Simple Yet Challenging Integral

Take , $I = \int_{0}^{\pi/4} \ln(1+\tan(x))$

Then, $I = \int_{0}^{\pi/4} \ln(1+\tan(\frac{\pi}{4}-x))$

$I = \int_{0}^{\frac{\pi}{4}} \ln(\frac{2}{1+\tan(x)})$

$I = \int_{0}^{\frac{\pi}{4}} \ln(2) - I$

$I = \frac{\pi \ln(2)}{8}$

$\int _{ 0 }^{ \pi /4 }{ \ln { (\tan { x+1)dx } } } = \int _{ 0 }^{ \pi /4 }{ (\ln { (\sin { x+\cos { x)-\ln { (cos{ x })))dx } } } } }$

$=\int _{ 0 }^{ \pi /4 }{ \ln { (\sin { x } +\cos { x)dx } } } -\int _{ 0 }^{ \pi /4 }{ \ln { (\cos { x)dx } } }$

Now we have:

$\sin { x } +\cos { x } =\sqrt { 2 } (\frac { \sqrt { 2 } }{ 2 } )(\sin { x } +\cos { x) }$

$=\sqrt { 2 } (\frac { \sqrt { 2 } }{ 2 } \sin { x } +\frac { \sqrt { 2 } }{ 2 } \cos { x) } = \sqrt { 2 } (\cos { \frac { \pi }{ 4 } \sin { x } } +\sin { \frac { \pi }{ 4 } } \cos { x) }$

$=\sqrt { 2 } \sin { (x+\frac { \pi }{ 4 }) }$

The integral then becomes:

$\int _{ 0 }^{ \pi /4 }{ \ln { (\sqrt { 2 } \sin { (x+\frac { \pi }{ 4 } } ) } } dx - \int _{ 0 }^{ \pi /4 }{ \ln { \cos { x } } } dx$

$=\int _{ 0 }^{ \pi /4 }{ \ln { \sqrt { 2 } } dx } + \int _{ \pi /2 }^{ \pi /4 }{ \ln { \sin { u } } } du - \int _{ 0 }^{ \pi /4 }{ \ln { \cos { x } } } dx$

Now consider $\int _{ 0 }^{ \pi /4 }{ \ln { \cos { x } } } dx$ ; make the substitution $x= \frac { \pi }{ 2 } -v$ and rewrite this integral as:

$-\int _{ \pi /2 }^{ \pi /4 }{ \ln { \sin { x } } } dx =\int _{ \pi /4 }^{ \pi /2 }{ \ln { \sin { x } } } dx$ .

This cancels, and we are left with $\int _{ 0 }^{ \pi /4 }{ \ln { \sqrt { 2 } dx } } = \frac { \ln { 2 } }{ 8 } \pi$ .

Since we want the answer in terms of $\pi,$ we can solve $\frac { \ln { 2 } }{ 8 }$ using a calculator and we get an answer of $0.087.$