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int_(0)^(pi/2) (cos^(2)x dx)/(cos^(2)x+4 sin^(2)x) |
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Answer» SOLUTION :`" LetI "= int_(0)^(pi//2) (cos^(2)x)/(cos^(2)x+4sin^(2)x)dx.....(1)` `=int_(0)^(pi//2) (cos^(2)x)/(4-3cos^(2)x)dx` `( :' sin^(2) x=1-cos^(2)x)` `=-(1)/(3)int_(0)^(pi//2) (4-3 cos^(2) x-4)/(4-3 cos^(2)x) dx` `=-(1)/(3) int_(0)^(pi//2) (1-(4)/(4-3 cos^(2)x))dx` `=-(1)/(3)[x]_(0)^(pi//2) +(4)/(3) int_(0)^(pi//2) (sec^(2)x)/(4sec^(2) x-3) dx` (Divide numeratornad denominator by `cos^(2)` x insecond integral) `=(pi)/(6)+(4)/(3) int_(0)^(pi//2) (sec^(2) x)/(4(1-tan^(2) x)-3)dx` `underset(" and" x=((pi)/(2)) rArrt=oo)underset("and" x=0 rArr t=0)("Let """ tan x=t)` `:. I= -(pi)/(6)+(4)/(3) int_(UU)^(oo) (DT)/(4(1+t^(2))-3)` `=-(pi)/(6)+(4)/(3) int_(0)^(oo) (dt)/(4t^(2)+1)` `=-(pi)/(6)+(4)/(3)xx(1)/(4) int_(0)^(oo)(dt)/(t^(2)+((1)/(2))^(2))` `=-(pi)/(6) +(1)/(3) .2 [tan^(-1) .((t)/(1))/((1)/(2))]_(0)^(oo)` `=-(pi)/(6)+(2)/(3) [tan^(-1) oo-tan^(-1) 0)` `=-(pi)/(6)+(2)/(3) ((pi)/(2)-0)=(pi)/(6)+(pi)/(3)=(pi)/(6)` |
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