Rozwiązanie
Rewrite the cosecant as csc(x)=sin(x)1:
∫csc(x)dx=∫sin(x)1dx
Rewrite the sine using the double angle formula sin(x)=2sin(2x)cos(2x):
∫sin(x)1dx=∫2sin(2x)cos(2x)1dx
Multiply the numerator and denominator by sec2(2x):
∫2sin(2x)cos(2x)1dx=∫2tan(2x)sec2(2x)dx
Let u=tan(2x).
Then du=(tan(2x))′dx=2sec2(2x)dx (steps can be seen »), and we have that sec2(2x)dx=2du.
The integral becomes
∫2tan(2x)sec2(2x)dx=∫u1du
The integral of u1 is ∫u1du=ln(∣u∣):
∫u1du=ln(∣u∣)
Recall that u=tan(2x):
ln(∣u∣)=ln(∣∣tan(2x)∣∣)
Dlatego,
∫csc(x)dx=ln(∣∣tan(2x)∣∣)
Dodaj stałą całkowania:
∫csc(x)dx=ln(∣∣tan(2x)∣∣)+C
Answer: ∫csc(x)dx=ln(∣∣tan(2x)∣∣)+C