sir, where is my mistake .please tell me

put secx=t

secxtanx dx=dt

dx=dt/secxtanx

now

root(secx-1)dt /secxtanx

root(secx-1)dt /secxroot(sec^2x-1)

root(secx-1)dt /secx[root(sec^x-1)(secx+1)]

root (secx-1) cancels then integral becomes

dt

now u can easily integrate it

put t-1=u^2

 

intg.  2du/u^2+1

2tan-1u    [tan inverse]

2tan-1(root(t-1))

2tan-1(root(secx-1)) ans.

any problems nudge me or rate me if useful

sorry that is 1/t

since it is 1^inf. form we use property

e^lim(t->0) {[a^t+1 - b^t+1]/(t+1)(a-b) - 1}*1/t

take LCM

now differentiating it ;since it is 0/0 form we get

e^ lim t->0 { a^(t+1).t.log a -b^(t+1).t.log b - a+b}/(2t+1)(a-b)

applying limit

e^ -(a-b)/(a-b)

e^-1

1/e  ans.

sir please tell if any mistake is there

edited

W=F.d  (dot product)

now F=3x^2 i + 2y j

d=(4 i +6 j )-  (2i+3j)

d=2i+3j

therefore

W=(3x^2i+2yj).(2i+3j)

W=6x^2+6y  ans.

sin(A+B) =sinAcosB +sinB cosA

sin(A-B) =sinAcosB -sinB cosA

cos(A+B) =cosAcosB-sinAsinB

cos(A-B) =cosAcosB+sinAsinB

which question??

i think i have done this

integrating by parts

I= root (1+x^3) int. 1dx - int. [d/dx(root(1+x^3).int. 1.dx]

I=x.root (1+x^3) - 3/2int. [x^3/root(1+x^3)]

take 1+x^3=t^2

3x^2dx=2t  dt

dx=2t/3x^2 dt

putting dx

I=x.root(1+x^3) - int. xdt

I=x.root(1+x^3) - int.(t^2-1)^1/3 dt

can anyone do after this.

 

USE PROPERTY

e^logx=x

now it is   int. e^loga^x dx

int. a^x.e^x dx

int. (ae)^x dx

its integration by using int. a^x dx= a^x/loga +c

(ae)^x / log ae +c ans. rate if useful

SIMPLE;

WRITE it as

integr. (e^logx^9 -e^logx^5)/(e^logx^2 - e^ logx^4) dx      [using property alogx =logx^a]

inte. (x^9 - x^5)/(x^2-x^4)dx                       [e^logx =x]

int.  -x^5(x^4-1)/(x^2-1)dx

int.  -x^3(x^2+1)(x^2-1)/(x^2-1)dx

int. - x^5dx - int. x^3

-x^6/6  - x^4/4  +c

rate if useful

ad and subtract 1

use  x^3+1=(x+1)(x^2-x+1)

do first integral by making whole sq. and other by partial fractions

any problems nudge me

else rate if useful

edited

yes ,it will  include.

emperical formula=MAB

put x=cos2@

dx=-sin2@/2d@

and apply the property

log (root2cos@ + root sin@) . -sin2@/2 d@

now it is log2(si(pi/4+@ ).-sin2@ d@

now try to integrate  .

any problems nudge me.

multiply and divide by root(cos2x) now write seperately  - integrate seperately for first part  take sin^2x common now take cot x =z dx=dz/-cosec^2x therefore its integral is [-log(cotx+root(cot^2x-1))] +c1  now integrate second one put cosx=z dx=dz/-sinz take -1/root2 common and form quad. final integral- (-1/root 2 log(cosx+root[cos^2x-1/2]) +c2 ans.[-log(cotx+root(cot^2x-1))]+(root 2 log(cosx+root[cos^2x-1/2]) +C

NOT MORRISON BOYD I THINK.I CONSIST MANY A THINGS THAT ARE OUT OF JEE SYLLBUS.GO FOR ARIHANT PUBLICATIONS