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Question Number 196496 by RoseAli last updated on 26/Aug/23

Answered by witcher3 last updated on 26/Aug/23

$$\int \frac{1}{{\mathrm{tg}}^2\left(\mathrm{x}\right).{\cos }^6\left(\mathrm{x}\right)}$$$$=\int \frac{1+{\mathrm{tg}}^2\left(\mathrm{x}\right)}{{\mathrm{tg}}^2\left(\mathrm{x}\right)}.{(1+{\mathrm{tg}}^2\left(\mathrm{x}\right))}^2\mathrm{dx}$$$$\mathrm{tg}\left(\mathrm{x}\right)=\mathrm{y}$$$$\iff \int \frac{{(1+{\mathrm{y}}^2)}^2}{{\mathrm{y}}^2}\mathrm{dy}.....$$

Commented by York12 last updated on 26/Aug/23

$$SirWhatCalculusbookdoyourecommend$$$$ForTheoryAndproblems$$

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