Question

Evaluate : \[ \int_{-\frac{\pi}{6}}^{\frac{\pi}{3}}(\sin|x|+\cos|x|)\,dx \]

Hint:

When an integral contains \(|x|\), always split the integral at \(x=0\). Replace \(|x|\) with \(-x\) for negative values and \(x\) for positive values before integrating.

Answer 1

Step 1: Understand the absolute value.
The expression contains \(|x|\). For negative values of \(x\): \[ |x|=-x \] For positive values: \[ |x|=x \] Thus the integral must be split at \(x=0\).
Step 2: Split the integral.
\[ \int_{-\pi/6}^{\pi/3}(\sin|x|+\cos|x|)dx \] \[ = \int_{-\pi/6}^{0}(\sin(-x)+\cos(-x))dx + \int_{0}^{\pi/3}(\sin x+\cos x)dx \] Step 3: Simplify using trigonometric identities.
We know \[ \sin(-x)=-\sin x \] \[ \cos(-x)=\cos x \] Thus \[ \int_{-\pi/6}^{0}(-\sin x+\cos x)dx + \int_{0}^{\pi/3}(\sin x+\cos x)dx \] Step 4: Evaluate the first integral.
\[ \int(-\sin x+\cos x)dx \] \[ =\cos x+\sin x \] Now evaluate from \(-\pi/6\) to \(0\): \[ (\cos0+\sin0)-(\cos(-\pi/6)+\sin(-\pi/6)) \] \[ (1+0)-\left(\frac{\sqrt3}{2}-\frac12\right) \] \[ 1-\frac{\sqrt3}{2}+\frac12 \] \[ =\frac{3}{2}-\frac{\sqrt3}{2} \] Step 5: Evaluate the second integral.
\[ \int(\sin x+\cos x)dx \] \[ =-\cos x+\sin x \] Evaluate from \(0\) to \(\pi/3\): \[ (-\cos\frac{\pi}{3}+\sin\frac{\pi}{3})-(-\cos0+\sin0) \] \[ \left(-\frac12+\frac{\sqrt3}{2}\right)-(-1) \] \[ =\frac12+\frac{\sqrt3}{2} \] Step 6: Add both parts.
\[ \left(\frac{3}{2}-\frac{\sqrt3}{2}\right)+\left(\frac12+\frac{\sqrt3}{2}\right) \] \[ =\frac{4}{2} \] \[ =2 \] Final Answer:
\[ \boxed{2} \]