Author Topic: Nisha's epic Spesh Thread (Read 2638 times) Tweet Share

nisha · « **on:** July 22, 2012, 08:22:01 pm »

Hey guys. Since its coming to the business end of the year, I have realised my list of questions is increasing, while help time from teachers and tutors is decreasing. So here, I am. Thanks in advance guys.

Questions:
1. Antidifferentiate the following:
sin^2(x)
----------
cos^2(x)

sin^2(x)
----------
cos^4(x)

2. Determine f(x) if f'(x)=sin(4x)cos(2x) and f(pi/3)=0

3. Calculate the area of the region enclosed by the graph of y=x/(x+1)^0.5 and the line y=x/2

4. Find the area enclosed by the curves y=xtan^-1(x) and y=tan^-1(x). Give the answer to two decimal places.

Bhootnike · « **Reply #1 on:** July 22, 2012, 08:34:56 pm »

for the first few,
$sin^2 (x) = \frac{1}{2} (1-cos(2x))$
and $cos^2 (x) = \frac{1}{2} (1+cos(2x))$

if that makes it any easier for you ?

pi · « **Reply #2 on:** July 22, 2012, 08:44:30 pm »

Quote from: Bhootnike on July 22, 2012, 08:34:56 pm

for the first few,
$sin^2 (x) = \frac{1}{2} (1-cos(2x))$
and $cos^2 (x) = \frac{1}{2} (1+cos(2x))$

if that makes it any easier for you ?

Try this:

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

For the next one, similar:
$\frac{\sin^2(x)}{\cos^4(x)} = ((\frac{\sin(x)}{\cos(x)})^2)(\sec^2(x)) = \tan^2(x)\sec^2(x)$
$\int(\tan^2(x)\sec^2(x))dx$
Let $u = \tan(x)$
etc.

Bhootnike · « **Reply #3 on:** July 22, 2012, 08:50:39 pm »

ohh
i just realised that the ----- meant a fraction lol.
i thought nisha was separating her questions with the --- :p

my bad!

b^3 · « **Reply #4 on:** July 22, 2012, 08:51:58 pm »

1a.
$\begin{alignedat}{1}\int\frac{\sin^{2}(x)}{\cos^{2}(x)}dx & =\int\tan^{2}(x)dx \\ & =\int\sec^{2}(x)-1dx \\ & =\tan(x)-x+C \end{alignedat}$
1b.
$\begin{alignedat}{1}\int\frac{\sin^{2}(x)}{\cos^{4}(x)}dx & =\int\sec^{2}(x)\tan^{2}(x)dx\end{alignedat}$
Then use the substitution $u=\tan(x)$

2. Hint $\sin(4x)=2\sin(2x)\cos(2x)$ .

3. Find the intersection points of the two.
$\begin{alignedat}{1}\frac{x}{\sqrt{x+1}} & =\frac{x}{2} \\ \frac{x^{2}}{x+1} & =\frac{x^{2}}{4} \\ 4x^{2} & =x^{3}+x^{2} \\ x^{3}-3x^{2} & =0 \\ x^{2}(x-3) & =0 \\ x=0, & x=3 \end{alignedat}$
(Best to try and draw it out first)
Area will be given by the integral from $0$ to $3$ of the top one ( $\frac{x}{\sqrt{x+1}}$ ) minus the bottom one ( $\frac{x}{2}$ )

This should help you below, but don't forget to finish it off.
$\begin{alignedat}{1}f(x) & =\int\frac{x}{\sqrt{x+1}}dx \\ \mathrm{Let\:}u=x+1 & ,\mathrm{then\:}x=u-1 \\ \frac{du}{dx}=1 \\ \int\frac{x}{\sqrt{x+1}}dx & =\int\frac{u-1}{u^{\frac{1}{2}}}du \\ & =\int u^{\frac{1}{2}}-u^{\frac{-1}{2}}du \end{alignedat}$

4. Find where they intersect.
$\begin{alignedat}{1}x\tan^{-1}(x) & =\tan^{-1}(x) \\ \tan^{-1}(x)\left(x-1\right) & =0 \\ \tan^{-1}(x)=0 & or\: x-1=0 \\ x=0 & x=1 \end{alignedat}$

For $x<1$ , $x\tan^{-1}(x)<\tan^{-1}(x)$ , so $\tan^{-1}(x)$ will be the curve that is above.
Area $=\begin{alignedat}{1}\int_{0}^{1}\tan^{1}(x)-x\tan^{1}(x)dx\end{alignedat}$
Since it has the 'to 2.d.p" part, it will be a calc question, as you can't do it by hand without using non-spesh methods.

Anyway hope that helps.

EDIT: Fixed LaTeX and beaten.

nisha · « **Reply #5 on:** July 23, 2012, 10:56:54 am »

Absolute legends! Thankyou.

I have found some more::

1. Prove the following identites:

sin(theta)
------------------- = tan(theta/2)
1+cos(theta)

2. Use Euler's methods with a step size of 0.1, to solve the differential equation dy/dx=e^-x , with initial condition y=1 at x=2. When x=2.2, determine the value of y, correct to 4 decimal places.

3. Find f'(x) if f(x)=sin(x^3/(cos(x^3))

Somye · « **Reply #6 on:** July 23, 2012, 03:18:17 pm »

1. Hint: let sin(theta) = 2cos(theta/2)sin(theta/2) and cos (theta) = 2cos^2(theta/2)

2. Apply your formula, y_n+1 = y_n +hf(x_n) where n will initially equal your initial conditions (2,1)

3. Derivative of $tan(x^3)$ is $3x^2sec^2(x^3)$ using chain rule

nisha · « **Reply #7 on:** July 29, 2012, 07:16:54 pm »

1. Find the antiderivative of
(a)
In(tan(x))
------------
sin(x)cos(x)

(b)
cos(4x)cos(8x)

(c) 6(tan(2x))^2(sec(2x))^4

(d) (tan(x/5)^4(sec(x/5))^4

(e)
4x-2
----------
x^2 +9

Q2. find g(x) if g'(x)= x^2 +1/(x^2 -2x-3) and g(4)=4-In(5) and state the domain of g(x)

Q3. Find the area between the curve, the x-axis and the given lines:
y=2xcos(x)^2, x=-pi/3 and x=0

Q4. Find the volume generated when the area bounded by the curve y=sec(x), the line x=pi/4 and the x and y-axes is rotated about the x-axis.

Q5. A model for a container is formed by rotating the area under yhe curve of y=2-(x^2/6) between x=-1 and x=1 about the x-axes. Find the volume of the container.

Q6. A hemisphere bowl of radius 10cm contains water to a depth of 5cm. What s the volume of water in the bowl?

Q7. A solid sphere of radius 6cm has a cylindrical hole of radius 1cm bored through the centre. What is the volume of the remainder of the sphere?

Q8. Find the volume of a truncated cone of height 10cm, a base radius of 5c and a top radius of 2cm?
--Sorry for the list, but i really am bad at this subject and need help in filing the holes in my understanding This is the first lot of questions i have after completing Integral calculus section.
thanks.

paulsterio · « **Reply #8 on:** July 29, 2012, 07:38:44 pm »

Quote from: nisha on July 29, 2012, 07:16:54 pm

--Sorry for the list, but i really am bad at this subject and need help in filing the holes in my understanding This is the first lot of questions i have after completing Integral calculus section.
thanks.

That's good, well in that case, I'll actually walk you through the questions rather than just giving you solutions.

Q1) a.

So pretty much what we have to do now is make a substitution. This comes from the instinctive recognition of the natural logarithmic function giving us the derivative over itself (i.e. it will give us $\frac{sec^2(x)}{tan(x)}$ )

So what we can do now is let $u = ln(tan(x))$

Now, $\frac{du}{dx}= \frac{sec^2(x)}{tan(x)}=\frac{1}{sin(x)cos(x)}$

So now it just becomes:

$\int{u}{.du}$

Now work it through.

Q1) b.

Let's play around a little bit here, hmmm, 4x and 8x, why don't we just have a little fun with double angle formulae:

$cos(A+B) = cos(A)cos(B) - sin(A)sin(B)$

$cos(A-B)=cos(A)cos(B)+sin(A)sin(B)$

Oh, isn't that convenient!

$cos(A+B)+cos(A-B) = 2cos(A)cos(B)$

So, $cos(A)cos(B) = \frac{1}{2}(cos(A+B) + cos(A-B))$

Now it's just a substitution A = 8x, B = 4x

So, the integral becomes:

$\int{cos(4x)cos(8x)}{.dx}=\int{\frac{1}{2}(cos(12x) + cos(4x))}{.dx}$

And that's easy from there.

Tip - play around with the trigonometric identities!

I'll do some more later

b^3 · « **Reply #9 on:** July 29, 2012, 08:05:20 pm »

1e) Firstly split the fraction up, then integrate each part seperately. For the first part we will need to sue a $u$ substituion, for the second part we need to make it into the form of $\frac{a}{x^{2}+a^{2}}$ so that it will result in the inverse tangent function when antiderived.
$\begin{alignedat}{1}\int\frac{4x-2}{x^{2}+9}dx & =\int\frac{4x}{x^{2}+9}dx-\int\frac{2}{x^{2}+9}dx \\ \mathrm{Let\:}u=x^{2}+9 \\ \frac{du}{dx}=2x \\ \int\frac{4x-2}{x^{2}+9}dx & =\int\frac{2}{u}du-\frac{2}{3}\int\frac{3}{x^{2}+9}dx \\ & =2\ln|u|-\frac{2}{3}\tan^{-1}(\frac{x}{3}) \\ & =2\ln(x^{2}+9)-\frac{2}{3}\tan^{-1}(\frac{x}{3})\:\left(\mathrm{remove\; the\: modulus\: as\:}x^{2}+9>0\right) \end{alignedat}$

3) Try a $u$ substitution with $u=x^{2}$ . (the derivative will get rid of the $2x$ ). Remember with $u$ substitutions its all about trying to make the derivate get rid of something so that you can get a function you can integrate easily.
Then go on to find the definite integral.

4) Volume of solid of rev around the x-axis will be given by $V=\pi\int y^{2}dx$
So you will end up with $\sec^{2}(x)$ which is the antiderivative of $tan(x)$ .

5) $V=\pi\int y^{2}dx$
Make sure you have $y^{2}$ , then expand it out, then integrate and sub in the terminals.

6, 7, 8 are similar, see if you can have a go at them now.

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