Evaluate ∫ sin 3x cos x dx.

Question 40

HardOriginal question

Evaluate $\int \sin 3x \cos x\,dx$ .

-\frac{\cos 3x}{3}\sin x + C

Correct answerYour answer

-\frac{\cos 4x}{8} - \frac{\cos 2x}{4} + C

Correct answerYour answer

\frac{\cos 4x}{8} + \frac{\cos 2x}{4} + C

Correct answerYour answer

-\frac{\cos 4x}{4} - \frac{\cos 2x}{2} + C

Correct answerYour answer

Show the answer and explanation

CorrectNot quiteAnswer B

Correct answer: B

Product to sum: $\sin 3x\cos x = \frac{1}{2}(\sin 4x + \sin 2x)$ .
$\frac{1}{2}\left(-\frac{\cos 4x}{4} - \frac{\cos 2x}{2}\right) = -\frac{\cos 4x}{8} - \frac{\cos 2x}{4}$ , plus $C$ .

Worth noting

The answer without the $\frac{1}{2}$ of the product-to-sum formula is twice too large.

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