Evaluate the integrals

$\int}_{0}^{\frac{\pi}{4}}\frac{{\mathrm{sec}}^{2}x}{{(1+7\mathrm{tan}x)}^{\frac{2}{3}}$

beljuA
2021-10-12
Answered

Evaluate the integrals

$\int}_{0}^{\frac{\pi}{4}}\frac{{\mathrm{sec}}^{2}x}{{(1+7\mathrm{tan}x)}^{\frac{2}{3}}$

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aprovard

Answered 2021-10-13
Author has **94** answers

To evaluate: ${\int}_{0}^{\frac{\pi}{4}}\frac{{\mathrm{sec}}^{2}x}{{(1+7\mathrm{tan}x)}^{\frac{2}{3}}}dx$

Let substitute$t=(1+7\mathrm{tan}x)$

$\frac{dt}{dx}=\frac{d}{dx}(1+7\mathrm{tan}x)$

$\frac{dt}{dx}=7{\mathrm{sec}}^{2}xdx$

$\frac{dt}{7}{\mathrm{sec}}^{2}xdx$

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,

$\int}_{0}^{\frac{\pi}{4}}\frac{{\mathrm{sec}}^{2}x}{{\left({(1+7\mathrm{tan}x)}^{\frac{2}{3}}\right)}^{\frac{2}{3}}}={\int}_{1}^{8}\frac{1}{{\left(t\right)}^{\frac{1}{{\left(t\right)}^{\frac{2}{3}}}}}\cdot \frac{dt}{7$

$=\frac{1}{7}{\int}_{1}^{8}{\left(t\right)}^{-\frac{2}{3}}dt$

$=\frac{1}{7}{\left[\frac{{t}^{-\frac{2}{3}+1}}{-\frac{2}{3}+1}\right]}_{1}^{8}$

$=\frac{1}{7}{\left[\frac{{t}^{\frac{1}{3}}}{\frac{1}{3}}\right]}_{1}^{8}$

$=\frac{3}{7}\left[{8}^{\frac{1}{3}-{1}^{\frac{1}{3}}}\right\}]$

$=\frac{3}{7}[2-1]$

$=\frac{3}{7}$

Hence required answer is$\frac{3}{7}$ .

Let substitute

Limit will also change accordingly.

When

when

Substituting the value,

Hence required answer is

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