To solve \(\int \tan(x) \, dx\), how must you first rewrite the integrand?
Rewrite \(\tan(x)\) as \(\frac{\sin(x)}{\cos(x)}\). The integral becomes \(\int \frac{\sin(x)}{\cos(x)} \, dx\).
Solve \(\int \tan(x) \, dx\).
Let \(u = \cos(x)\), so \(du = -\sin(x) \, dx\). The integral is \(\int \frac{-du}{u} = -\ln|u| + C\). Back-substituting gives \(-\ln|\cos(x)| + C\). This is also equivalent to \(\ln|\sec(x)| + C\).
Solve \(\int \cos(5x-2) \, dx\).
Let \(u = 5x-2\), so \(du = 5 \, dx\) and \(dx = \frac{1}{5}du\). The integral becomes \(\frac{1}{5} \int \cos(u) \, du = \frac{1}{5}\sin(u) + C\). The solution is \(\frac{1}{5}\sin(5x-2) + C\).
For an integral of the form \(\int \sin^n(x) \cos(x) \, dx\), what is the standard choice for \(u\)?
Let \(u = \sin(x)\). Then \(du = \cos(x) \, dx\), which simplifies the integral to \(\int u^n \, du\).
Solve \(\int \sin^4(x) \cos(x) \, dx\).
Using \(u = \sin(x)\) and \(du = \cos(x) \, dx\), the integral becomes \(\int u^4 \, du = \frac{u^5}{5} + C\). Back-substituting gives \(\frac{\sin^5(x)}{5} + C\).
For an integral of the form \(\int \cos^n(x) \sin(x) \, dx\), what is the standard choice for \(u\)?
Let \(u = \cos(x)\). Then \(du = -\sin(x) \, dx\), which simplifies the integral to \(\int -u^n \, du\).
What is a good choice for \(u\) in the integral \(\int \sec^2(x) \tan(x) \, dx\)?
Let \(u = \tan(x)\). The derivative is \(du = \sec^2(x) \, dx\), which is a factor in the integrand. This transforms the integral into \(\int u \, du\).
An alternative choice for \(u\) in \(\int \sec^2(x) \tan(x) \, dx\) is \(u = \sec(x)\). What is \(du\) in this case?
If \(u = \sec(x)\), then \(du = \sec(x)\tan(x) \, dx\). You can rewrite the integral as \(\int \sec(x) [\sec(x)\tan(x)] \, dx\) and substitute.
Solve \(\int x \cos(x^2) \, dx\).
Let \(u = x^2\), so \(du = 2x \, dx\) and \(x \, dx = \frac{1}{2} du\). The integral becomes \(\frac{1}{2} \int \cos(u) \, du = \frac{1}{2} \sin(u) + C\). The solution is \(\frac{1}{2} \sin(x^2) + C\).
How do you solve \(\int \sec(x) \, dx\)?
Multiply the integrand by \(\frac{\sec(x)+\tan(x)}{\sec(x)+\tan(x)}\). Then use u-substitution with \(u = \sec(x)+\tan(x)\). The result is \(\ln|\sec(x)+\tan(x)|+C\).
