The way I would normally do this problem is let u=(cosx), so du=sinxdx Then its just the Integral of u^2? No? With 1/sinx pulled outside of the integral. This is easy to integrate, but apparently is ...
Google has added a bunch of examples to the potentialAction.mathExpression-input section of the Math Solver help documentation. These examples explain how to handle derivatives, integrals, and limits ...
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