Question

\(\displaystyle \int \sec^{2}\!\left(\frac{17x}{23}\right)\,dx=\) ?

• A. \(k+\dfrac{17}{23}\tan\!\left(\dfrac{17x}{23}\right)\)
• B. \(k-\dfrac{17}{23}\tan\!\left(\dfrac{17x}{23}\right)\)
• C. \(k+\dfrac{23}{17}\tan\!\left(\dfrac{17x}{23}\right)\)
• D. \(k-\dfrac{23}{17}\tan\!\left(\dfrac{17x}{23}\right)\)

Hint:

Because \((\tan u)'=\sec^2 u\), the antiderivative is \(\tan u\) divided by \(u'\).

Answer 1

The Correct Option is C

\(\displaystyle \int \sec^{2}(ax)\,dx=\frac{1}{a}\tan(ax)+C\). With \(a=\dfrac{17}{23}\), \[ \int \sec^{2}\!\left(\tfrac{17x}{23}\right)\!dx =\frac{1}{17/23}\tan\!\left(\tfrac{17x}{23}\right)+k =\frac{23}{17}\tan\!\left(\tfrac{17x}{23}\right)+k. \]