# Math Help - Integration problem, please help.

1. ## Integration problem, please help.

the integral from pi/4 to pi/3 of (tanx/(cosx^2(6+tanx^2)^.5)dx

2. ## Re: Integration problem, please help.

If it's \displaystyle \begin{align*} \int_{\frac{\pi}{4}}^{\frac{\pi}{3}}{\frac{\tan{(x )}}{\cos^2{(x)}\sqrt{6 + \tan^2{(x)}}}} \end{align*}, notice that \displaystyle \begin{align*} \frac{d}{dx} \left[ \tan{(x)} \right] = \frac{1}{\cos^2{(x)}} \end{align*}, so a substitution of the form \displaystyle \begin{align*} u = \tan{(x)} \implies du = \frac{1}{\cos^2{(x)}}\,dx \end{align*} is appropriate. Notice that when \displaystyle \begin{align*} x = \frac{\pi}{4} , \, u = 1 \end{align*} and when \displaystyle \begin{align*} x = \frac{\pi}{3} , \, u = \sqrt{3} \end{align*}, and the integral becomes

\displaystyle \begin{align*} \int_{\frac{\pi}{4}}^{\frac{\pi}{3}}{\frac{\tan{(x )}}{\cos^2{(x)}\sqrt{6 + \tan^2{(x)}}}} = \int_1^{\sqrt{3}}{\frac{u}{\sqrt{6 + u^2}}\,du} \end{align*}

Can you go from here?