Question

Use the table of integrals at the back of the text to evaluate the integrals \(\displaystyle\int{8}{\sin{{\left({4}{t}\right)}}}{\sin{{\left({\frac{{{t}}}{{{2}}}}\right)}}}{\left.{d}{t}\right.}\)

Answer 1

Step 1
Let the given integral is,
\(\displaystyle\int{8}{\sin{{\left({4}{t}\right)}}}{\sin{{\left({\frac{{{t}}}{{{2}}}}\right)}}}{\left.{d}{t}\right.}\)
By using the formula,
\(\displaystyle{\sin{{\left({a}\right)}}}{\sin{{\left({b}\right)}}}={\frac{{-{\cos{{\left({a}+{b}\right)}}}+{\cos{{\left({a}-{b}\right)}}}}}{{{2}}}}\)
\(\displaystyle\int{8}{\left({\frac{{{\cos{{\left({4}{t}-{\frac{{{t}}}{{{2}}}}\right)}}}-{\cos{{\left({4}{t}+{\frac{{{t}}}{{{2}}}}\right)}}}}}{{{2}}}}\right)}{\left.{d}{t}\right.}\)
\(\displaystyle\Rightarrow{8}\int{\left({\frac{{{\cos{{\left({\frac{{{7}{t}}}{{{2}}}}\right)}}}-{\cos{{\left({\frac{{{9}{t}}}{{{2}}}}\right)}}}}}{{{2}}}}\right)}{\left.{d}{t}\right.}\)
Step 2
By separating the integrals,
\(\displaystyle\Rightarrow{\frac{{{8}}}{{{2}}}}\int{\left({\cos{{\left({\frac{{{7}{t}}}{{{2}}}}\right)}}}\right)}{\left.{d}{t}\right.}-\int{\left({\cos{{\left({\frac{{{9}{t}}}{{{2}}}}\right)}}}\right)}{\left.{d}{t}\right.}\)
Simplifying this,
\(\displaystyle\Rightarrow\int{8}{\sin{{\left({4}{t}\right)}}}{\sin{{\left({\frac{{{t}}}{{{2}}}}\right)}}}{\left.{d}{t}\right.}={4}{\left[{\frac{{{2}}}{{{7}}}}{\sin{{\left({\frac{{{7}{t}}}{{{2}}}}\right)}}}-{\frac{{{2}}}{{{9}}}}{\sin{{\left({\frac{{{9}{t}}}{{{2}}}}\right)}}}\right]}+{C}\)