integrate and explain please...
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sec(2x+3)tan(2x+3)=[1/cos(2x+3)][sin(2x+3)/cos(2x+3)]
=sin(2x+3)/cos²(2x+3)
put, cos(2x+3)=t
dt/dx=-2sin(2x+3)
sin(2x+3)dx=-dt/2
£-dt/2t²=-(-1/6t³)+c
1/6cos³(2x+3)+c
1/6sec³(2x+3)+c
=sin(2x+3)/cos²(2x+3)
put, cos(2x+3)=t
dt/dx=-2sin(2x+3)
sin(2x+3)dx=-dt/2
£-dt/2t²=-(-1/6t³)+c
1/6cos³(2x+3)+c
1/6sec³(2x+3)+c
ajayshotia:
but ur ans is also wrong
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