To integrate functions like \(\sin(ax+b)\), \(\cos(ax+b)\), \(\tan(ax+b)\), use substitution on the angle:

This introduces a factor of \(\frac{1}{a}\) in the answer.

\[ \int \sin(ax+b)\,dx = -\frac{1}{a}\cos(ax+b) + C \] \[ \int \cos(ax+b)\,dx = \frac{1}{a}\sin(ax+b) + C \] \[ \int \tan(ax+b)\,dx = -\frac{1}{a}\ln|\cos(ax+b)| + C \]

Example 1: \(\displaystyle \int \sin(5x)\,dx\)

Integrate \(\sin\) → \(-\cos\), then multiply by \(\frac{1}{5}\): \[ \int \sin(5x)\,dx = -\frac{1}{5}\cos(5x) + C \]

Example 2: \(\displaystyle \int \cos(3x+4)\,dx\)

\[ \int \cos(3x+4)\,dx = \frac{1}{3}\sin(3x+4) + C \]

Example 3: \(\displaystyle \int \tan(2x)\,dx\)

\[ \int \tan(2x)\,dx = -\frac{1}{2}\ln|\cos(2x)| + C \]

Example 4 (Definite): \(\displaystyle \int_{0}^{\pi/4} \sin(2x)\,dx\)

Use substitution or shortcut: \[ \int \sin(2x)\,dx = -\frac{1}{2}\cos(2x) \] \[ \left[-\frac{1}{2}\cos(2x)\right]_{0}^{\pi/4} = -\frac{1}{2}\left(\cos\frac{\pi}{2} - \cos 0\right) = -\frac{1}{2}(0 - 1) = \frac{1}{2} \]

• \(\displaystyle \int f'(x)\sin(f(x))\,dx = -\cos(f(x))+C\)
• \(\displaystyle \int f'(x)\cos(f(x))\,dx = \sin(f(x))+C\)
• \(\displaystyle \int f'(x)\tan(f(x))\,dx = -\ln|\cos(f(x))|+C\)

• Always multiply by \(\frac{1}{a}\) when the angle is \(ax+b\).
• For definite integrals, you may change limits OR substitute back at the end; both acceptable.
• Do not forget absolute values for logarithms with tan integrals.
• If a question says “Use substitution”, show the substitution even if you know the shortcut.
• Look for expressions like \( f'(x)\sin(f(x))\) — they are “instant substitution” problems.