∫dx1+x,x≥0 by trigonometric substitutionConsider 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=tan⁡θ,θ∈[0,π2) then x=tan2θ,dx=2tan⁡θsec2θ. Now, following your nose with right triangle trigonometry this leads directly to the solution I=2ln⁡(1+x)+c.

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Boehm98wy · Accepted Answer

I think you&#039;ve integrated the tangent incorrectly:∫2tan⁡θ,dθ=2log|sec⁡θ|+C=log⁡{(sec2θ)}+C=log⁡(1+tan2{θ})+C=log⁡(1+x)+CEdit per OP&#039;s comment: Starting from −2log⁡{cos⁡{θ}}=−log⁡{cos2{θ}} ,we have by Pythagoras cos2{θ}+sin2{θ}=1, and dividing by cos2θ gives1cos2{θ}=1+sin2{θ}cos2{θ}=1+tan2{θ}=1+xSo in fact cos2{θ}=11+x

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

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