1) I=∫dxcosx=∫cosxdxcos2x=∫cosxdx1−sin2xĐặt t=sinx⇒I=∫dt1−t2=∫(12(1−t)+12(1+t))=12[ln(1+t)−ln(1−t)]=12.ln1+t1−t=12.1+sinx1−sinx
1)
I=∫dxcosx=∫cosxdxcos2x=∫cosxdx1−sin2xĐặt
t=sinx⇒I=∫dt1−t2=∫(12(1−t)+12(1+t))$=\frac{1}{2}[ln(1+t)-ln(1-t)]
+c=\frac{1}{2}.ln\frac{1+t}{1-t}
+c=\frac{1}{2}.\frac{1+sinx}{1-sinx}
+c$