** I already found the Taylor series expansion to be f(x)=x-(x^2/2)+(x^3/6)-(x^4

Question

** I already found the Taylor series expansion to be f(x)=x-(x^2/2)+(x^3/6)-(x^4/12)

Assume that the integral In(1 + sin x) / x dx has true value 0.097545301 Approximate In(l + sin x) by its Taylor series expansion with the first nonzero term only, evaluate the integral In(1 + sin x) / x dx, and determine the truncation error in the result. Approximate In(l + sin x) by its Taylor series expansion with the first three nonzero terms, evaluate the integral In(1 + sin x) / x dx, and determine the truncation error in the result.

Explanation / Answer

Let's carefully go through the algebra.

Since ln(1 + t) = t - t^2/2 + t^3/3 - t^4/4 + ..., letting t = sin x yields
ln(1 + sin x) = sin x - (1/2) (sin x)^2 + (1/3) (sin x)^3 - (1/4)(sin x)^4 + ...


Now, use sin x = x - x^3/3! + ... [we don't need any higher degree terms]:

So, ln(1 + sin x)
= (x - x^3/3! + ...) - (1/2) (x - x^3/3! + ...)^2 + (1/3) (x - x^3/3! + ...)^3 - (1/4)(x - x^3/3! + ...)^4 + ...
= (x - x^3/3! + ...) - (1/2) (x^2 - (2/3!)x^4 + ...) + (1/3) (x^3 + ...) - (1/4)(x^4 + ...) + ...
= x - (1/2) x^2 + (1/6) x^3 - (1/12) x^4 + ...

I hope this helps!

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