Answer:
8. -⅓ cos³x + C
9. sec x + eˣ + C
Explanation:
8. ∫ sin x cos²x dx
If u = cos x, then du = -sin x dx.
∫ -u² du
-⅓ u³ + C
-⅓ cos³x + C
9. ∫ (sec x tan x + eˣ) dx
∫ sec x tan x dx + ∫ eˣ dx
These are both standard integrals:
sec x + eˣ + C
Question 8:
∫sinxcos²xdx
∫-u²du (let u = cos(x))
-∫u²du
-u(^2+1)/2+1
-cos^(2+1)(x)/2+1
-1/3cos³(x)
-1/3cos³(x) + C
Question 9:
∫(secxtanx + eˣ)dx
∫sec(x)tan(x)dx + ∫eˣdx (apply sum rule)
sec(x) + eˣ (sec(x)tan(x)dx = sec(x) and eˣdx = eˣ)
sec(x) + eˣ + C
Best of Luck!
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