The value of _-/6^/6(+4x)/(1-(|x|+/6)),dx is equal to:

Question

The value of $\displaystyle\int_{-\pi/6}^{\pi/6}\frac{\pi+4x}{1-\sin(|x|+\pi/6)}\,dx$ is equal to:

Accepted Answer

Split the numerator into even ($\pi$) and odd ($4x$) parts. Since the denominator $1-\sin(|x|+\pi/6)$ is even in $x$, the odd part integrates to $0$ over the symmetric interval, and

$$I = 2\pi\int_0^{\pi/6}\frac{dx}{1-\sin\left(x+\frac\pi6\right)} = 2\pi\int_0^{\pi/6}\frac{1+\sin\left(x+\frac\pi6\right)}{\cos^2\left(x+\frac\pi6\right)}\,dx$$

$$= 2\pi\int_0^{\pi/6}\left[\sec^2\left(x+\tfrac\pi6\right)+\tan\left(x+\tfrac\pi6\right)\sec\left(x+\tfrac\pi6\right)\right]dx$$

$$= 2\pi\Big[\tan\left(x+\tfrac\pi6\right)+\sec\left(x+\tfrac\pi6\right)\Big]_0^{\pi/6} = 4\pi$$

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