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 ...
Revise how to find the area above and below the x axis and the area between two curves by integrating, then evaluating from the limits of integration. Higher Maths - Applying integral calculus.
Brush up on your knowledge of using integration to calculate definite integrals for the Higher Maths exam.
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