Evaluating integrals dependent on a parameter:
Differentiate I with respect to the parameter within the sign an integrals taking variable of the integrand as constant. Now evaluate the integral so obtained as a function of the parameter then integrate then result of get I. Constant of integration can be computed by giving some arbitrary values to the parameter and the corresponding value of I.
The value`int_(0)^(pi//2)log(sin^(2)theta+k^(2)cos^(2)theta)d theta`, where `kge0,` is

Options

A. `pi log (1+k)+pilog2`
B. `plog(1+k)`
C. `pi log (1+k)-pi log 2`
D. `log (1+k)-log 2`

Answer

C

Explanation

Let `F(k)=int_(0)^(pi//2) In (sin^(2)theta+k^(2)cos^(2)theta) d theta`
`F'(k)=int_(0)^(pi//2)1/(sin^(2)theta+k^(2)cos^(2)theta) 2kcos^(2) theta d theta`
`=2kint_(0)^(pi//2) (cos^(2)theta)/(sin^(2) theta+k^(2)cos^(2)theta) d theta`
`=2k int_(0)^(pi//2) (d theta)/(tan^(2) theta +k^(2))`
`=2k int_(0)^(pi//2) (sec^(2)theta-tan^(2)theta)/(tan^(2) theta+k^(2)) d theta`
`=2kint_(0)^(oo) (dt)/(t^(2)+k^(2))-2kint_(0)^(pi//2) d theta +2k^(3) int_(0)^(pi//2) (d theta)/(tan^(2) theta +k^(2))`
`=2k 1/k "tan"^(-1) t/k|_(0)^(oo) -2k (pi)/2 +k^(2)F'(k)`
or `(1-k^(2))F'(k)=pi-kpi=pi(1-k)`
or `F'(k)=(pi)/(1+k)`
or `F(k)=pi log (1+k)+c`
For `k=1, F(1)=0` or `c=-pilog2`
or `F(k)=pilog(1+k)-pilog2`