0

I am trying to evaluate the following: $\int_0^\pi (5+3\cos(x))^{-3} \mathrm{d}x$ by using differentiation under the integral sign, however I cannot seem to find a suitable position to insert my '$A$' variable.

I have tried substituting the power of -3 for '$A$' however this left me with: $\int_0^\pi (5+3\cos(x))^{A}\ln(5+3\cos(x)) \mathrm{d}x$ which after attempting to solve by parts appears to be invalid as a $\sin(x)$ ends up on the bottom. I also tried inputting '$A$' in place of the $x$ in the $\cos(x)$ however it quickly became obvious to me that is probably not a valid thing to do because now I have lost the factor of $x$. Adding '$A$' in front of the function was of no help because I ended up with the same function as I started with.

Would someone be kind enough to point out where I should position my '$A$' variable in order for the equation to simplify - I'm not sure if the position must be found by trial and error or if there is some logic behind it. Many thanks!

Mark Schultz-Wu
  • 6,986
Tech
  • 78

1 Answers1

2

We can use differentiation under the integral sign to reduce it to an easier integral. Using the tangent half-angle substitution ($u = \tan(x/2)$) you can find $$\int_0^{\pi} \frac{1}{A + 3 \cos(x)} \, \mathrm{d}x = \frac{\pi}{\sqrt{A^2 - 9}}, \; \; |A| > 3.$$ Now differentiate twice with respect to $A$ and evaluate.

user320493
  • 86