Calculate-xtan-x-cos-4-x-dx-

Question Number 167773 by LEKOUMA last updated on 24/Mar/22

C a l c u l a t e

\int \frac{x \tan x}{\cos^{4} x} d x

Answered by qaz last updated on 25/Mar/22

I = \int \frac{xtan x}{\cos^{4} x} dx

= \int x (\tan^{3} + \tan x) d (\tan x)

= x (\frac{1}{4} \tan^{3} x + \frac{1}{2} \tan x) - \int (\frac{1}{4} \tan^{3} x + \frac{1}{2} \tan x) dx

= x (\frac{1}{4} \tan^{3} x + \frac{1}{2} \tan x) - I_{1} - I_{2}

I_{1} = \frac{1}{4} \int \tan^{3} xdx = \frac{1}{4} \int (\sec^{2} x - 1) \tan xdx = \frac{1}{8} \tan^{2} x + \frac{1}{4} lncos x + C_{1}

I_{2} = \frac{1}{2} \int \tan xdx = - \frac{1}{2} lncos x + C_{2}

\Rightarrow I = x (\frac{1}{4} \tan^{3} x + \frac{1}{2} \tan x) - \frac{1}{8} \tan^{2} x + \frac{1}{4} lncos x + C

Answered by MJS_new last updated on 25/Mar/22

\int \frac{x \tan x}{\cos^{4} x} d x = \int \frac{x \sin x}{\cos^{5} x} d x =

u^{'} = \frac{\sin x}{\cos^{5} x} \to u = \frac{1}{{4 \cos}^{4} x}

v = x \to v^{'} = 1

= \frac{x}{{4 \cos}^{4} x} - \frac{1}{4} \int \frac{d x}{\cos^{4} x} =

= \frac{x}{{4 \cos}^{4} x} - \frac{\sin x (1 + {2 \cos}^{2} x)}{{12 \cos}^{3} x} + C

Commented by peter frank last updated on 26/Mar/22

thanks