Question

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Answer 1

Answer:

Step-by-step explanation:

∫eˣ (2 + Sin2x)/Cos²x  dx

= ∫eˣ(2 + 2SinxCosx/Cos²x dx

= ∫2eˣ (1 + SinxCosx)/Cos²x dx

= 2∫eˣ (1/Cos²x +  SinxCosx/Cos²x) dx

= 2∫eˣ (Sec²x + tanx) dx

= 2∫eˣ (Tanx + Sec²x ) dx

The above matches the form ∫eˣ (f(x) + f ¹ (x)) dx which is equal to eˣf(x) + c

=> 2eˣTanx + c