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