## Calculus: Early Transcendentals 8th Edition

$$\int \frac{sin2x}{1+cos^{4}x}dx=-arctan(cos^{2}x)+C$$
Observe that: $$sin2x=2sinxcosx={(-cos^{2}x)}'$$ Hence: $$\int \frac{sin2x}{1+cos^{4}x}dx=-\int \frac{1}{1+(cos^{2}x)^{2}}d(cos^{2}x)$$ $$=-arctan(cos^{2}x)+C$$