the original question is ∫xarcsecx*dx, and i need to simplify the expression in my equation to get my final answer, which is ½( x² arcsec x -√(x²-1)) + C
u = x^2 - 1
du = 2xdx
(1/2) ∫ 1/√u du
(1/2) * 2 u^(1/2) + c
= √(x^2 - 1) + c
Cant simplify it to 1/√(x^2 - 1) because thats not what it equals
Integral of x dx/(x^2-1)^(0.5)
u = x^2 - 1, du = 2x dx, so the integrand becomes
(1/2) du / sqrt(u)
whose integral is sqrt(u) + C
= (x^2-1)^(0.5) + C
If you already used "u" in doing the earlier parts of the problem,
then of course use "w" or "z" or something!
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u = x^2 - 1
du = 2xdx
(1/2) ∫ 1/√u du
(1/2) * 2 u^(1/2) + c
= √(x^2 - 1) + c
Cant simplify it to 1/√(x^2 - 1) because thats not what it equals
Integral of x dx/(x^2-1)^(0.5)
u = x^2 - 1, du = 2x dx, so the integrand becomes
(1/2) du / sqrt(u)
whose integral is sqrt(u) + C
= (x^2-1)^(0.5) + C
If you already used "u" in doing the earlier parts of the problem,
then of course use "w" or "z" or something!