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

Evaluate

(a)  $\int \cos ^2 t \mathrm{~d} t$

(b)  $\int \sin ^2 t \mathrm{~d} t$

Step-by-Step
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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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