How do you integrate the following by using Euler's formula, without using integration by parts? I=∫3+4cosθ(3cosθ+4)2dθ
I did integrate it by parts, by writing the 3 in the numerator as 3sin2θ+3cos2θ, and then splitting the numerator.
But can it be solved by using complex numbers and the Euler's formula?
Answer
Hint
When you have an expression with a squared denominator, you could think that the solution is of the form I=∫3+4cosθ(3cosθ+4)2 dθ=a+bsinθ+ccosθ3cosθ+4 Differentiate the rhs and identify terms. You will get very simple results.
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