Evaluate: ∫π20cos2xdxcos2x+4sin2x
First approach :
∫π20cos2xdxcos2x+4(1−cos2x)
=∫π20cos2xdx4−3cos2x
=∫π2013{4−3cos2x−44−3cos2x}dx
=∫π2013{1−44−3cos2x}dx
=∫π20131dx−∫π20134sec2x4sec2x−3dx
=∫π20131dx−∫π20134sec2x4(1+tan2x)−3dx
Now I can easily put tanx=t and I get sec2xdx=dt
Second approach :
∫π20cos2xdxcos2x+4sin2x
Dividing numerator and denominator by cos2x we get :
=∫π20dx1+4tan2x
=∫π20dx(4){14+tan2x}
=∫π20dx(4){{12}2+(tanx)2}
Can we apply this formula of integral here :
∫1a2+x2dx=1atan−1xa
I tried but its not working here, I think doing some manipulation we can implement this here.. Please suggest thanks...
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