Integrate ~tan^2×sec^6×dx

Answer 1

Noting that sec²x = d/dx(tan(x)) and sec²x = 1+tan²x, you can rewrite the integral as

tan²(x)(1+tan²(x))²sec²x dx which is u²(1+u²)²du where u = tan(x)

You can multiply this all out and integrate the polynomial in u, then sub tan(x) back in for u.

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