First of all i would emphasize, that i'm not sure this is a valid method (the part involving the squareroot), so hopefully some would double check the answer
We want to solve
#I=int1/sqrt(cos^3(x)cos(x+a))dx#
Rewrite the integrand using the trigonmetric identity
#color(green)(cos(x+a)=cos(x)cos(a)-sin(x)sin(a))#
Therefore
#I=int1/sqrt(cos^3(x)(cos(x)cos(a)-sin(x)sin(a)))dx#
#=int1/sqrt(cos^4(x)(cos(a)-tan(x)sin(a))dx#
The next step* involves some problem,
we will extract #cos^4(x)# from the squareroot,
but the remember the term we extract,
might be negative,
but hopefully we can make up for this in the end
#I=int1/(cos^2(x)sqrt((cos(a)-tan(x)sin(a)))dx#
#=intsec^2(x)/sqrt((cos(a)-tan(x)sin(a)))dx#
Make a substitution remember #color(blue)cos(a)# and #color(blue)sin(a)# are constants
Let #u=cos(a)-sin(a)tan(x)=>(du)/dx=-sin(a)sec^2(x)#
#I=-int1/(sin(a)sqrt(u))du#
#=-csc(a)int1/sqrt(u)du#
#=-2csc(a)sqrt(u)+C#
Substitute #u=cos(a)-sin(a)tan(x)#
#I=-2csc(a)sqrt(cos(a)-sin(a)tan(x))+C#
To make up for the step*,we would rewrite the answer,
to include the squareroot, which cause the problem
Notice
#sqrt(cos(a)-sin(a)tan(x))#
#=sec^2(x)color(green)sqrt(cos^4(x)(cos(a)-tan(x)sin(a))#
Therefore
#I=-2csc(a)sec^2(x)sqrt(cos^4(x)(cos(a)-sin(a)tan(x)))+C#
#=-2csc(a)sec^2(x)sqrt(cos^3(x)cos(x+a))+C#