How do you find the definite integral for: #dx /(a cos^2x + b sin^2 x)^2# for the intervals #[0, pi]#?
1 Answer
Jun 4, 2018
Assuming
Explanation:
Let
#I=int_0^pidx/(acos^2x+bsin^2x)^2#
From symmetry of the squared functions:
#I=2int_0^(pi/2)dx/(acos^2x+bsin^2x)^2#
Multiply numerator and denominator by
#I=2int_0^(pi/2)(sec^4x)/(a+btan^2x)^2dx#
Assuming
#I=2/(ab)^(3/2)int_0^(pi/2)(b+atan^2theta)/sec^2thetad theta#
Simplify:
#I=2/(ab)^(3/2)int_0^(pi/2)(bcos^2theta+asin^2theta)d theta#
Apply the double-angle formula for
#I=1/(ab)^(3/2)int_0^(pi/2){(a+b)-(a-b)cos2theta}d theta#
Integrate directly:
#I=1/(ab)^(3/2)[(a+b)theta-1/2(a-b)sin2theta]_0^(pi/2)#
Hence:
#I=(a+b)/(ab)^(3/2)pi/2#