sec-4-x-cot-4-x-dx-

Question Number 196487 by RoseAli last updated on 25/Aug/23

\int (\sec^{4} x - \cot^{4} x) d x

Answered by MM42 last updated on 26/Aug/23

I_{1} = \int {s e c}^{4} x d x = \int (1 + {t a n}^{2} x) (1 + {t a n}^{2} x) d x

l e t t a n x = u \Rightarrow

I_{1} = \int (1 + u^{2}) d u = u + \frac{1}{3} u^{3} + c_{1} = t a n x + \frac{1}{3} {t a n}^{3} x + c_{1}

I_{2} = \int {c o t}^{4} x d x = \int ({c o t}^{4} x + {c o t}^{2} x - {c o t}^{2} x - 1 + 1) d x

= \int [{c o t}^{2} x (1 + {c o t}^{2} x) - (1 + {c o t}^{2} x) + 1] d x

= - \frac{1}{3} {c o t}^{3} x + c o t x + x + c_{2}

\Rightarrow I = t a n x - c o t x + \frac{1}{3} ({t a n}^{3} x + {c o t}^{3} x) - x + c ✓

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