The correct option is
D 2tan−1√tan2x2(1+cosα)+cosα−1√2+CConsider the given integral.
I=∫√1−cosxcosα−cosxdx
I=∫√1−cosx√cosα−cosxdx
I=∫
⎷1−1−tan2x21+tan2x2
⎷cosα−1−tan2x21+tan2x2dx
Put t=tanx2
t=tanx2dt=sec2x2×12dxdx=2sec2x2dtdx=21+tan2x2dtdx=21+t2dt
Therefore,
I=∫√1−1−t21+t2√cosα−1−t21+t221+t2dt
I=232∫t(t2+1)√cosα+t2cosα−1+t2dt
I=232∫t(t2+1)√t2(cosα+1)+cosα−1dt
Let, m=√t2(1+cosα)+cosα−1
dt=√t2(1+cosα)+cosα−1t(cosα+1)dm
Therefore,
I=232∫1m2+2dm
Put the value of m, we get
I=2tan−1√t2(1+cosα)+cosα−1√2+C
Put the value of t, we get
I=2tan−1√tan2x2(1+cosα)+cosα−1√2+C
I=2tan−1√tan2x2(1+cosα)+cosα−1√2+C
Hence, this is the value of the given integral.