Evaluate the following integral ∫_0^(π⁄4)sin⁡2θ/(〖sin〗^4 θ+〖cos〗^4 θ) dθ

∫_0^(π⁄4)sin⁡2θ/(〖sin〗^4 θ+〖cos〗^4 θ) dθ      
=∫_0^(π⁄4)(2 sin⁡θ  cos⁡θ)/(〖sin〗^4 θ+〖cos〗^4 θ) dθ  
→ divide the numerator and denominator by  〖cos〗^4 θ    
∴I=∫_0^(π⁄4)(2 tan⁡θ 〖sec〗^2 θ)/(〖tan〗^4 θ+1) dθ  
Let:
u=〖tan〗^2 θ→du=2 tan⁡θ  〖sec〗^2⁡θ dθ,θ=0→u=0,θ=π/4→u=1
∴I=∫_0^1du/(u^2+1)=├ tan^(-1)⁡u ┤|_0^1=π/4