Evaluate ∫ sin xcos x dx.

Indefinite Integrals · Question 29 of 90

Question 29

MediumOriginal question

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

\frac{\sin^{2} x}{2} + C

Correct answerYour answer

\sin^{2} x + C

Correct answerYour answer

\frac{\cos^{2} x}{2} + C

Correct answerYour answer

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

Correct answerYour answer

Show the answer and explanation

CorrectNot quiteAnswer A

Correct answer: A

Put $t = \sin x$ , $dt = \cos x\,dx$ : $\int t\,dt = \frac{t^{2}}{2}$ .
$= \frac{\sin^{2} x}{2} + C$ . (Through $\sin x\cos x = \frac{1}{2}\sin 2x$ one gets $-\frac{\cos 2x}{4} + C$ , the same up to the constant.)

Worth noting

$\frac{\cos^{2} x}{2}$ has the sign wrong: its derivative is $-\sin x\cos x$ .

Ten questions from Indefinite Integrals, at your own pace.

Ten questions from Indefinite Integrals, at your own pace.