Section 7.2 : Integrals Involving Trig Functions
9. Evaluate \( \displaystyle \int_{1}^{3}{{\sin \left( {8x} \right)\sin \left( x \right)\,dx}}\).
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Start SolutionThere really isn’t all that much to this problem. All we have to do is use the formula given in this section for reducing a product of a sine and a cosine into a sum. Doing this gives,
\[ \int_{1}^{3}{{\sin \left( {8x} \right)\sin \left( x \right)\,dx}} = \int_{1}^{3}{{\frac{1}{2}\left[ {\cos \left( {8x - x} \right) - \cos \left( {8x + x} \right)} \right]\,dx}} = \frac{1}{2}\int_{1}^{3}{{\cos \left( {7x} \right) - \cos \left( {9x} \right)\,dx}}\] Show Step 2Now all we need to do is evaluate the integral.
\[\begin{align*}\int_{1}^{3}{{\sin \left( {8x} \right)\sin \left( x \right)\,dx}} & = \left. {\frac{1}{2}\left[ {\frac{1}{7}\sin \left( {7x} \right) - \frac{1}{9}\sin \left( {9x} \right)} \right]} \right|_1^3\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{1}{{14}}\sin \left( {21} \right) - \frac{1}{{18}}\sin \left( {27} \right) - \frac{1}{{14}}\sin \left( 7 \right) + \frac{1}{{18}}\sin \left( 9 \right) = - 0.0174}}\end{align*}\]Make sure your calculator is set to radians if you computed a decimal answer!