Question:
$\int \tan ^{2}(2 x-3) d x$
Solution:
Let $I=\int \tan ^{2}(2 x-3) d x$
$=\int \tan ^{2}(2 x-3) d x$
$=\int \sec ^{2}(2 x-3)-1 d x$
Let $2 x-3=t d x=d t / 2$
$=\frac{1}{2} \int \sec ^{2} t-1 d t$
$=\frac{1}{2} \tan t-x$
Substitute the value of $t$
Hence, $I=\frac{1}{2} \tan (2 x-3)-x+C$
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