Question

\(\displaystyle \int \sin\!\left(\frac{3x}{4}\right)\,dx=\) ?

• A. \(k-\dfrac{3}{4}\cos\!\left(\dfrac{3x}{4}\right)\)
• B. \(k+\dfrac{3}{4}\cos\!\left(\dfrac{3x}{4}\right)\)
• C. \(k-\dfrac{4}{3}\cos\!\left(\dfrac{3x}{4}\right)\)
• D. \(k+\dfrac{4}{3}\cos\!\left(\dfrac{3x}{4}\right)\)

Hint:

\(\displaystyle \int \sin(ax)\,dx=-\frac{1}{a}\cos(ax)+C\).

Answer 1

The Correct Option is C

Reason. Use the chain rule in reverse (substitution).
Let \(u=\dfrac{3x}{4}\Rightarrow du=\dfrac{3}{4}dx\Rightarrow dx=\dfrac{4}{3}du\). Then \[ \int \sin\!\left(\tfrac{3x}{4}\right)\,dx=\int \sin u\cdot \frac{4}{3}\,du =\frac{4}{3}\int \sin u\,du =\frac{4}{3}(-\cos u)+k =-\frac{4}{3}\cos\!\left(\tfrac{3x}{4}\right)+k. \] Thus the antiderivative is \(k-\dfrac{4}{3}\cos\!\left(\dfrac{3x}{4}\right)\).