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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.

111+x2sinx dx\int _ { - 1 } ^ { 1 } \sqrt { 1 + x ^ { 2 } } \sin x\ d x

Solution

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Answered 1 year ago
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Note that we can rewrite the integral as

101+x2sinxdx+011+x2sinxdx\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=dxu = -x, du = -dx:

101+x2sinxdx=101+(u)2sin(u)du\int_{-1}^{0}\sqrt{1+x^2}\sin x\, dx= -\int_{1}^{0}\sqrt{1+(-u)^2}\sin(-u)\, du

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