Question

\(\int\sin2x\ cosx\ dx=\)

• A. \(\frac{-1}{3}cos^3x+C\)
• B. \(\frac{-2}{3}cos^3x+C\)
• C. \(\frac{2}{3}cos^3x+C\)
• D. \(\frac{1}{3}cos^3x+C\)
• E. \(\frac{-4}{3}cos^3x+C\)

Answer 1

The Correct Option is B

We are given the integral \( \int \sin(2x) \cos(x) \, dx \) and are asked to find the solution. 

We can solve this using a trigonometric identity and substitution. First, use the identity for \( \sin(2x) \):

\( \sin(2x) = 2 \sin(x) \cos(x) \),

so the integral becomes:

\( \int 2 \sin(x) \cos^2(x) \, dx \).

Now, we can use the substitution \( u = \cos(x) \), so that \( du = -\sin(x) \, dx \). The integral becomes:

\( -2 \int u^2 \, du \).

The integral of \( u^2 \) is \( \frac{u^3}{3} \), so we have:

\( -\frac{2}{3} u^3 + C \).

Substituting \( u = \cos(x) \) back, we get:

\( -\frac{2}{3} \cos^3(x) + C \).

The correct answer is \( \frac{-2}{3} \cos^3(x) + C \).