HSC 2015 MX2 Integration Marathon (archive) (3 Viewers)

Ekman · May 5, 2015

Re: MX2 2015 Integration Marathon

Axio said:
$\int\frac{sin8x}{sinx}dx$

$\int\frac{sin8x}{sinx}dx = \int \frac{2\sin{4x}\cos{4x}}{\sin{x}} dx = \int 8\cos{x}\cos{2x}\cos{4x} dx$
$\\=2\int \cos{x} + \cos{3x} + \cos{5x} +\cos{7x} dx = 2\left[\sin{x} + \frac{\sin{3x}}{3} + \frac{\sin{5x}}{5} + \frac{\sin{7x}}{7}\right] +c$

leehuan · May 10, 2015

Re: MX2 2015 Integration Marathon

Please help...

$\int { x\sin { \left( \ln { x } \right) } dx }$

Sy123 · May 10, 2015

Re: MX2 2015 Integration Marathon

leehuan said:
View attachment 32053
Please help...

$\int { x\sin { \left( \ln { x } \right) } dx }$

Integrate by parts, letting $u = \sin \ln x$ to get

$I = \frac{x^2}{2} \sin \ln x - \frac{1}{2} \int x \cos \ln x \ \mathrm{d}x$

Integrate by parts again, letting $u = \cos \ln x$

$I = \frac{x^2}{2} \sin \ln x - \frac{1}{2} \left(\frac{x^2}{2} \cos \ln x - \frac{1}{2} \int x \sin \ln x \ \mathrm{d}x \right)$

Notice that the integral now is the same integral originally, so let that equal I, then collect like terms:

$I = \frac{x^2}{5} (2 \sin \ln x - \cos \ln x) + c$

photastic · May 10, 2015

Re: MX2 2015 Integration Marathon

Sy123 said:
Integrate by parts, letting $u = \sin \ln x$ to get

$I = \frac{x^2}{2} \sin \ln x - \frac{1}{2} \int x \cos \ln x \ \mathrm{d}x$

Integrate by parts again, letting $u = \cos \ln x$

$I = \frac{x^2}{2} \sin \ln x - \frac{1}{2} \left(\frac{x^2}{2} \cos \ln x - \frac{1}{4} \int x \sin \ln x \ \mathrm{d}x \right)$

Notice that the integral now is the same integral originally, so let that equal I, then collect like terms:

$I = \frac{x^2}{3} (2 \sin \ln x - \cos \ln x) + c$

Shouldn't the answer be

$I = \frac{x^2}{5} (2 \sin \ln x - \cos \ln x) + c$

Sy123 · May 10, 2015

Re: MX2 2015 Integration Marathon

photastic said:
Shouldn't the answer be

$I = \frac{x^2}{5} (2 \sin \ln x - \cos \ln x) + c$

Yea whatever it might be

mrpotatoed · May 13, 2015

Re: MX2 2015 Integration Marathon

1/[1+(cosx)^2]

InteGrand · May 13, 2015

Re: MX2 2015 Integration Marathon

mrpotatoed said:
1/[1+(cosx)^2]

$$\frac{1}{1+\cos^2 x}=\frac{\sec^2 x}{\sec^2 x + 1}$ (multiplying top and bottom by $\sec^2 x$)$

$$ = \frac{\sec^2 x}{2+\tan^2 x}$ (since $\sec^2 x = \tan^2 x + 1$)$

$\Rightarrow \int \frac{1}{1+\cos^2 x} \text{ d}x=\int \frac{\sec^2 x}{2+\tan^2 x}\text{ d}x$

$=\frac{1}{\sqrt{2}}\tan^{-1} \left( \frac{\tan x}{\sqrt{2}}\right) + c$

leehuan · May 15, 2015

Re: MX2 2015 Integration Marathon

$\int { \ln { x } \sin ^{ -1 }{ x } dx }$

braintic · May 15, 2015

Re: MX2 2015 Integration Marathon

leehuan said:
$\int { \ln { x } \sin ^{ -1 }{ x } dx }$

View attachment ln.arcsin.pdf

leehuan · May 16, 2015

Re: MX2 2015 Integration Marathon

$\int { { e }^{ 3x+{ e }^{ x } }dx }$

Drsoccerball · May 16, 2015

Re: MX2 2015 Integration Marathon

leehuan said:
$\int { { e }^{ 3x+{ e }^{ x } }dx }$

let $u=e^x$ and then use IBP twice to get:
$e^{e^x}e^{2x} -2e^xe^{e^x} +2e^{e^x}$

Ekman · May 16, 2015

Re: MX2 2015 Integration Marathon

Next Question:

$\int_{0}^{\frac{\pi}{2}} \frac{1}{\sin{x}+\cos{x}} dx$

Edit: Try doing it without using t-method

Drsoccerball · May 16, 2015

Re: MX2 2015 Integration Marathon

Ekman said:
Next Question:

$\int_{0}^{\frac{\pi}{2}} \frac{1}{\sin{x}+\cos{x}} dx$

Edit: Try doing it without using t-method

Ill give others a chance before posting my solution

Ekman · May 16, 2015

Re: MX2 2015 Integration Marathon

Here's a better one:

$\int_{0}^{\pi} \frac{\sin{2015x}}{\sin{x}} dx$

leehuan · May 16, 2015

Re: MX2 2015 Integration Marathon

$\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { 1 }{ \sin { x } +\cos { x } } dx } \\ =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { 1 }{ \sqrt { 2 } \sin { \left( x+\frac { \pi }{ 4 } \right) } } dx } \\ =\frac { 1 }{ \sqrt { 2 } } \int _{ 0 }^{ \frac { \pi }{ 2 } }{ \csc { \left( x+\frac { \pi }{ 4 } \right) } dx }$

$=-\frac { 1 }{ \sqrt { 2 } } \int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { -\csc { \left( x+\frac { \pi }{ 4 } \right) } \cot { \left( x+\frac { \pi }{ 4 } \right) } -\csc ^{ 2 }{ \left( x+\frac { \pi }{ 4 } \right) } }{ \csc { \left( x+\frac { \pi }{ 4 } \right) + } \cot { \left( x+\frac { \pi }{ 4 } \right) } } dx }$

$=-\frac { 1 }{ \sqrt { 2 } } { \left[ \ln { \left| \csc { \left( x+\frac { \pi }{ 4 } \right) + } \cot { \left( x+\frac { \pi }{ 4 } \right) } \right| } \right] }_{ 0 }^{ \frac { \pi }{ 2 } }\\ =-\frac { 1 }{ \sqrt { 2 } } \ln { \left| \sqrt { 2 } -1 \right| } +\frac { 1 }{ \sqrt { 2 } } \ln { \left| \sqrt { 2 } +1 \right| }$

$=\frac { 1 }{ \sqrt { 2 } } \ln { \left( 2\sqrt { 2 } +3 \right) } $ from rationalising the denominator$$

leehuan · May 16, 2015

Re: MX2 2015 Integration Marathon

${ I }_{ n }=\int _{ 0 }^{ \pi }{ \frac { \sin { \left( nx \right) } }{ \sin { x } } dx } \quad n\in \mathbb Z$

$=\int _{ 0 }^{ \pi }{ \frac { \sin { \left( \left( n-1 \right) x \right) } \cos { x } }{ \sin { x } } dx } +\int _{ 0 }^{ \pi }{ \frac { \cos { \left( \left( n-1 \right) x \right) } \sin { x } }{ \sin { x } } dx }$

$2\sin { \left( \left( n-1 \right) x \right) } \cos { x } =\sin { \left( \left( n-1 \right) x+x \right) } +\sin { \left( \left( n-1 \right) x-x \right) } \\ \sin { \left( \left( n-1 \right) x \right) } \cos { x } =\frac { 1 }{ 2 } \left( \sin { \left( nx \right) } +\sin { \left( \left( n-2 \right) x \right) } \right)$

${ I }_{ n }=\frac { 1 }{ 2 } \int _{ 0 }^{ \pi }{ \frac { \sin { \left( \left( n-2 \right) x \right) } }{ \sin { x } } dx } +\frac { 1 }{ 2 } \int _{ 0 }^{ \pi }{ \frac { \sin { \left( nx \right) } }{ \sin { x } } dx } +\int _{ 0 }^{ \pi }{ \cos { \left( \left( n-1 \right) x \right) } dx }$

${ I }_{ n }=\frac { 1 }{ 2 } { I }_{ n-2 }+\frac { 1 }{ 2 } { I }_{ n }+{ \left[ \frac { \sin { \left( \left( n-1 \right) x \right) } }{ n-1 } \right] }_{ 0 }^{ \pi }$

$\frac { 1 }{ 2 } { I }_{ n }=\frac { 1 }{ 2 } { I }_{ n-2 }+\left( \frac { \sin { \left( \left( n-1 \right) \pi \right) } }{ n-1 } -\frac { \sin { \left( \left( n-1 \right) \left( 0 \right) \right) } }{ n-1 } \right)$

${ I }_{ n }={ I }_{ n-2 }+0\\ { I }_{ n }={ I }_{ n-2 }$

$Using this reduction formula for n=2015, 1007 times, gives us$

${ I }_{ 2015 }={ I }_{ 1 }\\ =\int _{ 0 }^{ \pi }{ \frac { \sin { x } }{ \sin { x } } dx } \\ =\int _{ 0 }^{ \pi }{ dx } \\ ={ \left[ x \right] }_{ 0 }^{ \pi }\\ =\pi$

leehuan · May 17, 2015

Re: MX2 2015 Integration Marathon

$\int { \frac { \tan { \left( 2x \right) } }{ \tan { x } } dx }$

Ekman · May 17, 2015

Re: MX2 2015 Integration Marathon

leehuan said:
$\int { \frac { \tan { \left( 2x \right) } }{ \tan { x } } dx }$

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

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

$2I= \int \frac{2 + 2(\cos^{2}(x)-\sin^{2}(x))}{\cos^{2}(x)-\sin^{2}(x)}dx$

$\int \frac{2}{\cos^{2}(x)-\sin^{2}(x)} dx = \int \frac{2\sec^{2}(x)}{1-\tan^{2}(x)}dx = ln\left(\frac{1+\tan(x)}{1-\tan(x)}\right)$

$\therefore 2I= 2x + ln\left(\frac{1+\tan(x)}{1-\tan(x)}\right) +c$

kawaiipotato · May 17, 2015

Re: MX2 2015 Integration Marathon

Ekman said:
$\int \frac{\tan(2x)}{\tan(x)} dx= \int \frac{2}{1-\tan^{2}(x)} dx = \int \frac{2\cos^{2}(x)}{\cos^{2}(x) - \sin^{2}(x)} dx$

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

$2I= \int \frac{2 + 2(\cos^{2}(x)-\sin^{2}(x))}{\cos^{2}(x)-\sin^{2}(x)}dx$

$\int \frac{2}{\cos^{2}(x)-\sin^{2}(x)} dx = \int \frac{2\sec^{2}(x)}{1-\tan^{2}(x)}dx = ln\left(\frac{1+\tan(x)}{1-\tan(x)}\right)$

$\therefore 2I= 2x + ln\left(\frac{1+\tan(x)}{1-\tan(x)}\right) +c$

How did you get those expressions for the integral in the first line ?

Ekman · May 17, 2015

Re: MX2 2015 Integration Marathon

kawaiipotato said:
How did you get those expressions for the integral in the first line ?

$\tan(2x) = \frac{2\tan(x)}{1-\tan^{2}(x)}$ And so the tan at the top will cancel with the denominator. For converting it into the sin and cos fraction, just make the tan^2x in the denominator into sin^2x/cos^2x, then just simplify

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