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You dont really need it in that form but if you want, you can do this: \int (\frac{1}{1-t}+\frac{1}{1+t})dt =\ln(\frac{1+t}{1-t}) =\ln[\frac{(1+t)^2}{1-t^2}] =\ln(\frac{1+t^2}{1-t^2}+\frac{2t}{1-t^2}) =\ln(\sec\Theta +\tan\Theta ) However, you cant always expect the...

let t=tan@/2 dt/d@=(1/2)sec^2(@/2)=(1/2)(1+t^2) d@=2dt/(1+t^2) \int \sec\Theta d\Theta \int (\frac{1+t^2}{1-t^2})(\frac{2dt}{1+t^2}) \int \frac{2dt}{(1-t)(1+t)} Then procede with partial fractions.

I guess you just know from experience. You should also learn the integral of cosec@ which uses the same method. If you're unsure though, you could still use the t formula as last resort

\int \sec\Theta d\Theta =\int \frac{\sec\Theta (\sec\Theta +\tan\Theta )}{\sec\Theta +\tan\Theta }d\Theta =\ln(\sec\Theta +\tan\Theta ) as the derivative of demoninator is the numberator. Alternatively, you could just use the t-formula.

let x=cos@ dx=-sin@d@ x=0, @=pi/2 x=1, @=0 \int_{0}^{1}\frac{x^7dx}{\sqrt{1-x^2}} \int_{\pi /2}^{0}\frac{\cos^7\Theta (-\sin\Theta d\Theta )}{sin\Theta } \int_{0}^{\pi /2}\cos^7\Theta d\Theta then use the reduction formula from the previous part.

can't you just go y=tx from the parametric equation and as t approaches -1, asymptote is y=-x ?

yeh ok i get it. Not sure where the question is from, it was just from a worksheet i happen to have.

The question says "resistance is proportional to velocity". This implies R=kv. If we choose downwards as positive then F=mg-kv=ma (F being total force and "a" being acceleration) rearranging the equation, v(dv/dx)=a=g-(kv/m) Isnt that right? yeh, "m" is mass by the way.

A particle is moving vertically downward in a medium which exerts a resistance to the motion whithc is proportional to the speed of the particle. The particle is released from rest at O and at time t its position is at a distance x below O and its speed is v. isnt the equation of motion...

how would you find out what happens when x goes to infinity?

you sure you can accurately sketch it using cartesian form? How do find asymptotes?

Excel and Success One aren't that great for maths. If you want some real stuff, get Coroneos.

oh right, its cause the circles are concentric. i get it

4 boys and 4 girls sit around 2 concentic circles such that there are 4 in each circle. In how many ways can they be seated if boys and girls sit around different circles. Can someone tell me whats wrong with this: If boys in inner circle, 3! ways to seat boys, 3! ways to seat girls, but...

x=\int_{0}^{t}vdt Does this apply for any function v? Also, what is the proof/explanation to this? Thanks

if its between 300 and 500, its gotta be 3 digits so you can only pick from 10 to 20 (which is 11/20)

i dont know of another method at the momment, but i guess you can go: \sin8a\sin2a =(2\sin4a\cos4a)(2\sin a\cos a) =(2\cos4a\sin a)(2\sin4a\cos a) =(\sin(4a+a)-\sin(4a-a))(\sin(4a+a)+\sin(4a-a)) =\sin^25a-\sin^23a pretty much same as sums/products

\sin^25a-\sin^23a =(\sin5a-\sin3a)(\sin5a+\sin2a) Using sums to products (which you can derive): =(2\cos4a\sin a)(2\sin4a\cos a) =(2\sin4a\cos4a)(2\sin a\cos a) =\sin8a\sin2a

P(cp,\frac{c}{p}) Q(cq,\frac{c}{q})\Rightarrow (\frac{cp}{2},\frac{2c}{p}) The coordinates of midpoint M is given by: x=\frac{cp+\frac{cp}{2}}{2}=\frac{3cp}{4} y=\frac{\frac{c}{p}+\frac{2c}{p}}{2}=\frac{3c}{2p} xy=(\frac{3cp}{4})(\frac{3c}{2p})=\frac{9c^2}{8}=d^2 which is in form of...

If w=cis(2pi/8)=cis(pi/4). Prove (1-w)(1-w^2)(1-w^3)...(1-w^7)=8