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Question 13.5: Evaluate (a) ∫ cos² t dt (b) ∫ sin² t dt...

Evaluate

(a)  \int \cos ^2 t \mathrm{~d} t

(b)  \int \sin ^2 t \mathrm{~d} t

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Powers of trigonometric functions, for example  \sin ^2 t,  do not appear in the table of standard integrals. What we must attempt to do is rewrite the integrand to obtain a standard form.

(a) From Table 3.1

\cos ^2 t=\frac{1+\cos 2 t}{2}

and so

\begin{aligned}\int \cos ^2 t \mathrm{~d} t & =\int \frac{1+\cos 2 t}{2} \mathrm{~d} t \\& =\int \frac{1}{2} \mathrm{~d} t+\int \frac{\cos 2 t}{2} \mathrm{~d} t \\& =\frac{t}{2}+\frac{\sin 2 t}{4}+c\end{aligned}

(b)

\begin{array}{rlrl}\int \sin ^2 t \mathrm{~d} t & =\int 1-\cos ^2 t \mathrm{~d} t & & \text { using the trigonometric identities } \\& =\int 1 \mathrm{~d} t-\int \cos ^2 t \mathrm{~d} t & & \text { using linearity } \\& =t-\left(\frac{t}{2}+\frac{\sin 2 t}{4}+c\right) & & \text { using part (a) } \\& =\frac{t}{2}-\frac{\sin 2 t}{4}+k &\end{array}
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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