∫1+2sinycosydy=∫16(9x+x)(4+9+x)dx
4+9+x=t
29+x1×2xdx=1dx
21ℓn∣1+2siny∣=∫16t4dt+C
21ℓn∣1+2siny∣=41ℓn4+9+x+C
21ℓn(2siny+1)=41ℓn∣4+9+x∣+C
Substituting (256,2π)
21ℓn3=21ℓn3+CC=0
Substituting (49,α)
21ℓn(2sinα+1)=41ℓn8
ℓn(2sinα+1)=21ℓn8
ℓn(2sinα+1)=ℓn22
2sinα+1=22
2sinα=22−1