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How do I integrate [math]\int \frac{dx}{sin^2x+tan^2x}[/math]?
Okay. And how can we integrate that?
We can rewrite the integral as [math]\int \frac{cos^2(x) dx}{sin^2(x)(1+cos^2(x))}.[/math] Let [math]u=cos(x)[/math]. Then [math]du=-sin(x) dx[/math]. We can write the integral as [math]-\int \frac{-du}{u^2(1-u^2)}=-\int\left(\frac{1}{u^2}-\frac{1}{1-u^2}\right)\frac{du}{u}[/math].