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Integrate the function
$\sin (a x+b) \cos (a x+b)$
$\sin (a x+b) \cos (a x+b)$
Solution:
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Verified Answer
Let $\sin (a x+b)=t \Rightarrow \cos (a x+b) d x=d t$
$\begin{aligned}
&\therefore \int \sin (a x+b) \cos (a x+b) d x=\frac{1}{a} \int t d t \\
&=\frac{1}{a} \cdot \frac{t^2}{2}+\mathrm{C}=\frac{1}{2 a} \sin ^2(a x+b)+\mathrm{C}
\end{aligned}$
$\begin{aligned}
&\therefore \int \sin (a x+b) \cos (a x+b) d x=\frac{1}{a} \int t d t \\
&=\frac{1}{a} \cdot \frac{t^2}{2}+\mathrm{C}=\frac{1}{2 a} \sin ^2(a x+b)+\mathrm{C}
\end{aligned}$
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