\(\displaystyle\int{\frac{{{\left.{d}{x}\right.}}}{{{1}+{x}}}},{x}\geq{0}\) by trigonometric substitution Consider the integral

Lucian Ayers

Lucian Ayers

Answered question

2022-03-31

dx1+x,x0 by trigonometric substitution
Consider the integral I=dx1+x,x[0,). A standard treatment using the substitution u=1+x directly gives the result ln(1+x)+c. Now consider doing this with the trigonometric substitution x=tanθ,θ[0,π2) then x=tan2θ,dx=2tanθsec2θ. Now, following your nose with right triangle trigonometry this leads directly to the solution I=2ln(1+x)+c.

Answer & Explanation

Boehm98wy

Boehm98wy

Beginner2022-04-01Added 18 answers

I think you've integrated the tangent incorrectly:
2tanθ,dθ=2log|secθ|+C=log{(sec2θ)}+C=log(1+tan2{θ})+C=log(1+x)+C
Edit per OP's comment: Starting from 2log{cos{θ}}=log{cos2{θ}} ,we have by Pythagoras cos2{θ}+sin2{θ}=1, and dividing by cos2θ gives
1cos2{θ}=1+sin2{θ}cos2{θ}=1+tan2{θ}=1+x
So in fact cos2{θ}=11+x

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