Question 13.6: Find (a) ∫ sin 2t cos t dt (b) ∫ sin mt sin nt dt, where m a......
Find
(a) \int \sin 2 t \cos t \mathrm{~d} t
(b) \int \sin m t \sin n t \mathrm{~d} t, where m and n are constants with m ≠ n
Step-by-Step
(a) Using the identities in Table 3.1 we find
2 \sin A \cos B=\sin (A+B)+\sin (A-B)
hence \sin 2 t \cos t can be written \frac{1}{2}(\sin 3 t+\sin t). Therefore,
\begin{aligned}\int \sin 2 t \cos t \mathrm{~d} t & =\int \frac{1}{2}(\sin 3 t+\sin t) \mathrm{d} t \\& =\frac{1}{2}\left(\frac{-\cos 3 t}{3}-\cos t\right)+c \\& =-\frac{1}{6} \cos 3 t-\frac{1}{2} \cos t+c\end{aligned}
(b) Using the identity 2 \sin A \sin B=\cos (A-B)-\cos (A+B), we find
\sin m t \sin n t=\frac{1}{2}\{\cos (m-n) t-\cos (m+n) t\}
Therefore,
\begin{aligned}\int \sin m t \sin n t \mathrm{~d} t & =\int \frac{1}{2}\{\cos (m-n) t-\cos (m+n) t\} \mathrm{d} t \\& =\frac{1}{2}\left\{\frac{\sin (m-n) t}{m-n}-\frac{\sin (m+n) t}{m+n}\right\}+c\end{aligned}
| Table 3.1 Common trigonometric identities. |
| \begin{aligned}& \tan A=\frac{\sin A}{\cos A} \\& \sin (A \pm B)=\sin A \cos B \pm \sin B \cos A \\& \cos (A \pm B)=\cos A \cos B \mp \sin A \sin B \\& \tan (A \pm B)=\frac{\tan A \pm \tan B}{1 \mp \tan A \tan B} \\& 2 \sin A \cos B=\sin (A+B)+\sin (A-B) \\& 2 \cos A \cos B=\cos (A+B)+\cos (A-B) \\& 2 \sin A \sin B=\cos (A-B)-\cos (A+B) \\& \sin ^2 A+\cos ^2 A=1 \\& 1+\cot ^2 A=\operatorname{cosec^ 2} A \\& \tan ^2 A+1=\sec ^2 A \\& \cos 2 A=1-2 \sin ^2 A=2 \cos ^2 A-1=\cos ^2 A-\sin ^2 A \\& \sin 2 A=2 \sin A \cos A \\& \sin ^2 A=\frac{1-\cos 2 A}{2} \\& \cos ^2 A=\frac{1+\cos 2 A}{2}\end{aligned} |
| Note: \sin ^2 A is the notation used for (\sin A)^2. Similarly \cos ^2 A means (\cos A)^2. |
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