The value of $\int\limits_{-\pi/6}^{\pi/6}\left(\frac{\pi+4x^{11}}{1-\sin(|x|+\pi/6)}\right) dx$ is equal to:

Question

Accepted Answer

B. 1. **Split the numerator into even ($\pi$) and odd ($4x^{11}$) parts**, and pair $x$ with $-x$ over the symmetric interval — since $|x|=|-x|$, the denominator is the same for both.
$$I=\int_0^{\pi/6}\left[\frac{\pi+4x^{11}}{1-\sin(x+\pi/6)}+\frac{\pi-4x^{11}}{1-\sin(x+\pi/6)}ight]dx$$

2. **The odd terms cancel**, leaving twice the constant-numerator term.
$$I=2\pi\int_0^{\pi/6}\frac{dx}{1-\sin(x+\pi/6)}$$

3. **Rationalize** by multiplying numerator and denominator by $1+\sin	heta$, using $1-\sin^2	heta=\cos^2	heta$.
$$\frac{1}{1-\sin	heta}=\frac{1+\sin	heta}{\cos^2	heta}=\sec^2	heta+	an	heta\sec	heta$$

4. **Substitute this identity** with $	heta=x+\pi/6$.
$$I=2\pi\int_0^{\pi/6}\left[\sec^2\left(x+\frac{\pi}{6}ight)+	an\left(x+\frac{\pi}{6}ight)\sec\left(x+\frac{\pi}{6}ight)ight]dx$$

5. **Integrate directly**, since $\frac{d}{dx}(	an	heta)=\sec^2	heta$ and $\frac{d}{dx}(\sec	heta)=\sec	heta	an	heta$.
$$I=2\pi\left[	an\left(x+\frac{\pi}{6}ight)+\sec\left(x+\frac{\pi}{6}ight)ight]_0^{\pi/6}$$

6. **Evaluate at the limits** (argument becomes $\pi/3$ at $x=\pi/6$, and $\pi/6$ at $x=0$).
$$	an\frac{\pi}{3}+\sec\frac{\pi}{3}=\sqrt3+2,\qquad 	an\frac{\pi}{6}+\sec\frac{\pi}{6}=\sqrt3$$

7. **Subtract and multiply by $2\pi$.**
$$I=2\pi\left[(\sqrt3+2)-\sqrt3ight]=2\pi(2)=4\pi$$

Hence, the answer is **Option B**.

The value of ∫−π/6π/6(π+4x111−sin⁡(∣x∣+π/6))dx\int\limits_{-\pi/6}^{\pi/6}\left(\frac{\pi+4x^{11}}{1-\sin(|x|+\pi/6)}\right) dx−π/6∫π/6​(1−sin(∣x∣+π/6)π+4x11​)dx is equal to:

The value of $\int\limits_{-\pi/6}^{\pi/6}\left(\frac{\pi+4x^{11}}{1-\sin(|x|+\pi/6)}\right) dx$ is equal to: