Tricky trigonometric integral

Using the sum-to-product identity for cosine ,

$\cos(t) + \cos(2t) = 2\cos(\frac{3t}{2})\cos(\frac{t}{2})$

Thus,

$f(x) = 3\int_0^{x}\frac{\cos(t) + \cos(2t)}{\cos(\frac{t}{2})} dt = 6\int_0^{x} \cos(\frac{3t}{2}) dt = 4\sin(\frac{3x}{2})$ $f(\pi) = -4$

Note that $\cos2t + \cos t \; = \; 2\cos^2t + \cos t - 1 \; = \; (\cos t + 1)(2\cos t - 1) \; = \;2\cos^2\tfrac12t(2\cos t - 1)$ so that $g(t) \; =\; \frac{\cos2t + \cos t}{\cos\frac12t} \; = \; 4\cos\tfrac12t \cos t - 2\cos\tfrac12t \; = \; 2(\cos\tfrac32t + \cos\tfrac12) - 2\cos\tfrac12t \; =\; 2\cos\tfrac32t$ and hence $f(x) \; = \;3 \int_0^x g(t)\,dt \; = \; 4\sin\tfrac32x$ Thus $f(\pi) = \boxed{-4}$ .