Question
Integrate the functions in Exercises:
$\tan^2(2\text{x}-3)$
$\tan^2(2\text{x}-3)$
✓
Answer
$\int\tan^2(2\text{x}-3)\text{ dx}=\int\big\{\sec^2(\text{2x}-3)-1\big\}\text{ dx}$
$=\int\sec^2(2\text{x}-3)\text{ dx }-\int1\text{ dx}$
$=\frac{\tan(2\text{x}-3)}{2}-\text{x}+\text{c}$
$=\frac{1}{2}\tan(2\text{x}-3 )+\text{c} $
$=\int\sec^2(2\text{x}-3)\text{ dx }-\int1\text{ dx}$
$=\frac{\tan(2\text{x}-3)}{2}-\text{x}+\text{c}$
$=\frac{1}{2}\tan(2\text{x}-3 )+\text{c} $
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