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Question

# Evaluate each integral using any algebraic method or trigonometric identity you think is appropriate. When necessary, use a substitution to reduce it to a standard form.$\int _ { - 1 } ^ { 1 } \sqrt { 1 + x ^ { 2 } } \sin x\ d x$

Solution

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Note that we can rewrite the integral as

$\int_{-1}^{0}\sqrt{1+x^2}\sin x\, dx + \int_{0}^{1}\sqrt{1+x^2}\sin x\, dx$

In the first term, we let $u = -x, du = -dx$:

$\int_{-1}^{0}\sqrt{1+x^2}\sin x\, dx= -\int_{1}^{0}\sqrt{1+(-u)^2}\sin(-u)\, du$

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