# Thread: HSC 2018 MX2 Integration Marathon

1. ## Re: HSC 2018 MX2 Integration Marathon

only requires u substitution.
$\int \frac{1}{\sqrt{x^2+x}}$

2. ## Re: HSC 2018 MX2 Integration Marathon

\begin{aligned} &\int \frac{1}{\sqrt{x^2+x}} \,dx \\ &= \int \frac{1}{\sqrt{\left(x+\frac{1}{2}\right)^2 - \frac{1}{4}}}\,dx\end{aligned}

\begin{aligned}\mathrm{Let}\,\, x + \frac{1}{2} &= \frac{1}{2} \sec \theta \\ x &= \frac{1}{2}(\sec \theta - 1) \end{aligned}

Justifying this substitution:

$x = (1/2)(\sec \theta - 1)$ has a solution for $\theta$ only when $x \geq 0$ and $x \leq -1$.

The domain of the integrand is $x > 0$ and $x < -1$.

$\theta$ does not exist only when $-1 < x < 0$. But since this range is undefined in the integrand to begin with, the substitution is valid.

So,

$dx = \frac 1 2 \sec \theta \tan \theta\, d\theta$

Substituting, we get

\begin{aligned}&\int \frac{\frac 1 2 \sec \theta \tan \theta}{\sqrt{\frac 1 4 \sec^2 \theta - \frac 1 4}} \,d\theta \\ &=\int \frac{\frac 1 2 \sec \theta \tan \theta}{\frac 1 2 \sqrt{\tan^2 \theta}} \,d\theta \\ &= \int \sec \theta \frac{ \tan \theta}{|\tan \theta|} \,d\theta \\ &= \frac{ \tan \theta}{|\tan \theta|} \int \sec \theta \,d\theta \\ &= \left(\frac{ \tan \theta}{|\tan \theta|} \right)\log |\sec \theta + \tan \theta| + C\end{aligned}

Not sure if that second last step was valid though.

(edit: oops forgot to sub back in)

And $\sec \theta = 2x +1$ so

$\cos \theta = \frac{1}{2x + 1}$

and by considering the appropriate right angled triangle, $\tan \theta = 2 \sqrt{x^2+x}$. Since that's always positive, the sign function on the left can be eliminated.

So we have

$\log|2x + 1 + 2\sqrt{x^2+x}| + C$

3. ## Re: HSC 2018 MX2 Integration Marathon

an easier way is to just use the substitution u = sqrt(x)
the integrand will be 2/sqrt(u^2+1), which is trivial to integrate (those usual log ones)

4. ## Re: HSC 2018 MX2 Integration Marathon

correct but the two tan signs are generally said to cancel each other out; although i understand it can vary from +-1. that being said if u look at example at the integral tanx/|tanx| it is a horizontal line being split up into positive and negative in pi/2 segments at a time and hence u should not be allowed to take it out theoretically (although area stays the same; definite integral get affected from my understanding)but discontinuous graphs in integration are probably beyond the scope of even ext2(when dealing with indefinite integrals btw; or i havent been taught it yet). Its easier just to cancel the tan out imo at this level.

im not completely sure but in the solution i got had the tan cancelled out

5. ## Re: HSC 2018 MX2 Integration Marathon

is anyone else seeing a bunch of extra addition symbols lodged between every character or is that just me

6. ## HSC 2018 MX2 Integration Marathon

Originally Posted by jathu123
an easier way is to just use the substitution u = sqrt(x)
the integrand will be 2/sqrt(u^2+1), which is trivial to integrate (those usual log ones)
may as well just jump ahead and multiply by $\frac{\sqrt{x} + \sqrt{x+1}}{\sqrt{x} + \sqrt{x+
1}}$

7. ## Re: HSC 2018 MX2 Integration Marathon

is anyone else seeing a bunch of extra addition symbols lodged between every character or is that just me
Yeah I'm seeing that and it's really confusing

Sent from my iPhone using Tapatalk

8. ## Re: HSC 2018 MX2 Integration Marathon

How to do this question https://imgur.com/a/jrRra

9. ## Re: HSC 2018 MX2 Integration Marathon

is it 13/15 pi u^2 ??

10. ## Re: HSC 2018 MX2 Integration Marathon

Originally Posted by CapitalSwine
How to do this question https://imgur.com/a/jrRra
We can get the volume of the solid by taking the volume of $x = y^2-2$ from 0 to 1 (rotated about the y-axis) and then subtracting the volume of the cylinder inside, which has radius and height 1.

\begin{aligned} V &= \pi \int_0^1 (y^2-2)^2\, dy - \pi(1)^2(1) \\ &= \pi\left[\frac 1 5 y^5 - \frac 4 3 y^3 + 4y\right]^1_0 - \pi \\ &= \pi \left(\frac 1 5 - \frac 4 3 + 4-1\right) \\ &= \left(\frac{28}{15} \pi \right)\mathrm{u^3}\end{aligned}

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