1Ani 3rd one is simple yaar...........
divide the numerator and denomenator by cos6x.........
then the question becomes.........
∫sec6/(1+tan6x)
=∫dt(1+t4)/(1+t6)
Rest should be able to do by urselves........
let t=tanx
dt=sec2xdx
1)
A ray of light is coming along the line
y = b
from the positive direction of x axis and strikes a concave mirror whose intersection with xy plane is a parabola
y2 = 4ax .
If a and b are positive, then the equation of the reflected ray is :
(a) y-2ab=\frac{2b}{b^2+1}(x-ab^2)
(b) y-2ab=\frac{2b}{b^2-1}(x-ab^2)
(c) y-2ab=\frac{-2b}{b^2-1}(x-ab^2)
(d) None\; of\; these
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34 Answers
11ani bhai
ab dhyaan se dekh
jab x ---->∞ then the other constant will be having no importance
that's why -1/4 is not taken into account
1
1u can write num as sin^2x+cos^2x
nw deno is (sin^2x)^3+(cos^2x)^3 =(sin^2x+cos^2x)(sin^4x+cos^4x+sin^2xcos^2x)
so nw itz dx/(sin^4x+cos^4x+sin^2xcos^2x)
=dx/(1-sin^2xcos^2x)
multiply n divide by sec^4x
n subsitute tanx=t
so ur guy bcums
(t^2+1)dt/[(t^2+1)^2 - t^2]
nw multiply n divide by 2 so u can write ur num as
(t^2+1-t) +(t^2+1+t)
n ur deno is (t^2+1+t)*(t^2+1-t)
so nw ur int is
dt/(t^2+1-t) +dt/(t^2+1+t)
u can do dah easily
11OHK, mani .....i think i got it:
The entire thing will tend to ∞......so x - x -1/2
and so ans is -1/2???