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Evaluate Integral: (sinx+sinxtan^2x)/sec^2x) dx on the integral 0 to pi/3?

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  • pc-5 answered 2 years ago
The integrand can be simplified.

sinx(1+tan^2x)/sec^2x = sinx

The integral of sinx is -cosx.

So the value of the integral is -cos0 + cos(pi/3) = -1 + 1/2 = -1/2.
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Evaluate Integral: (sinx+sinxtan^2x)/sec^2x) dx on the integral 0 to pi/3?
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