HSC 2017 MX2 Integration Marathon (archive) (1 Viewer)
- Thread starter leehuan
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stupid_girl
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Re: HSC 2017 MX2 Integration Marathon
Try this one.
)
Try this one.
pikachu975
Premium Member
Re: HSC 2017 MX2 Integration Marathon
2/3 * d(lnsqrt(sin3x)) = cot3x dx
Therefore integrand becomes 2/3 * sec^2 (lnsqrt(sin3x)) * d(lnsqrt(sin3x))
Integrate:
= 2/3 * tan(lnsqrt(sin3x)) + C
d(lnsqrt(sin3x)) = 3/2 * cot3x * dxTry this one.
2/3 * d(lnsqrt(sin3x)) = cot3x dx
Therefore integrand becomes 2/3 * sec^2 (lnsqrt(sin3x)) * d(lnsqrt(sin3x))
Integrate:
= 2/3 * tan(lnsqrt(sin3x)) + C
pikachu975
Premium Member
Re: HSC 2017 MX2 Integration Marathon
 By inspection $ \frac{\left(x^2-2\right)}{\left(x^4+5x^2+4\right)} $ is the derivative of $ tan^{-1}\left(x+\frac{2}{x}\right) $ hence the integral becomes: $ \\ \\ \quad \frac{d\left(tan^{-1}\left(x+\frac{2}{x}\right)\right)}{tan^{-1}\left(x+\frac{2}{x}\right)} $ which equals $ ln\left|tan^{-1}\left(x+\frac{2}{x}\right)\right|+C $ )
 let$ \quad u = lnx , \quad e^u du = dx \\ $ Integral becomes $ \int \:\left(\frac{1}{u}+lnu\right)e^udu $ and applying reverse product rule you get \quad xln(lnx) + C. )
long time no integrals
here's a couple of good ones to try out!
\tan^{-1}\left ( x+\frac{2}{x} \right )}$ d$x \\ \\ \\ \int e^x\left ( \frac{1-x}{1+x^2} \right ) ^2$ d$x \\ \\ \\ \int \frac{\sin x - \cos x}{\sqrt{\sin(2x)}}$ d$x \\ \\ \\\int \left ( \frac{1}{\ln x}+\ln \left ( \ln x \right ) \right )$ d$x )
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Re: HSC 2017 MX2 Integration Marathon
^2 &= \frac{(1 - x)^2}{(1 + x^2)^2} = \frac{1 + x^2}{(1 + x^2)^2} - \frac{2x}{(1 + x^2)^2} = \frac{1}{1 + x^2} - \frac{2x}{(1 + x^2)^2} = f(x) + f'(x),\end{align*})
 = \frac{1}{1 + x^2}.)
^2 \, dx &= \int e^x (f(x) + f'(x)) \, dx = e^x f(x) + C = \frac{e^x}{1 + x^2} + C.\end{align*})
long time no integrals
here's a couple of good ones to try out!
Re: HSC 2017 MX2 Integration Marathon
 + 2\sin x \cos x - 1 = (\cos x + \sin x)^2 - 1.\end{align*})
^2 - 1}} \, dx.\end{align*})
 \, dx $ so$)
long time no integrals
here's a couple of good ones to try out!
pikachu975
Premium Member
Re: HSC 2017 MX2 Integration Marathon
Integrate 1/[(sinx)sqrt(sin2x)]
Integrate 1/[(sinx)sqrt(sin2x)]
dan964
what
Re: HSC 2017 MX2 Integration Marathon
2017 thread is now closed.
The new 2018 marathon has been created.
dan
2017 thread is now closed.
The new 2018 marathon has been created.
dan
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