∫1/(sin^2x+sinx+1)
itz sin square x +sinx+1
tryin cinc couple o dayz
wa cod b substitn >>>>>>>>
39first write everything in terms of `x/2` [submultiple angles].then divide throughout by cos4x/2.then take tanx/2=t.thats is. (that is cosx/2 to the power of 4.)
1well....
consider the denominator...
sin^2x+sinx+1... break it into two complex factors...
sin^2x+sinx+1 = [sinx + (1/2 + i√3/2)][sinx + (1/2 - i√3/2)]
now,,,
I = ∫dx/[sinx + (1/2 + i√3/2)][sinx + (1/2 - i√3/2)]
do partials....
I = (1/i√3) ∫[sinx + (1/2 + i√3/2) - sinx - (1/2 - i√3/2)] dx/ [sinx + (1/2 + i√3/2)][sinx + (1/2 - i√3/2)]
I= (1/i√3) ∫dx/[sinx + (1/2 - i√3/2)] - (1/i√3) ∫dx/[sinx + (1/2 + i√3/2)]
now solvable :)
1if 0 ∫ Λ/2 sinx = k(means area of the curve from 0 to Λ/2 uia is k)
Whats the value of
a 0∫Λ/4 (1 +tanx)
62If i understood ur question correctly!
This seems to be quite easy abhishke!
0∫Π/2 sinx dx = 1 (Why?) so k=1
0∫Λ/4 (1 +tanx) dx = x+sec2x from 0 to pi/4
=pi/4+1/2 - 1 = pi/4-1/2
