Calculusf(x):
I have not been to the thread since morning...kudos to master laplacian,master echibuzor,master factorial(who used logic to abuse me on the quiz thread),my main oga rhydex,master chiboy and master smurfy...for not allowing the clinic to be silent as mortuary...i respect you all.

statusnet: Integrate ^{(tan^2) x} / _tan2x dx

smurfy:

Thanks Laplacian. Thanks for this method.

Tell me, assuming you got something like dC/C = 1/2(dr/r), what would you have said about P%?

benbuks: Find the anti-derivative of (cosx +1)^-2

statusnet: Integrate ^{(tan^2) x} / _tan2x dx

benbuks: Find the anti-derivative of (cosx +1)^-2

benbuks: Find the anti-derivative of (cosx +1)^-2

smurfy:

Here goes...

tan2x = (2tanx)/(1-tanx^2(x).

So the expression [tan^2(x)]/tan2x becomes [tan^2(x) * (1 - tan^2(x)]/2tanx.

This gives 1/2[tanx - tan^3(x)].

Time to integrate.

Integral of tanx is ln(secx). [I'm gonna import this into my final answer, so take note.]

Now let's integrate tan^3(x).

tan^3(x) = tan^2(x) * tanx = [(sec^2(x) - 1] * tanx.

This finally gives sec^2(x) * tanx - tanx

Put u = tanx, then du = sec^2(x) dx.

So $udu - $tanx dx
= 1/2(u)^2 - ln(secx)
= 1/2[tan^2(x) - ln(secx)]

Therefore, 1/2($tanx - 1/2[$tan^3(x) gives

1/2ln(secx) - 1/2[tan^2(x) - ln(secx) + c

= 1/2ln(secx) + 1/2ln(secx) - 1/2tan^2(x) + c

= ln(secx) - 1/2tan^2(x) + c

Laplacian:
observe that;
cosx+1=2cos^2(x/2)
so
(cosx+1)^-1=1/2*sec^2(x/2)

y=(cosx+1)^-2=1/4*sec^4(x/2)

integration by part gives;

y=1/4*[2sec^2(x/2)tan(x/2)-
§2sec^2(x/2)tan^2(x/2)dx]

or
y=1/4*[2sec^2(x/2)tan(x/2)-
§2sec^2(x/2){sec^2(x/2)-1}dx]

or
y=1/4*[2sec^2(x/2)tan(x/2)-
8y+4tan(x/2)]

or by collecting 'y' together,
3y=1/4*[2sec^2(x/2)tan(x/2)+4tan(x/2)]
the required integration is therefore
y=1/12*[2sec^2(x/2)+4]*tan(x/2)
or
y=[6tan(x/2)+2tan^3(x/2)]/12+C

echibuzor:
I aint gonna lie, you lost me from the first line...

echibuzor:
I aint gonna lie, you lost me from the first line...

echibuzor:
I aint gonna lie, you lost me from the first line...

statusnet:

Thanks but it's ln (secx) -1/4*tan^2x + c

ameer!:
Prove dat
cosecX-1/cosecX+1=[1-tan(x/2)/1+tan(x/2)]^2. . .any1 dat cn prove dis will have a special reward.I promise.

benbuks: Find the anti-derivative of (cosx +1)^-2

smurfy:

Here goes...

The last digit can be arranged in 3 ways, since it must end with 1, 3 or 5. Each of the first two digits can be arranged in 5 ways, since each digit can be used more than once. The answer therefore is 5 * 5 * 3 = 75 ways.

Laplacian:
pls recheck ur solution

factorial1: Play with the questions below:
1. If the radius of a sphere is increased from 10cm to 10.1cm, what is the approximate increase in the surface area?

2. The height of a cylinder is 10cm and its radius is 4cm. Find the approximate increase in volume when the radius increases to 4.02cm

3. An error of 3% is made in measuring the radius of a sphere. Find the percentage error in volume?

4. The side of a square is 5cm. How much will the area of the square increase when the side expands by 0.01cm? And lastly

5. If x=tany; find dy/dx?

factorial1: Play with the questions below:
1. If the radius of a sphere is increased from 10cm to 10.1cm, what is the approximate increase in the surface area?

2. The height of a cylinder is 10cm and its radius is 4cm. Find the approximate increase in volume when the radius increases to 4.02cm

3. An error of 3% is made in measuring the radius of a sphere. Find the percentage error in volume?

4. The side of a square is 5cm. How much will the area of the square increase when the side expands by 0.01cm? And lastly

factorial1: Play with the questions below:
1. If the radius of a sphere is increased from 10cm to 10.1cm, what is the approximate increase in the surface area?

2. The height of a cylinder is 10cm and its radius is 4cm. Find the approximate increase in volume when the radius increases to 4.02cm

3. An error of 3% is made in measuring the radius of a sphere. Find the percentage error in volume?

4. The side of a square is 5cm. How much will the area of the square increase when the side expands by 0.01cm? And lastly

5. If x=tany; find dy/dx?

smurfy: How many even numbers greater than 4000 can be formed using the digits 0, 1, 2, 3, 4, 5?

Richiez: We need Generals who will help us with some very short yet very technical questions which we are going to use for the final round of the nairaland math quiz....oya jackpot, ortarico, calculusfx, rhydex, ositadima and the rest pm me if you're ready with the question. I need the on or before 4:00pm 2mao
Thanks Generals!

Ortarico:

I've been too busy these days, but I'll see what I can do before then, even if 5 Sir.

Ortarico:

Hmmmmm. . . . Let's go:

1. Let A = 4r2 be the surface area of the sphere in cm2 . Then dA = 8r. The small dr change in A when the radius changes from 10cm to 10.1cm is then given by small change in A dA (small change in r) dr r=10 = 8 10 0.1 = 8. Thus the increase in the surface area is approximately 8 cm2


2. The volume of the cylinder (in cm2 ) is given by V = r2 h where h is the height and r is the radius (in cm). Thus dV = 2rh. Hence the change in V when the radius changes dr from 4cm to 4.02cm and the height is 10cm is given by change in V dV (change in r) dr r=4 = 2 4 h 0.02 = 2 4 10 0.02 40 = . 25 Thus the volume increases by 40 25 cm3

3. For a sphere, the volume V = (4/3) pi r^3
Take log, log V = 3 log r + log[(4/3) pi]
Differentiate both sides
dV/V = 3 dr/r
If dr/r = 0.03 = 3%, the dV/V is 0.09 or 9%.
You may also find it arithmetically.
Error in V,
eV = abs(V-V0)/V0 = abs( r^3/r0^3 – 1) = abs[(1+- 0.03)^3 -1 ]
Take 1+0.03, then eV = 1.03^3 – 1 = 0.093 = 9.3%
Take 1-0.03, then eV = 1 – 0.97^3 = 0.087 = 8.7%

4. x = tan(y)
y = arctan(x)
dy/dx = 1/(x² + 1)