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Using properties evaluate the following definite integrals, evaluate the following: int_(-pi/2)^(pi/2) sin^2x dx

Answer»

SOLUTION :`sin^2(-x) = (-SINX)^2 = sin^2x`
THEREFORE `sin^2x` is an even FUNCTION
therefore `int_(-pi/2)^(pi/2) sin^2 x dx = 2int_0^(pi/2) sin^2x dx`
=`2 int_0^(pi/2) (1-cos 2x)/2 dx`
=`[x-(sin2x)/2]_0^(pi/2)`
`(pi/2 -sinpi/2) -(0-sin0/2) = pi/2`


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