Calculus - Integration by parts (HL)

Evaluate the indefinite integral $\int x \cos(x) \, dx$.

Let $u = x$ and $dv = \cos(x) \, dx$. Then $du = dx$ and $v = \int \cos(x) \, dx = \sin(x)$. Using the formula $\int u \, dv = uv - \int v \, du$: $\int x \cos(x) \, dx = x \sin(x) - \int \sin(x) \, dx$ $\int x \cos(x) \, dx = x \sin(x) - (-\cos(x)) + C$ $\int x \cos(x) \, dx = x \sin(x) + \cos(x) + C$

We use the LIATE rule where $x$ is Algebraic ($A$) and $\cos(x)$ is Trigonometric ($T$). Since $A$ comes before $T$, we set $u = x$.

Find $\int \ln(x) \, dx$.

Let $u = \ln(x)$ and $dv = 1 \, dx$. Then $du = \frac{1}{x} \, dx$ and $v = x$. Using the formula: $\int \ln(x) \, dx = x \ln(x) - \int x \cdot \frac{1}{x} \, dx$ $\int \ln(x) \, dx = x \ln(x) - \int 1 \, dx$ $\int \ln(x) \, dx = x \ln(x) - x + C$

Even though there isn't a visible product, we treat $\ln(x)$ as $\ln(x) \times 1$. We choose $u = \ln(x)$ because it is Logarithmic ($L$).

Evaluate $\int x^2 e^x \, dx$.

Let $u = x^2$ and $dv = e^x \, dx$. Then $du = 2x \, dx$ and $v = e^x$. $\int x^2 e^x \, dx = x^2 e^x - \int 2x e^x \, dx$ Now apply integration by parts again to $\int 2x e^x \, dx$. Let $u_1 = 2x$ and $dv_1 = e^x \, dx$. Then $du_1 = 2 \, dx$ and $v_1 = e^x$. $\int 2x e^x \, dx = 2xe^x - \int 2e^x \, dx = 2xe^x - 2e^x$ Substitute back into the original equation: $\int x^2 e^x \, dx = x^2 e^x - (2xe^x - 2e^x) + C$ $\int x^2 e^x \, dx = e^x(x^2 - 2x + 2) + C$

This is an example of repeated integration by parts. Since the power of $x$ is 2, the process is applied twice until the algebraic term becomes a constant.