1.

Evaluate: \(\int\frac{sin2x}{1 + cos^2x}dx\)

Answer»

\(\int\frac{sin2x}{1 + cos^2x}dx\)

\(\int\frac{-dt}{t}\)

= -logt + c 

= -log(1 + cos2x) + c 

put 1 + cos2x = t2 

cosx(-sinx) dx = dt 

-sin2x dx = dt 

sin2xdx = -dt



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