Question
Find the integrals of the functions in Exercises:
$\tan^32\text{x}\sec2\text{x}$
$\tan^32\text{x}\sec2\text{x}$
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Answer
$\tan^32\text{x }\sec2\text{x}=\tan^22\text{x }\tan2\text{x }\sec2\text{x}$
$=(\sec^22\text{x}-1)\tan2\text{x }\sec2\text{x}$
$=\sec^22\text{x }\tan2\text{x }\sec2\text{x}-\tan2\text{x }\sec2\text{x}$
$\therefore\int\tan^32\text{x }\sec2\text{x}\text{ dx}$
$=\int\sec^22\text{x }\tan2\text{x }\sec2\text{x}\text{ dx}-\int\tan2\text{x }\sec2\text{x}\text{ dx}$
$=\int\sec^22\text{x }\tan2\text{x }\sec2\text{x}\text{ dx}-\frac{\sec2\text{x}}{2}+\text{C}$
$\text{Let }\sec2\text{x}=\text{t}$
$\therefore2\sec2\text{x }\tan2\text{x}\text{ dx}=\text{dt}$
$\therefore\int\tan^32\text{x }\sec2\text{x}\text{ dx}\text=\frac{1}{2}\int\text{t}^2\text{dt}-\frac{\sec2\text{x}}{2}+\text{C}$
$=\frac{\text{t}^3}{6}-\frac{\sec2\text{x}}{2}+\text{C}$
$=\frac{(\sec2\text{x})^3}{6}-\frac{\sec2\text{x}}{2}+\text{C}$
$=(\sec^22\text{x}-1)\tan2\text{x }\sec2\text{x}$
$=\sec^22\text{x }\tan2\text{x }\sec2\text{x}-\tan2\text{x }\sec2\text{x}$
$\therefore\int\tan^32\text{x }\sec2\text{x}\text{ dx}$
$=\int\sec^22\text{x }\tan2\text{x }\sec2\text{x}\text{ dx}-\int\tan2\text{x }\sec2\text{x}\text{ dx}$
$=\int\sec^22\text{x }\tan2\text{x }\sec2\text{x}\text{ dx}-\frac{\sec2\text{x}}{2}+\text{C}$
$\text{Let }\sec2\text{x}=\text{t}$
$\therefore2\sec2\text{x }\tan2\text{x}\text{ dx}=\text{dt}$
$\therefore\int\tan^32\text{x }\sec2\text{x}\text{ dx}\text=\frac{1}{2}\int\text{t}^2\text{dt}-\frac{\sec2\text{x}}{2}+\text{C}$
$=\frac{\text{t}^3}{6}-\frac{\sec2\text{x}}{2}+\text{C}$
$=\frac{(\sec2\text{x})^3}{6}-\frac{\sec2\text{x}}{2}+\text{C}$
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