Integration by Parts: To solve the integral of x^2 cos(2x+1) with respect to x, we will use integration by parts, which is based on the product rule for differentiation and is given by the formula: ∫u dv = uv - ∫v du We need to choose u and dv such that the resulting integral is simpler than the original one. Let's choose: u = x^2 (which means du = 2x dx) dv = cos(2x+1) dx (which means v = ∫cos(2x+1) dx)Finding v: Now we need to find v by integrating dv. To integrate cos(2x+1), we can use a substitution method. Let w = 2x+1, which implies dw = 2 dx or dx = dw/2. ∫cos(2x+1) dx = ∫cos(w) (dw/2) = (1/2) ∫cos(w) dw The integral of cos(w) with respect to w is sin(w), so we have: v = (1/2) sin(w) = (1/2) sin(2x+1)Applying Integration by Parts: Now we can apply the integration by parts formula: ∫x^2 cos(2x+1) dx = uv - ∫v du = x^2 * (1/2) sin(2x+1) - ∫(1/2) sin(2x+1) * 2x dx = (1/2) x^2 sin(2x+1) - ∫x sin(2x+1) dxSecond Integration by Parts: The remaining integral, ∫x sin(2x+1) dx, still requires integration by parts. We will again choose new u and dv: u = x (which means du = dx) dv = sin(2x+1) dx (which means v = -1/2 cos(2x+1))Combining Results: Applying integration by parts to the remaining integral: ∫x sin(2x+1) dx = uv - ∫v du = x * (-1/2) cos(2x+1) - ∫(-1/2) cos(2x+1) dx = (-1/2) x cos(2x+1) + (1/4) sin(2x+1)Now we can combine the results from the two applications of integration by parts: ∫x^2 cos(2x+1) dx = (1/2) x^2 sin(2x+1) - [(-1/2) x cos(2x+1) + (1/4) sin(2x+1)] = (1/2) x^2 sin(2x+1) + (1/2) x cos(2x+1) - (1/4) sin(2x+1) + C where C is the constant of integration.

Resources

Plans

Resources

Testimonials

Plans

AI tutor

Welcome to Bytelearn!

Let’s check out your problem:

$\int x^{2} \cos (2 x+1) d x$

View step-by-step help

Home

Math Problems

Algebra 2

Simplify radical expressions involving fractions

Full solution

\int x^{2} \cos (2 x+1) d x

Integration by Parts: To solve the integral of $x^2 \cos(2x+1)$ with respect to $x$ , we will use integration by parts, which is based on the product rule for differentiation and is given by the formula: $\newline$ $\int u \, dv = uv - \int v \, du$ $\newline$ We need to choose $u$ and $dv$ such that the resulting integral is simpler than the original one. $\newline$ Let's choose: $\newline$ $u = x^2$ (which means $du = 2x \, dx$ ) $\newline$ $dv = \cos(2x+1) \, dx$ (which means $v = \int\cos(2x+1) \, dx$ )
Finding $v$ : Now we need to find $v$ by integrating $dv$ . To integrate $\cos(2x+1)$ , we can use a substitution method. Let $w = 2x+1$ , which implies $dw = 2 dx$ or $dx = \frac{dw}{2}$ . $\newline$ $\int \cos(2x+1) dx = \int \cos(w) \left(\frac{dw}{2}\right) = \left(\frac{1}{2}\right) \int \cos(w) dw$ $\newline$ The integral of $\cos(w)$ with respect to $w$ is $v$ $0$ , so we have: $\newline$ $v$ $1$
Applying Integration by Parts: Now we can apply the integration by parts formula: $\newline$ $\int x^2 \cos(2x+1) \, dx = uv - \int v \, du$ $\newline$ = $x^2 \cdot (\frac{1}{2}) \sin(2x+1) - \int (\frac{1}{2}) \sin(2x+1) \cdot 2x \, dx$ $\newline$ = $(\frac{1}{2}) x^2 \sin(2x+1) - \int x \sin(2x+1) \, dx$
Second Integration by Parts: The remaining integral, $\int x \sin(2x+1) \, dx$ , still requires integration by parts. We will again choose new $u$ and $dv$ : $u = x$ (which means $du = dx$ ) $dv = \sin(2x+1) \, dx$ (which means $v = -\frac{1}{2} \cos(2x+1)$ )
Combining Results: Applying integration by parts to the remaining integral: $\newline$ $\int x \sin(2x+1) \, dx = uv - \int v \, du$ $\newline$ = $x \cdot (-\frac{1}{2}) \cos(2x+1) - \int(-\frac{1}{2}) \cos(2x+1) \, dx$ $\newline$ = $(-\frac{1}{2}) x \cos(2x+1) + (\frac{1}{4}) \sin(2x+1)$
Combining Results: Applying integration by parts to the remaining integral: $\newline$ $\int x \sin(2x+1) \, dx = uv - \int v \, du$ $\newline$ = $x \cdot (-\frac{1}{2}) \cos(2x+1) - \int (-\frac{1}{2}) \cos(2x+1) \, dx$ $\newline$ = $(-\frac{1}{2}) x \cos(2x+1) + (\frac{1}{4}) \sin(2x+1)$ Now we can combine the results from the two applications of integration by parts: $\newline$ $\int x^2 \cos(2x+1) \, dx = (\frac{1}{2}) x^2 \sin(2x+1) - [(-\frac{1}{2}) x \cos(2x+1) + (\frac{1}{4}) \sin(2x+1)]$ $\newline$ = $(\frac{1}{2}) x^2 \sin(2x+1) + (\frac{1}{2}) x \cos(2x+1) - (\frac{1}{4}) \sin(2x+1) + C$ $\newline$ where $C$ is the constant of integration.