Integrate : 1/√(cos^3xcos(x+a)) ?

1 Answer
Feb 15, 2018

#I=-2csc(a)sec^2(x)sqrt(cos^3(x)cos(x+a))+C#

Explanation:

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#