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Integrate the function
$\tan ^2(2 x-3)$
$\tan ^2(2 x-3)$
Solution:
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Verified Answer
$\int \tan ^2(2 x-3) d x=\int\left[\sec ^2(2 x-3)-1\right] d x=I$
Put $2 x-3=t$, so that $2 d x=d t$
$\therefore \mathrm{I}=\frac{1}{2} \int \sec ^2 t d t-x+\mathrm{C}$
$=\frac{1}{2} \tan t-x+\mathrm{C}=\frac{1}{2} \tan (2 x-3)-x+\mathrm{C}$
Put $2 x-3=t$, so that $2 d x=d t$
$\therefore \mathrm{I}=\frac{1}{2} \int \sec ^2 t d t-x+\mathrm{C}$
$=\frac{1}{2} \tan t-x+\mathrm{C}=\frac{1}{2} \tan (2 x-3)-x+\mathrm{C}$
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