Question-14479

Question Number 14479 by tawa tawa last updated on 01/Jun/17

Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 01/Jun/17

l n (1 + {t g}^{2} x) = l n \frac{1}{{c o s}^{2} x} = - {l n c o s}^{2} x = - 2 l n c o s x .

a^{\frac{3}{2} {l o g}_{\sqrt{a}}^{(s e c x)}} = a^{\frac{3}{2} (- 2 {l o g}_{a} c o s x)} = a^{{l o g}_{a}^{\frac{1}{{c o s}^{3} x}}} = \frac{1}{{c o s}^{3} x}

I = \int - \frac{2 l n c o s x}{{c o s}^{3} x} d x

u = l n c o s x \Rightarrow d u = \frac{- s i n x}{c o s x} d x

\Rightarrow I = \int \frac{2 s i n x}{{c o s}^{4} x} d x = \frac{- 2}{5 {c o s}^{5} x} + C .

Commented by tawa tawa last updated on 01/Jun/17

God bless you sir .