Question

\(\displaystyle \int \cos\!\left(\frac{7x}{9}\right)\,dx=\) ?

• A. \(k+\sin\!\left(\dfrac{7x}{9}\right)\)
• B. \(\dfrac{7}{9}+\sin\!\left(\dfrac{7x}{9}\right)+k\)
• C. \(\dfrac{9}{7}\sin\!\left(\dfrac{7x}{9}\right)+k\)
• D. \(k+\cos\!\left(\dfrac{7x}{9}\right)\)

Hint:

Differentiate to check: \(\dfrac{d}{dx}\big(\tfrac{1}{a}\sin(ax)\big)=\cos(ax)\).

Answer 1

The Correct Option is C

\(\displaystyle \int \cos(ax)\,dx=\frac{1}{a}\sin(ax)+C\). Here \(a=\dfrac{7}{9}\), hence \[ \int \cos\!\left(\tfrac{7x}{9}\right)dx=\frac{1}{7/9}\sin\!\left(\tfrac{7x}{9}\right)+k =\frac{9}{7}\sin\!\left(\tfrac{7x}{9}\right)+k. \]