Question

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∫ sec³x tan³x dx

Explanation:

Step1: Rewrite the integrand

Use the identity \(\tan^{2}x=\sec^{2}x - 1\).
\(\int\sec^{3}x\tan^{3}x dx=\int\sec^{3}x\tan^{2}x\tan xdx=\int\sec^{3}x(\sec^{2}x - 1)\tan xdx\)

Step2: Use substitution

Let \(u = \sec x\), then \(du=\sec x\tan xdx\).
The integral becomes \(\int u^{2}(u^{2}-1)du=\int(u^{4}-u^{2})du\)

Step3: Integrate term - by - term

\(\int(u^{4}-u^{2})du=\frac{u^{5}}{5}-\frac{u^{3}}{3}+C\)

Step4: Substitute back

Substitute \(u = \sec x\) back into the result.

Answer:

\(\frac{\sec^{5}x}{5}-\frac{\sec^{3}x}{3}+C\)