Example about \( {sec}^{4}x\)

Example:

Compute , $\int{{{\sec }^{4}}x\,dx}$

Solution: we know that $\int{{{\sec }^{4}}x\,dx}=\int{{{\sec }^{2}}x{{\sec }^{2}}x\,dx}$ but ${{\sec }^{2}}x=1+{{\tan }^{2}}x$

So $\int{{{\sec }^{4}}x\,dx}=\int{\left( 1+{{\tan }^{2}}x \right){{\sec }^{2}}x\,dx}$ take $u=\tan x\Rightarrow du={{\sec }^{2}}x\,dx$

Thus $\int{{{\sec }^{4}}x\,dx}=\int{\left( 1+{{u}^{2}} \right)du}=u+\frac{1}{3}{{u}^{3}}+c=\tan x+\frac{1}{3}{{\tan }^{3}}x+c$ 

We will test our blogger now :



tex Code :


Evaluate $\int \sqrt{1+x^2} \; \frac{\ln(1+x^2) - 2 \ln x}{x^4} \;dx $

Answer:

$\int \sqrt{1+x^2} \;\frac{\ln(1+x^2) - 2 \ln x}{x^4} \;dx $

$= \int \sqrt{1+x^2} \; \ln\left(\frac{1+x^2}{x^2}\right) \frac{1}{x^4} \;dx $

$= \int \sqrt{1+\frac{1}{x^2}} \; \ln\left(1 + \frac{1}{x^2}\right) \frac{1}{x^3} \;dx $

Substitute $u = 1 + \frac{1}{x^2}$ and $du = \frac{-2}{x^3} \; dx$ or $\frac{-1}{2} \;du= \frac{1}{x^3} \; dx$ to get

$= \int \sqrt{1+\frac{1}{x^2}} \; \ln\left(1 + \frac{1}{x^2}\right) \frac{1}{x^3} \;dx $

=$ \frac{-1}{2} \int \sqrt{u}\;\ln u \;du $

Substitute $df = \sqrt{u} = u^{\frac{1}{2}} \; du$ or, $f = \frac{u^ { \frac{3}{2} }} { \frac{3}{2} } = \frac{2}{3} u^{\frac{3}{2}}$ and  $ g = \ln u $ or, $dg = \frac{1}{u} \; du$ to get

$ \frac{-1}{2} \int \sqrt{u}\;\ln u \;du $

$= \frac{-1}{2} \int g \; df $

$ = \frac{-1}{2} \left( gf - \int f \; dg \right)$

$= \frac{-1}{2} \left( \frac{2}{3} u^{\frac{3}{2}} \ln u - \int \frac{2}{3} u^{\frac{3}{2}} \frac{1}{u} \; du \right)$

$= \frac{-1}{3} u^{\frac{3}{2}} \ln u + \frac{1}{3} \int u^{\frac{1}{2}} \; du $

$= \frac{-1}{3} u^{\frac{3}{2}} \ln u  + \frac{1}{3} \frac{u^{\frac{3}{2}}}{\frac{3}{2}} +C $

$= \frac{-1}{3} u^{\frac{3}{2}} \ln u + \frac{2}{9} u^{\frac{3}{2}} + C $

$= \frac{-1}{3} \left(1 + \frac{1}{x^2} \right)^{\frac{3}{2}} \ln \left(1 + \frac{1}{x^2} \right) + \frac{2}{9} \left(1 + \frac{1}{x^2} \right)^{\frac{3}{2}} + C $