Answer to Question #324833 in Calculus for peaceboy

Question

Integrate :

(1/2*sin2x-cos^2x)/(sin^2x-cos^2x)

Answer 1

The integral has the form: $\int\frac{\frac12sin(2x)-cos^2x}{sin^2x-cos^2x}dx$. In order to compute the integral we will use the substitution: $z=tan(x)$. Namely, we have: $\int\frac{\frac12sin(2x)-cos^2x}{sin^2x-cos^2x}dx=\int\frac{sin(x)cos(x)-cos^2x}{sin^2x-cos^2x}dx=$

$=\int\frac{sin(x)cos(x)-cos^2x}{tan^2x-1}d(tan(x))$. Rewrite the numerator as: .$sin(x)cos(x)-cos^2x=cos^2x(tan(x)-1)=\frac{tan(x)-1}{tan^2(x)+1}$. Thus, after substitution $z=tan(x)$, we receive: $\int\frac{1}{(z+1)(z^2+1)}dz=\frac12\int(\frac{1}{z+1}+\frac{-z+1}{z^2+1})dz=\frac12\int\frac{dz}{z+1}-\frac12\int\frac{z\,dz}{z^2+1}+\frac12\int\frac{dz}{z^2+1}=\frac12ln(z+1)-\frac14ln(z^2+1)+\frac12arctan(z)+C=\frac12ln(tan(x)+1)-\frac14ln(tan^2(x)+1)+\frac{x}{2}+C$