Question:
$\displaystyle \int \sqrt{1 + \sin\left(\frac{x}{8}\right)} \, dx =$
$\displaystyle \int \sqrt{1 + \sin\left(\frac{x}{8}\right)} \, dx =$
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Use $1+\sin\theta$ identity to convert root into linear trig form.
Updated On: Apr 24, 2026
- $16\sin\left(\frac{x}{32}\right) - 16\cos\left(\frac{x}{32}\right) + C$
- $16\sin\left(\frac{x}{16}\right) - 16\cos\left(\frac{x}{16}\right) + C$
- $16\sin\left(\frac{x}{32}\right) + 16\cos\left(\frac{x}{32}\right) + C$
- $16\sin\left(\frac{x}{16}\right) + 16\cos\left(\frac{x}{16}\right) + C$
- $8\sin\left(\frac{x}{16}\right) - 8\cos\left(\frac{x}{16}\right) + C$
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The Correct Option is B
Solution and Explanation
Concept:
• Use identity: \[ 1 + \sin\theta = \left(\sin\frac{\theta}{2} + \cos\frac{\theta}{2}\right)^2 \]
Step 1: Apply identity
\[ \sqrt{1 + \sin\left(\frac{x}{8}\right)} = \sin\left(\frac{x}{16}\right) + \cos\left(\frac{x}{16}\right) \]
Step 2: Integrate
\[ \int \left[\sin\left(\frac{x}{16}\right) + \cos\left(\frac{x}{16}\right)\right] dx \] \[ = -16\cos\left(\frac{x}{16}\right) + 16\sin\left(\frac{x}{16}\right) + C \] Final Conclusion:
\[ = 16\sin\left(\frac{x}{16}\right) - 16\cos\left(\frac{x}{16}\right) + C \]
• Use identity: \[ 1 + \sin\theta = \left(\sin\frac{\theta}{2} + \cos\frac{\theta}{2}\right)^2 \]
Step 1: Apply identity
\[ \sqrt{1 + \sin\left(\frac{x}{8}\right)} = \sin\left(\frac{x}{16}\right) + \cos\left(\frac{x}{16}\right) \]
Step 2: Integrate
\[ \int \left[\sin\left(\frac{x}{16}\right) + \cos\left(\frac{x}{16}\right)\right] dx \] \[ = -16\cos\left(\frac{x}{16}\right) + 16\sin\left(\frac{x}{16}\right) + C \] Final Conclusion:
\[ = 16\sin\left(\frac{x}{16}\right) - 16\cos\left(\frac{x}{16}\right) + C \]
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