Roman C. answered • 02/16/17
Tutor
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Masters of Education Graduate with Mathematics Expertise
∫ 1/√(52x - 1) dx
= (1/ln 5)∫ 5x(ln 5)/[5x√((5x)2 - 1)] dx Let sec θ = 5x,sec θ tan θ dθ= 5x(ln 5) dx
= (1/ln 5)∫ (sec θ tan θ)/[(sec θ)√(sec2 θ - 1)] dθ
=(1/ln 5)∫ (tan θ)/(tan θ) dθ
=(1/ln 5)∫ dθ
= θ/(ln 5) + C
= (sec-1 5x)/(ln 5) + C
Roman C.
tutor
They are equivalent. This is because:
sec-1 u = cos-1(1/u) = π/2 - sin-1 (1/u)
This identity is justified by the relationships between the trigonometric functions.
When applying this to my answer, and absorbing π/(2 ln 5) into the constant C, you get your answer.
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02/17/17
Essie S.
When I did it, I got (-1/ln5) * arcsin(5^-x)+C . Is this the equivalent to your answer?
02/17/17