Question
grade \[ \int_{0}^{\pi / 6} 9 \sec ^{2} x d x \]
Ask by Powell Powell. in the United States
Mar 11,2025
Upstudy AI Solution
Tutor-Verified Answer
Answer
The integral of \(9\sec^2(x)\) from \(0\) to \(\frac{\pi}{6}\) is \(3\sqrt{3}\).
Solution
Calculate or simplify the expression \( \int_{0}^{\pi / 6} 9 \sec^{2}(x) dx \).
Evaluate the integral by following steps:
- step0: Evaluate using formulas and rules:
\(\int_{0}^{\frac{\pi }{6}} 9\sec^{2}\left(x\right) dx\)
- step1: Evaluate the integral:
\(\int 9\sec^{2}\left(x\right) dx\)
- step2: Use properties of integrals:
\(9\times \int \sec^{2}\left(x\right) dx\)
- step3: Evaluate the integral:
\(9\tan\left(x\right)\)
- step4: Return the limits:
\(\left(9\tan\left(x\right)\right)\bigg |_{0}^{\frac{\pi }{6}}\)
- step5: Calculate the value:
\(3\sqrt{3}\)
The integral of \(9\sec^2(x)\) from \(0\) to \(\frac{\pi}{6}\) is \(3\sqrt{3}\).
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Simplify this solution The Deep Dive
To evaluate the integral \( \int_{0}^{\pi / 6} 9 \sec^2 x \, dx \), we start by recalling that the antiderivative of \( \sec^2 x \) is \( \tan x \). Therefore, we can rewrite the integral as: \[ \int 9 \sec^2 x \, dx = 9 \tan x + C \] Now, we need to evaluate this from \( 0 \) to \( \frac{\pi}{6} \): \[ \left[ 9 \tan x \right]_{0}^{\pi/6} = 9 \tan\left(\frac{\pi}{6}\right) - 9 \tan(0) \] The value of \( \tan\left(\frac{\pi}{6}\right) \) is \( \frac{1}{\sqrt{3}} \) and \( \tan(0) = 0 \). Thus, we have: \[ 9 \left(\frac{1}{\sqrt{3}}\right) - 9(0) = \frac{9}{\sqrt{3}} = 3\sqrt{3} \] So, the value of the integral is: \[ \int_{0}^{\pi / 6} 9 \sec ^{2} x \, dx = 3\sqrt{3} \]
