Evaluate the following integrals:
Assume sin2x + tanx - 5 =t
d(tanx + sin2x - 5) =dt
(2cos2x + sec2x)dx = dt
Put t and dt in given equation we get
⇒ 
= 
But t = sin2x + tanx - 5
= ln|sin2x + tanx - 5| + c.
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