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junior college 2 | H2 Maths
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Date Posted: 3 years ago
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1a)
f(x) = (1 + sinx)/cosx = 1/cosx + sinx/cosx
= secx + tanx
f'(x) = secx tanx + sec²x = secx (secx + tanx)
= secx f(x)
So, secx = f'(x)/f(x)
∫ secx dx = ∫ f'(x)/f(x) dx
= ln ∣ f(x) ∣ + c
= ln ∣ secx + tanx ∣ + c
1b)
Recall that d/dx tan-¹(f(x)) = f'(x) / (1 + f(x)²)
So, we try to rewrite the expression into that manner.
∫ 4x³/(a⁸ + x⁸) dx
= ∫ 4x³/(a⁸(1 + x⁸/a⁸)) dx
= ∫ (4x³/a⁸)/(1 + (x⁴/a⁴)²) dx
= 1/a⁴ ∫ (4x³/a⁴)/(1 + (x⁴/a⁴)²) dx
= 1/a⁴ tan-¹(x⁴/a⁴) + c
(or tan-¹(x⁴/a⁴) / a⁴ + c )
Here your f(x) = x⁴/a⁴ (or 1/a⁴ x⁴) , f'(x) = 4x³/a⁴ ( or 1/a⁴ (4x³) )
f(x) = (1 + sinx)/cosx = 1/cosx + sinx/cosx
= secx + tanx
f'(x) = secx tanx + sec²x = secx (secx + tanx)
= secx f(x)
So, secx = f'(x)/f(x)
∫ secx dx = ∫ f'(x)/f(x) dx
= ln ∣ f(x) ∣ + c
= ln ∣ secx + tanx ∣ + c
1b)
Recall that d/dx tan-¹(f(x)) = f'(x) / (1 + f(x)²)
So, we try to rewrite the expression into that manner.
∫ 4x³/(a⁸ + x⁸) dx
= ∫ 4x³/(a⁸(1 + x⁸/a⁸)) dx
= ∫ (4x³/a⁸)/(1 + (x⁴/a⁴)²) dx
= 1/a⁴ ∫ (4x³/a⁴)/(1 + (x⁴/a⁴)²) dx
= 1/a⁴ tan-¹(x⁴/a⁴) + c
(or tan-¹(x⁴/a⁴) / a⁴ + c )
Here your f(x) = x⁴/a⁴ (or 1/a⁴ x⁴) , f'(x) = 4x³/a⁴ ( or 1/a⁴ (4x³) )
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