calculate-C-dt-1-tan-2-t-using-x-tan-t-

Question Number 169251 by mathocean1 last updated on 26/Apr/22

c a l c u l a t e C = \int \frac{d t}{1 + {t a n}^{2} (t)} u s i n g x = t a n (t)

Answered by thfchristopher last updated on 27/Apr/22

Let x = \tan t

d x = \sec^{2} t d t

d x = (1 + x^{2}) d t

\frac{d x}{1 + x^{2}} = d t

∴ \int \frac{d t}{1 + \tan^{2} t}

= \int \frac{d x}{{(x^{2} + 1)}^{2}}

= \frac{x}{2 (x^{2} + 1)} + \frac{1}{2} \int \frac{d x}{x^{2} + 1} (reduction formula)

= \frac{\tan t}{{2 \sec}^{2} t} + \frac{1}{2} \int d t

= \frac{1}{2} (\sin t \cos t + t) + c