# Evaluate the integrals \int_0^{\pi/4}\frac{\sec^2x}{(1+7\tan x)^{2/3}}ZS

Evaluate the integrals
${\int }_{0}^{\frac{\pi }{4}}\frac{{\mathrm{sec}}^{2}x}{{\left(1+7\mathrm{tan}x\right)}^{\frac{2}{3}}}$
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aprovard
To evaluate: ${\int }_{0}^{\frac{\pi }{4}}\frac{{\mathrm{sec}}^{2}x}{{\left(1+7\mathrm{tan}x\right)}^{\frac{2}{3}}}dx$
Let substitute $t=\left(1+7\mathrm{tan}x\right)$
$\frac{dt}{dx}=\frac{d}{dx}\left(1+7\mathrm{tan}x\right)$
$\frac{dt}{dx}=7{\mathrm{sec}}^{2}xdx$
$\frac{dt}{7}{\mathrm{sec}}^{2}xdx$
Limit will also change accordingly.
When $x\to 0$ then $t\to 1$
when $x\to \frac{\pi }{4}$ then $t\to 8$
Substituting the value,
${\int }_{0}^{\frac{\pi }{4}}\frac{{\mathrm{sec}}^{2}x}{{\left({\left(1+7\mathrm{tan}x\right)}^{\frac{2}{3}}\right)}^{\frac{2}{3}}}={\int }_{1}^{8}\frac{1}{{\left(t\right)}^{\frac{1}{{\left(t\right)}^{\frac{2}{3}}}}}\cdot \frac{dt}{7}$
$=\frac{1}{7}{\int }_{1}^{8}{\left(t\right)}^{-\frac{2}{3}}dt$
$=\frac{1}{7}{\left[\frac{{t}^{-\frac{2}{3}+1}}{-\frac{2}{3}+1}\right]}_{1}^{8}$
$=\frac{1}{7}{\left[\frac{{t}^{\frac{1}{3}}}{\frac{1}{3}}\right]}_{1}^{8}$
$=\frac{3}{7}\left[{8}^{\frac{1}{3}-{1}^{\frac{1}{3}}}\right\}\right]$
$=\frac{3}{7}\left[2-1\right]$
$=\frac{3}{7}$
Hence required answer is $\frac{3}{7}$.