Question

find the integrals of tanx×secx​

Answer 1

Answer: -secx+c

Explanation:

$\int\ {tanxsecx} \, dx$

$\int\frac{sinx}{cos^{2}x} \, dx$

suppose ,cosx=z

-sinx dx =dz

⇒$-\int\frac{1}{z^2} \, dz$

⇒-{$-\frac{1}{z}$}+c

⇒$\frac{1}{cosx}$+c

⇒secx+c