Integrating Trigonometric Functions by Recognition

Example 1

Find the derivative of \( \sin (2x-5) \) and use this result to deduce \( \displaystyle \int 10 \cos (2x-5) dx \).

\( \begin{align} \displaystyle \frac{d}{dx} \sin (2x-5) &= \cos (2x-5) \times \frac{d}{dx} (2x-5) &\color{green}{\text{Chain Rule}} \\ &= \cos (2x-5) \times 2 \\ &= 2 \cos (2x-5) \\ \sin (2x-5) &= \int 2 \cos (2x-5) dx \\ \int 10 \cos (2x-5) dx &= 5 \int 2 \cos (2x-5) dx \\ \require{AMSsymbols} \therefore \int 10 \cos (2x-5) dx &= 5 \sin (2x-5) + C \end{align} \)

Example 2

Find the derivative of \( x \cos x \) and use this result to find \( \displaystyle \int x \sin x dx \).

\( \begin{align} \displaystyle \frac{d}{dx} x \cos x &= x \times \frac{d}{dx} \cos x + \frac{d}{dx} x \times \cos x &\color{green}{\text{Product Rule}} \\ &= x \times (- \sin x) + 1 \times \cos x \\ &= -x \sin x + \cos x \\ x \cos x &= \int (-x \sin x + \cos x) dx \\ x \cos x &= – \int x \sin x dx + \int \cos x dx \\ \int x \sin x dx &= \int \cos x dx – x \cos x \\ \require{AMSsymbols} \therefore \int x \sin x dx &= \sin x – x \cos x + C &\color{green}{\int \cos x dx = \sin x + C} \end{align} \)

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