Calculus question - Proving the formula for tan?

Question

I don't get why the integral of 1/(t^2+a) = 1/(a)^1/2 arctan(t/ (a)^1/2))+C. I proved it about a week ago in the library when there was nothing else to do but now I can't find that scrap of paper!

Answer 1

∫ 1/( t² + a ) dt

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Let √(a)·tan(x) = t

Then √(a)·sec²(x)dx = dt

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Substitute for t and dt:

→ ∫ 1/( [√(a)·tan(x)]² + a ) √(a)·sec²(x)dx

= ∫ 1/( a·tan²(x) + a ) √(a)·sec²(x)dx

Factor out 1/a and combine it with the √(a):

= ∫ (1/a) 1/( tan²(x) + 1 ) √(a)·sec²(x)dx

= 1/√(a) ∫ 1/( tan²(x) + 1 ) sec²(x)dx

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Use the Pythagorean trigonometric identity:

sin²(x) + cos²(x) = 1

sin²(x)/cos²(x) + cos²(x)/cos²(x) = 1/cos²(x)

tan²(x) + 1 = sec²(x)

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Change tangent into secant, reduce, and integrate:

→ 1/√(a) ∫ 1/( sec²(x) ) sec²(x)dx

= 1/√(a) ∫ 1 dx

= x / √(a) + C

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Solve for x in the previous substitution equation:

√(a)·tan(x) = t

x = arctan[ t/√(a) ]

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Reverse the substitution:

→ arctan[ t/√(a)] /√(a) + C