We are currently struggling to cover the operational costs of Xtremepapers, as a result we might have to shut this website down. Please donate if we have helped you and help make a difference in other students' lives!
Click here to Donate Now (View Announcement)
u r asked to find the coordinates means the value of x and y. u found x=2. now keep it in the equation to find y. it will be ln2 +1
tangent passes through point P and through the origin. at point P the x coordinate is "p". at origin it is 0. the tangent will have same gradient at both the points. take out the gradient of P by diffrenciating the equation given.
can u post the link plx.
q7.
(tanx^n+2 tanx^n)dx substitute tanx = u
(tanx^n+2 tanx^n)dx substitute tanx = u
(u^n+2 + u^n)dx now du = sec^2x dx
so when substituting du will give
(u^n+2 + u^n)/sec^2x du
now use identity sec^2x = 1+ tan^2x .. means (1 + U^2)
substitute this value of sec^2x in the eq. above
(u^n+2 + u^n)/ (1 + U^2)du now take U^n in numenator
U^n (U^2 + 1 )/ (1 + U^2)du now cxancel (U^2 +1)
u wil get a simple eq. (U^n)du
inegrate it
u^n+1 / n+1 now convert back into original by substiuting tanx
tanx^n+1/ n+1 put the limits
tan(45)^n+1/n+1 tan45 = 1 so... 1^n+1 = 1
it wil be as follows 1/ n+1
ur welcum
iib) (tan9 + tan7) + (tan5+tan3) + 4(tan7+tan5)how about part b)ii its hard
+ tan(n+2).
sorry i dint get yuiib) (tan9 + tan7) + (tan5+tan3) + 4(tan7+tan5)
we have the integration for tan
so we will open all these according to the formula. we wrote 4(tan7+tan5) cox there are 5tan7 and 5tan5. and only one of them are used with tan9 and tan3 respectively
Do you mean 6(ii)?
yes 6 ii
For almost 10 years, the site XtremePapers has been trying very hard to serve its users.
However, we are now struggling to cover its operational costs due to unforeseen circumstances. If we helped you in any way, kindly contribute and be the part of this effort. No act of kindness, no matter how small, is ever wasted.
Click here to Donate Now