Question

Integrate $x\cos(2x^2 + 7) dx$.

• A. $(1/4)\sin(2x^2 + 7) + C$
• B. $(1/4)\cos(2x^2 + 7) + C$
• C. $((\sin \theta) / 4(x^2 + 7)) + C$
• D. $\sin(2x^2 + 7) + C$

Answer 1

Answer: (A)

$(1/4)\sin(2x^2 + 7) + C$

Answer 2

To integrate the given function $x\cos(2x^2 + 7) dx$, we can use the method of substitution.

Let $u = 2x^2 + 7$. Then, differentiate $u$ with respect to $x$:

$\frac{du}{dx} = \frac{d}{dx}(2x^2 + 7)$

$\frac{du}{dx} = 4x$

From this, we can express $x dx$ in terms of $du$:

$du = 4x dx$

$x dx = \frac{1}{4} du$

Now, substitute $u$ and $x dx$ into the original integral:

$\int x\cos(2x^2 + 7) dx = \int \cos(u) \left(\frac{1}{4} du\right)$

$= \frac{1}{4} \int \cos(u) du$

The integral of $\cos(u)$ with respect to $u$ is $\sin(u)$. So,

$= \frac{1}{4} \sin(u) + C$

Finally, substitute back $u = 2x^2 + 7$:

$= \frac{1}{4} \sin(2x^2 + 7) + C$

Therefore, the correct answer is $(1/4)\sin(2x^2 + 7) + C$.