integration help (MX2) (1 Viewer)

Carrotsticks · Apr 21, 2012

raufmalik said:
I used to do this way....but i saw they make a U substitution in past papers

You can jump from the question to the (correct) answer with no penalty.

umm what · Apr 21, 2012

someone do this plzzz

jet · Apr 21, 2012

Just be careful writing the letter 'x' because it doesn't always look like an x (more than once I thought it was an 'n').

umm what · Apr 21, 2012

even if we dnt show working?

Carrotsticks said:
You can jump from the question to the (correct) answer with no penalty.

Carrotsticks · Apr 21, 2012

deswa1 said:
About this, in a test, how many lines would you recommend doing it in. Do you think markers wouldn't like it in just one line or is that fine?

Your markers will not like it (mine hated it), but they have no reason to penalise you.

deswa1 · Apr 21, 2012

raufmalik said:
View attachment 24973
someone do this plzzz

Which one? Do you want e,f or both?

umm what · Apr 21, 2012

both plz

deswa1 said:
Which one? Do you want e,f or both?

someth1ng · Apr 21, 2012

Can someone help me with:

Integrate: dx/(4+5sin^2(x)) and Integrate (0 to pi): sinx/(1+sinx)

jet · Apr 21, 2012

$\begin{align*}\int \sqrt{\frac{5 - x}{5 + x}} \, dx &=\int \sqrt{\frac{\left ( 5 - x \right )^2}{\left ( 5 - x \right ) \left ( 5 + x \right )}} \, dx \\ &= \int \frac{5 - x}{\sqrt{25 - x^2}} \, dx \\ &= \int \frac{5}{\sqrt{25 - x^2}} \, dx - \int \frac{x}{\sqrt{25 - x^2}} \, dx\end{align*}$

And you should be able to finish from there. My reasoning: the question told me to make the numerator rational. I know that for a square root to be rational, the argument needs to be a square. So if I multiply the top and bottom by the numerator, the square root cancels on the numerator and I can finish the integral.

$\begin{align*}\int \frac{dx}{x^2 \sqrt{4 - x^2}} \\ \text{Let } x = 2\cos\theta \text{. Then } dx = -2\sin\theta \, d\theta \\ \therefore \int \frac{dx}{x^2 \sqrt{4 - x^2}} &= \int \frac{-2\sin\theta}{4\cos^2\theta \sqrt{4 - 4\cos^2\theta}} \, d\theta \\ &= \int \frac{-\sin\theta}{4\cos^2\theta \sqrt{\sin^2\theta}} \, d\theta \\ &= \int -\frac{\sin\theta}{4\cos^2\theta\sin\theta}\, d\theta \\ &= -\frac{1}{4}\int \frac{1}{\cos^2\theta} \, d\theta \\ &= -\frac{1}{4}\int \sec^2\theta \, d\theta \\ &= -\frac{1}{4}\tan\theta + C\end{align*}$
And then you just transform back to x. To do this you reason as follows:

$x = 2 \cos\theta \text{ so } \cos\theta = \frac{x}{2} \\ \text{So we can draw a right-angled triangle with one angle } \theta \text{.} \\ \text{The adjacent has length } x \text{, the hypotenuse has length } 2 \text{.} \\ \text{This means the opposite has length } \sqrt{4 - x^2} \text{.} \\ \text{So means that } \tan\theta = \frac{\sqrt{4 - x^2}}{x} \text{ and } \int \frac{dx}{x^2 \sqrt{4 - x^2}} = -\frac{\sqrt{4 - x^2}}{4x} + C$

Carrotsticks · Apr 21, 2012

someth1ng said:
Can someone help me with:

Integrate: dx/(4+5sin^2(x)) and Integrate (0 to pi): sinx/(1+sinx)

Make your own thread, best to not hijack somebody else's.

deswa1 · Apr 21, 2012

For f, try the substituion x=2sin(theta) because that will give you the oppurtunity to get rid of the square roots when you simplify.

umm what · Apr 21, 2012

i can do the rest; but how did u get (5-x)(5+x) at the bottom>

jetblack2007 said:
$\begin{align*}\int \sqrt{\frac{5 - x}{5 + x}} \, dx &=\int \sqrt{\frac{\left ( 5 - x \right )^2}{\left ( 5 - x \right ) \left ( 5 + x \right )}} \, dx \\ &= \int \frac{5 - x}{\sqrt{25 - x^2}} \, dx \\ &= \int \frac{5}{\sqrt{25 - x^2}} \, dx - \int \frac{x}{\sqrt{25 - x^2}} \, dx\end{align*}$

And you should be able to finish from there. My reasoning: the question told me to make the numerator rational. I know that for a square root to be rational, the argument needs to be a square. So if I multiply the top and bottom by the numerator, the square root cancels on the numerator and I can finish the integral.

SpiralFlex · Apr 21, 2012

someth1ng said:
Can someone help me with:

Integrate: dx/(4+5sin^2(x)) and Integrate (0 to pi): sinx/(1+sinx)

First one use t substitute with t = tan x. Or use normal substitution. (Cbb, just typed a long latex proof)

Second one:

$\int \frac{\sin x}{1+\sin x}*\frac{1-\sin x}{1-\sin x}\; dx$

$=\int \frac{\sin x -\sin^2 x}{\cos^2 x}\;dx$

$=\int (\tan x \sec x -(\sec^2 x -1))\;dx$

$=\sec x -\tan x +x$

Do limits yourself.

Off to play team fortress!

seanieg89 · Apr 21, 2012

you lost a factor of cos(x) spiral.

deswa1 · Apr 21, 2012

raufmalik said:
i can do the rest; but how did u get (5-x)(5+x) at the bottom>

He multiplied the top and bottom by the conjugate (5-x)

someth1ng · Apr 21, 2012

@Spiral, I sort of had the idea of t sub but t=tan(theta/2) but tan(pi/2) is undefined...

nightweaver066 · Apr 21, 2012

someth1ng said:
Can someone help me with:

Integrate: dx/(4+5sin^2(x)) and Integrate (0 to pi): sinx/(1+sinx)

1. Use double angle formula to change sin^2x in terms of cos2x
Use t-substitution

2. Multiply top and bottom by (1 - sinx), split the integral and you'll end up with something nice.

Edit: lol. everyone posts too quickly.

SpiralFlex · Apr 21, 2012

seanieg89 said:
you lost a factor of cos(x) spiral.

Yeah I realised, argh. I keep doing that.

umm what · Apr 21, 2012

http://community.boredofstudies.org/showthread.php?s=&goto=lastpost&threadid=281674

mh1228cz · Apr 21, 2012

raufmalik said:
is that right???

wow, very neat writing.
the result is correct, but I think you can save some steps so that you can save some time. especially when you r having an exam.
I think you can do more excercise to get use to this type of question.

integration help (MX2) (1 Viewer)

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