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in Mathematics by (179 points)

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1 Answer

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by (63.2k points)
edited by

Solution:

We have:

I=∫cosx/1+sin2xdx

Let sinx=tanθ. This may look like a wild substitution, but the goal is to get the denominator of the fraction to 1+tan2θ=sec2θ. Note that differentiating sinx=tanθ gives cosxdx=sec2θdθ. It's also helpful that cosxdx is already present in the integrand. Substituting these in:

I=∫sec2θ/1+tan2θdθ=∫dθ = θ+C

Since sinx=tanθ, we see that θ=arctan(sinx):

I=arctan(sinx)+C

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