Calculus - Integration by substitution

Evaluate the indefinite integral $\int 3x^2(x^3 + 5)^4 dx$.

Let $u = x^3 + 5$. Then $\frac{du}{dx} = 3x^2$, which implies $du = 3x^2 dx$. Substitute $u$ and $du$ into the integral: $\int u^4 du$ Integrate with respect to $u$: $\frac{u^5}{5} + C$ Substitute back $u = x^3 + 5$: $\frac{(x^3 + 5)^5}{5} + C$

We identify $x^3+5$ as the inner function because its derivative $3x^2$ is present as a factor in the integrand.

Find the exact value of $\int_{0}^{1} x e^{x^2} dx$.

Let $u = x^2$. Then $du = 2x dx$, so $x dx = \frac{1}{2} du$. Change the limits of integration: When $x = 0$, $u = 0^2 = 0$. When $x = 1$, $u = 1^2 = 1$. Substitute into the integral: $\int_{0}^{1} e^u \cdot \frac{1}{2} du = \frac{1}{2} \int_{0}^{1} e^u du$ Evaluate the integral: $\frac{1}{2} [e^u]_{0}^{1} = \frac{1}{2} (e^1 - e^0)$ Final Answer: $\frac{1}{2}(e - 1) \text{ or } \frac{e-1}{2}$

Since the derivative of $x^2$ is $2x$, and we have an $x$ term, we use $u=x^2$. Note the change of limits to keep the integral entirely in terms of $u$.

Evaluate $\int \tan(x) dx$.

Rewrite $\tan(x)$ as $\frac{\sin(x)}{\cos(x)}$: $\int \frac{\sin(x)}{\cos(x)} dx$ Let $u = \cos(x)$. Then $du = -\sin(x) dx$, so $\sin(x) dx = -du$. Substitute: $\int \frac{-1}{u} du = -\ln|u| + C$ Substitute back $u = \cos(x)$: $-\ln|\cos(x)| + C$ Using log properties, this can also be written as: $\ln|\sec(x)| + C$

This example uses the $\frac{f'(x)}{f(x)}$ pattern where the numerator is the derivative of the denominator (with a sign change).