Evaluate the following integrals : ∫(cos4x-cos2x)/(sin4x-sin2x) dx ​

Question

Evaluate the following integrals :

\(\int\)\(\frac{cos4x-cos2x}{sin4x-sin2x}\)dx

Answer 1

Assume,

sin4x – sin2x = t 

d(sin4x – sin2x) = dt 

(cos4x – cos2x)dx = dt 

Put t and dt in given equation we get 

⇒ ∫\(\frac{dt}{t}\)

= ln|t| + c.

But,

t = sin4x – sin2x 

= ln| sin4x – sin2x | + c.