Mathematics: Post your doubts here!

[Rutzaba]

How to Integrate Sec(x) ??

not in our course... but theres a formula for this
ln |sec x + tan x| + C

 

[TERMINATOR]

all regards to Talha Khatri

JazaakAllah Rutzaba and Talha Khatri.

 

[sitooon]

not in our course... but theres a formula for this
ln |sec x + tan x| + C

Thanks

 

[sitooon]

not in our course... but theres a formula for this
ln |sec x + tan x| + C

Can you help me with this ( Total 2 marks )
How to simplify -e^-y = 0.5e^2x - 1.5 -----> y=ln(2/(3-e^2x ))
How to draw Re z =1 on argand diagram !

 

[Rutzaba]

Can you help me with this ( Total 2 marks )
How to simplify -e^-y = 0.5e^2x - 1.5 -----> y=ln(2/(3-e^2x ))
How to draw Re z =1 on argand diagram !

-e^-y = 0.5e^2x - 1.5
2x - 1.5 <--- this whole thing is in power?

 

[Rutzaba]

http://www.mathcentre.ac.uk/resources/Engineering maths first aid kit/latexsource and diagrams/7_3.pdf

http://mathworld.wolfram.com/ComplexNumber.html

 

[ZaqZainab]

in uae? u have em evrywhere dear

where exactly should i go to get them? like an ordinary book shop does not

 

[Rutzaba]

where exactly should i go to get them? like an ordinary book shop does not

the place where u get ur copurse books
also u can buy it online

 

[Heyyy]

Yes,please aage bhi kardo...
Nahin samj ae :/

agey karun?

 

[ZaqZainab]

the place where u get ur copurse books
also u can buy it online

i got them from school
online is a long process

 

[Rutzaba]

Yes,please aage bhi kardo...
Nahin samj ae :/

let u^2 =x
that means dx/du =2u that means differentiating u^2 woth respect to u
dx= 2u du
next thing is to change limits...
if u^2 =x
when upper limit of x=p^2 then replace x with p^2
u^2 = p^2
new upper limit =p
and since o will remain o
the new limits are p and o
itna agya?

 

[sitooon]

-e^-y = 0.5e^2x - 1.5
2x - 1.5 <--- this whole thing is in power?

No only the 2x
What about complex no , q
0.5e^(2x) - 1.5

 

[Rutzaba]

No only the 2x
What about complex no , q
0.5e^(2x) - 1.5

ive seriously frgtten comp num

 

[sitooon]

ive seriously frgtten comp num

Np , try the other one

 

[sitooon]

ive seriously frgtten comp num

Np , try the other one

 

[Rutzaba]

Np , try the other one

srry brain not working atm

 

[sitooon]

srry brain not working atm

Thats fine

 

[Rutzaba]

Thats fine

srry

 

[Lostsoul]

We dont get the course books ANYWHERE in Tanzania. Have tried my best since I tried my A Levels

 

[Rutzaba]

Yes,please aage bhi kardo...
Nahin samj ae :/

let u^2 =x
that means dx/du =2u that means differentiating u^2 woth respect to u
dx= 2u du
next thing is to change limits...
if u^2 =x
when upper limit of x=p^2 then replace x with p^2
u^2 = p^2
new upper limit =p
and since o will remain o
the new limits are p and o

then u have to integrate cos of root of u^2
root will cut the sqr u will get
integral of cos u dx
now look at the third line of this solution u will find dx= 2u du
integral of cosu 2u du for the limit p and 0
integrating by parts
u= 2u dv= cos u
u'=2 v= sinu
2u sin u- integral of 2sin u
2usinu -( -2cosu)
2usinu + 2cos u... apply limits p and 0
2p sinp + 2cosp - ( o +2 cos0)
2psin p +2cosp -2(1)=1 as this is also equal to the area of the shaded region...
2p sinp= 3-2cosp
sinp= 3-2cosp / 2p