Integrating Trigonometric Expressions Involving Sine and Cosine

When dealing with the integral of expressions involving sine and cosine, such as:

[ I int frac{dx}{sin x - acos x} ]

The process can become complex. In this article, we'll explore a method to integrate the given expression using the Weierstrass substitution.

Step-by-Step Approach Using Weierstrass Substitution

To integrate the expression ( int frac{dx}{sin x - acos x} ), we can use a substitution method involving trigonometric identities. The Weierstrass substitution is particularly useful in this context.

Substitution

Let ( t tanleft(frac{x}{2}right) ). This substitution leads to the following identities:

( sin x frac{2t}{1 t^2} ) ( cos x frac{1 - t^2}{1 t^2} ) ( dx frac{2}{1 t^2} dt )

Rewriting the Integral

Let's substitute these identities into the integral:

[ sin x - acos x frac{2t}{1 t^2} - aleft(frac{1 - t^2}{1 t^2}right) ] [ frac{2t - a(1 - t^2)}{1 t^2} ] [ frac{2t at^2 - a}{1 t^2} ]

Substituting back into the integral:

[ int frac{dx}{sin x - acos x} int frac{frac{2}{1 t^2} dt}{frac{2t at^2 - a}{1 t^2}} ] [ int frac{2}{2t at^2 - a} dt ]

Completing the Square

Complete the square for the quadratic in the denominator:

[ 2t at^2 - a aleft(t^2 frac{2}{a}t - 1right) ] [ aleft(left(t frac{1}{a}right)^2 - left(frac{1}{a}right)^2 - 1right) ]

This can be rewritten as:

[ t^2 frac{2}{a}t - frac{a^2 1}{a^2} left(t frac{1}{a}right)^2 - frac{a^2 1}{a^2} ]

Now, the integral can be tackled using the arctangent formula or a standard integral form depending on whether the expression under the square root is positive or zero.

Final Integration

The final result will depend on the values of ( a ) and ( b ). For specific values, you can further simplify and evaluate the integral.

Case Analysis

Depending on the values of ( a ) and ( b ), there are two cases to consider:

Case I: ( a^2 - b^2 - 1 > 0 )

[ I frac{1}{a - b} int frac{2 dt}{left(frac{t}{a - b}right)^2 - left(frac{sqrt{a^2 - b^2 - 1}}{a - b}right)^2} ] [ frac{2}{sqrt{a^2 - b^2 - 1}(a - b)} tan^{-1}left(frac{frac{t}{a - b}}{frac{sqrt{a^2 - b^2 - 1}}{a - b}}right) C ] [ boxed{frac{2}{sqrt{a^2 - b^2 - 1}}tan^{-1}left(frac{a - btanleft(frac{x}{2}right)}{sqrt{a^2 - b^2 - 1}}right) C} ]

Case II: ( a^2 - b^2 - 1

[ I frac{1}{a - b} int frac{2 dt}{left(frac{t}{a - b}right)^2 - left(frac{sqrt{1 - b^2 a^2}}{a - b}right)^2} ] [ frac{2}{2sqrt{1 - b^2 a^2}(a - b)} lnleft(frac{frac{t}{a - b} - frac{sqrt{1 - b^2 a^2}}{a - b}}{frac{t}{a - b} frac{sqrt{1 - b^2 a^2}}{a - b}}right) ] [ boxed{frac{1}{sqrt{1 - b^2 a^2}}lnleft(frac{a - btanleft(frac{x}{2}right) - 1 sqrt{1 - b^2 a^2}}{a - btanleft(frac{x}{2}right) 1 - sqrt{1 - b^2 a^2}}right) C} ]

In conclusion, by using the Weierstrass substitution and completing the square, we can integrate expressions involving sine and cosine with a methodical approach. The final result depends on the values of ( a ) and ( b ), and the specific case must be considered to simplify and evaluate the integral.