The value of the integral _( )/(6)^( )/(3)(4- ^2x ^4x)dx is :

Question

The value of the integral $\int _{\frac{\pi }{6}}^{\frac{\pi }{3}}(\frac{4-{\csc }^{2}x}{{\cos }^{4}x})dx$ is :

Accepted Answer

$\begin{gathered} x=2t\Rightarrow dx=2dt \ I={\int }_{\infty}^{0}\frac{ln2t}{4{t}^{2}+4}(2dt)=\frac{1}{2}{\int }_{\infty}^{0}\frac{ln2+lnt}{{t}^{2}+1}dt \ =\frac{1}{2}{\int }_{\infty}^{0}\frac{ln2}{{t}^{2}+1}dt+\frac{1}{2}{\int }_{\infty}^{0}\frac{lnt}{{t}^{2}+1}dt \end{gathered}$

= ${[\frac{\ln 2}{2}{	an }^{-1}t]}_{\infty}^{0}+{I}_{1}$

$=\frac{ln2}{2}\cdot \frac{\pi }{2}+{I}_{1}$

${I}_{2}=\frac{1}{2}{\int }_{\infty}^{0}\frac{lnt}{{t}^{2}+1}dt$

$t=\frac{1}{u}\Rightarrow dt=-\frac{du}{{u}^{2}}$

$\begin{gathered} =\frac{1}{2}{\int }_{0}^{\infty}\frac{ln(1/u)}{\frac{1}{{u}^{2}}+1}(-\frac{du}{{u}^{2}}) \ Add\Rightarrow {I}_{1}=0 \ I=\frac{\pi ln2}{4} \end{gathered}$

Solutions · 1