Inte-great! returns

Calculus Level 2

$\large \int _{ 1 }^{ {e}^{\pi /4} }{ \cfrac { \tan ( \ln { x } ) }{ x } } \, dx = \, \dfrac{A}{B} \, \ln C$

Given $A$ , $B$ and $C$ are positive integers satisfying the equation above, with $A$ and $B$ coprime and $C$ minimised, find $A+B+C$ .

1 solution

Aditya Kumar
Jul 7, 2016

$\displaystyle I=\int _{ 1 }^{ { e }^{ \frac { \pi }{ 4 } } }{ \frac { \tan { \left( \ln { \left( x \right) } \right) } }{ x } dx }$

Substitute: $\ln(x)=t$ . On differentiating both the sides w.r.t. $t$ , we have $\frac{1}{x}dx=dt$ .

$\displaystyle I=\int _{ 0 }^{ \frac { \pi }{ 4 } }{ \tan { \left( t \right) } dt }$ .

We use the result of the indefinite integral: $\displaystyle \ln { \left( \sec { \left( t \right) } \right) } +C$

$I={ \left\{ \ln { \left( \sec { \left( t \right) } \right) } \right\} }_{ 0 }^{ \frac { \pi }{ 4 } }$

Therefore, $\boxed{I=\frac { 1 }{ 2 } \ln { \left( 2 \right) } }$ .

Isn't the antiderivative of tanx = -ln(cosx) also?

J D - 4 years, 11 months ago

-ln(cosx)=ln(secx). Use the property of lnx=-ln(1/x)

Aditya Kumar - 4 years, 11 months ago

I think you'd get a different answer that way

J D - 4 years, 11 months ago

You may have to read logarithms .

Aditya Kumar - 4 years, 11 months ago