 Integration Marathon

dan964 said:
what substitution did you use for the second line

Border flip

i.e.
Definite integral rule that f(x)dx = f(a-x)dx

Note complementary angle identities

InteGrand · Nov 29, 2015

Re: MX2 2016 Integration Marathon

dan964 said:
what substitution did you use for the second line

$He used the rule $\int ^\alpha _0 f(x) \text{ d}x = \int _0 ^\alpha f\left( \alpha -x \right) \text{ d}x$.$

Drsoccerball · Nov 29, 2015

Re: MX2 2016 Integration Marathon

omegadot said:
$\noindent \textbf{Follow-up Question}$

$$\noindent Find $\int \frac{a \cos x + b \sin x}{\cos x + \sin x} \, dx.$

$$ Divide I by $ cosx $ and let $ u=tanx$

$I= \int{\frac{a+ub}{(1+u^2)(1+u)}}du$

$$ By separating into partial fractions and subbing $ u=-1,0 $ and comparing $ u^2$

$I= \int{\frac{(\frac{a-b}{2})}{1+u}+\frac{\frac{(b-a)u}{2} + \frac{(a+b)}{2}}{1+u^2}}du$

$I=\frac{a-b}{2}ln|1+u| + \frac{b-a}{4}ln|1+u^2| +\frac{a+b}{2}tan^{-1}u + \epsilon$

$I=\frac{a-b}{2}ln|1+tanx| + \frac{b-a}{4}ln|1+tan^2x| +\frac{a+b}{2}x + \epsilon$

Paradoxica · Nov 29, 2015

Re: MX2 2016 Integration Marathon

omegadot said:
$\noindent \textbf{Follow-up Question}$

$$\noindent Find $\int \frac{a \cos x + b \sin x}{\cos x + \sin x} \, dx.$

Drsoccerball said:
$$ Divide I by $ cosx $ and let $ u=tanx$

$I= \int{\frac{a+ub}{(1+u^2)(1+u)}}du$

$$ By separating into partial fractions and subbing $ u=-1,0 $ and comparing $ u^2$

$I= \int{\frac{(\frac{a-b}{2})}{1+u}+\frac{\frac{(b-a)u}{2} + \frac{(a+b)}{2}}{1+u^2}}du$

$I=\frac{a-b}{2}ln|1+u| + \frac{b-a}{4}ln|1+u^2| +\frac{a+b}{2}tan^{-1}u + \epsilon$

$I=\frac{a-b}{2}ln|1+tanx| + \frac{b-a}{4}ln|1+tan^2x| +\frac{a+b}{2}x + \epsilon$

$C'mon Doc, you can do better than that :P$

$I = \frac{1}{2}\int \frac{a\cos{x} + b\sin{x} + a\sin{x} + b\cos{x} + a\cos{x} + b\sin{x} - a\sin{x} - b\cos{x}}{\cos{x} + \sin{x}} dx$

$= \frac{1}{2} \int \frac{(a+b)(\cos{x} + \sin{x}) + (a-b)(\cos{x}-\sin{x})}{\cos{x} + \sin{x}} dx$

$=\frac{1}{2} \int (a+b) + (a-b)\frac{-\sin{x} + \cos{x}}{\cos{x} + \sin{x}} dx$

$= \frac{a+b}{2}x + \frac{a-b}{2}\ln{(\cos{x} + \sin{x})} +C$

InteGrand · Nov 29, 2015

Re: MX2 2016 Integration Marathon

omegadot said:
$\noindent \textbf{Follow-up Question}$

$$\noindent Find $\int \frac{a \cos x + b \sin x}{\cos x + \sin x} \, dx.$

$\noindent \textbf{ALTERNATE METHOD}$

$\noindent The goal is to rewrite $a \cos x + b \sin x$ in the form $A \left( \cos x + \sin x\right)+B \left( \cos x + \sin x\right)^\prime$ so that we can easily compute the integral as it will be in the form $\int \left(A + B\frac{f^\prime (x)}{f(x)} \right)\text{ d}x$, where $f(x) = \cos x + \sin x$.$

$\noindent Note that $\left( \cos x + \sin x\right)^\prime = -\sin x + \cos x$. So writing $a \cos x + b \sin x \equiv A \left(\cos x + \sin x \right)+B \left( -\sin x + \cos x \right)$, we have$

$\begin{align*}a \cos x + b \sin x &\equiv A\cos x + A\sin x- B\sin x +B \cos x \\ \Rightarrow a\cos x + b\sin x &\equiv \left(A+B \right)\cos x + \left(A-B\right)\sin x. \end{align*}$

$\noindent Equating coefficients of $\sin x$ and $\cos x$ gives us the following system:$

$\noindent $\left\{\begin{matrix} A+B=a \quad (1) & \\ A-B=b \quad (2)& \end{matrix}\right. .$$

$\noindent Adding $(1)$ and (2) gives $2A = a+b\Rightarrow A = \frac{a+b}{2}$. Subtracting (2) from (1) gives $2B = a-b\Rightarrow B = \frac{a-b}{2}$. So we have$

$\begin{align*}\int \frac{a \cos x + b \sin x}{\cos x + \sin x} \text{ d}x &= \int \left( A + B\frac{f^\prime (x)}{f(x)} \right)\text{ d}x, \quad \text{where }f(x) = \cos x + \sin x \\ &= Ax + B\ln \left|f(x) \right| + \mathcal{C} \\ &= \frac{a+b}{2} x + \frac{a-b}{2} \ln \left|\cos x + \sin x \right|+\mathcal{C} \end{align*}$

Edit: done above.

Drsoccerball · Nov 29, 2015

Re: MX2 2016 Integration Marathon

Where's my mistake D:

dan964 · Nov 29, 2015

Re: MX2 2016 Integration Marathon

Paradoxica said:
$C'mon Doc, you can do better than that :P$

$I = \frac{1}{2}\int \frac{a\cos{x} + b\sin{x} + a\sin{x} + b\cos{x} + a\cos{x} + b\sin{x} - a\sin{x} - b\cos{x}}{\cos{x} + \sin{x}} dx$

$= \frac{1}{2} \int \frac{(a+b)(\cos{x} + \sin{x}) + (a-b)(\cos{x}-\sin{x})}{\cos{x} + \sin{x}} dx$

$=\frac{1}{2} \int (a+b) + (a-b)\frac{-\sin{x} + \cos{x}}{\cos{x} + \sin{x}} dx$

$= \frac{a+b}{2}x + \frac{a-b}{2}\ln{(\cos{x} + \sin{x})} +C$

which then can be used to evaluate the integral using the limits from earlier.

InteGrand · Nov 29, 2015

Re: MX2 2016 Integration Marathon

Drsoccerball said:
Where's my mistake D:

Not necessarily any mistake, try showing your answer differs from mine and Paradoxica's by a constant, which it looks like it does from a glance. Otherwise, you may have just made a silly mistake in arithmetic.

Edit: yeah, your answer simplifies to ours quickly by using log. laws. Convert the 1 + tan² x to sec² x and use log. laws.

Drsoccerball · Nov 29, 2015

Re: MX2 2016 Integration Marathon

InteGrand said:
Not necessarily any mistake, try showing your answer differs from mine and Paradoxica's by a constant, which it looks like it does from a glance. Otherwise, you may have just made a silly mistake in arithmetic.

I like Paradoxica's method

dan964 · Nov 29, 2015

Re: MX2 2016 Integration Marathon

InteGrand said:
$He used the rule $\int ^\alpha _0 f(x) \text{ d}x = \int _0 ^\alpha f\left( \alpha -x \right) \text{ d}x$.$

I got that he used that substitution from the fact of complementary angles
I was just trying to think where the limits came from. Ah I see I forget to account for the dx as well which comes out negative which flips the limits back to what they are originally.

InteGrand · Nov 29, 2015

Re: MX2 2016 Integration Marathon

Drsoccerball said:
I like Paradoxica's method

$\noindent It's the good method for doing these ones where it's a fraction with a linear combination of $\sin x$ and $\cos x$ on both the numerator and denominator. You rewrite the numerator as a linear combination of the denominator and the denominator's derivative and after that it is easy. The hardest part is just finding the right linear combination to take, which can be done by solving a linear system of equations as I showed (and this isn't hard to do of course, perhaps tedious a little; but much less tedious than partial fractions or $t$-method for this!).$

HSC 2016 MX2 Integration Marathon (archive) (8 Viewers)

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