Question

Integrate : 1/√(cos^3xcos(x+a)) ?

Answer 1

`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`