Integrate it!

Calculus Level 3

$\textbf{Find the following integral}$

$\int_{0}^{1} \frac{ln(1+ x)}{1 + x^{2}} dx$

The answer is 0.272.

1 solution

Nikhil Kashyap
Mar 16, 2014

$Put\quad x=tan\theta \\ dx={ \sec ^{ 2 }{ \theta } }d\theta \\ I=\int _{ 0 }^{ \frac { \pi }{ 4 } }{ \ln { (1+tan\theta )d\theta } } \\ I=\int _{ 0 }^{ \frac { \pi }{ 4 } }{ ln(1+tan( } \frac { \pi }{ 4 } -\theta ))d\theta \\ I=\int _{ 0 }^{ \frac { \pi }{ 4 } }{ \ln { (1+\frac { 1-tan\theta }{ 1+tan\theta } } } )d\theta \\ I=(\int _{ 0 }^{ \frac { \pi }{ 4 } }{ (\ln { 2)d\theta } ) } -I\\ 2I=\int _{ 0 }^{ \frac { \pi }{ 4 } }{ \ln { 2 } } d\theta \\ { \boxed{ \ I=\frac { \pi }{ 8 } \ln { 2 } } }$

oooooooooooooooooooooo........ good approach

Mayank Holmes - 7 years, 2 months ago

BUT tan(theta) DOESN'T EQUAL tan(pi/4 - theta)

Amr El-Sayed - 7 years, 1 month ago

Note: To get from the 3rd line to the fourth line, he is not saying that $\tan \theta = \tan ( \pi/4 - \theta$ .

Instead, he did a substitution of $\theta = \pi/4 - \alpha$ , to conclude that

$\int_0^{ \frac{\pi}{4} } \ln ( 1 + \tan \theta ) \, d \theta = \int_{ \frac{\pi}{4} } ^ 0 \ln ( 1 + \tan ( \pi/4 - \alpha ))\, - d \alpha \\ = \int_0 ^ { \frac \pi { 4} } \ln ( 1 + \tan ( \pi - \alpha ) ) \, d \alpha.$

Calvin Lin Staff - 6 years, 7 months ago

thanx.i saw bt can't remembr d soln m8hod.

saumik dey - 7 years, 2 months ago

It should be 22.5*0.6931=15.59

Amartya Anshuman - 7 years, 2 months ago

How

Arijit Banerjee - 7 years, 1 month ago

any other solutions

Tim Hooper - 6 years, 9 months ago

How $1 + \tan( \theta ) = 1 + tan( \pi/4 - \theta )$ ?

Carlos David Nexans - 6 years, 10 months ago

property of definite integrals, $\int _{ b }^{ a }{ f(x) } { =\int _{ b }^{ a }{ f(a+b-x) } }$

Aman Gautam - 6 years, 7 months ago

Integrate it!

The answer is 0.272.

1 solution

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