Definite Integral Challenge

[anemone]

Evaluate \(\displaystyle \int_2^4 \frac{\sqrt{\ln (9-x)}\,dx}{\sqrt{\ln (9-x)}+\sqrt{\ln (x+3)}}\)

 

[MarkFL]

My solution:

Spoiler

We are given to evaluate:

\(\displaystyle I=\int_2^4 \frac{\sqrt{\ln (9-x)}\,dx}{\sqrt{\ln (9-x)}+\sqrt{\ln (x+3)}}\)

Let:

\(\displaystyle u=x-3\,\therefore\,du=dx\)

and we have:

\(\displaystyle I=\int_{-1}^{1}\frac{\sqrt{\ln(6-u)}}{\sqrt{\ln(6-u)}+\sqrt{\ln(6+u)}}\,du\)

Using the property of definite integrals:

\(\displaystyle \int_{-a}^a f(x)\,dx=\int_0^a \left[f(x)+f(-x) \right]\,dx\)

we may write:

\(\displaystyle I=\int_{0}^{1}\frac{\sqrt{\ln(6-u)}}{\sqrt{\ln(6-u)}+\sqrt{\ln(6+u)}}+\frac{\sqrt{\ln(6+u)}}{\sqrt{\ln(6+u)}+\sqrt{\ln(6-u)}}\,du\)

\(\displaystyle I=\int_{0}^{1}\frac{\sqrt{\ln(6-u)}+\sqrt{\ln(6+u)}}{\sqrt{\ln(6-u)}+\sqrt{\ln(6+u)}}\,du\)

\(\displaystyle I=\int_{0}^{1}\,du=_0^1=1-0=1\)

 

[anemone]

My solution:

Spoiler

We are given to evaluate:

\(\displaystyle I=\int_2^4 \frac{\sqrt{\ln (9-x)}\,dx}{\sqrt{\ln (9-x)}+\sqrt{\ln (x+3)}}\)

Let:

\(\displaystyle u=x-3\,\therefore\,du=dx\)

and we have:

\(\displaystyle I=\int_{-1}^{1}\frac{\sqrt{\ln(6-u)}}{\sqrt{\ln(6-u)}+\sqrt{\ln(6+u)}}\,du\)

Using the property of definite integrals:

\(\displaystyle \int_{-a}^a f(x)\,dx=\int_0^a \left[f(x)+f(-x) \right]\,dx\)

we may write:

\(\displaystyle I=\int_{0}^{1}\frac{\sqrt{\ln(6-u)}}{\sqrt{\ln(6-u)}+\sqrt{\ln(6+u)}}+\frac{\sqrt{\ln(6+u)}}{\sqrt{\ln(6+u)}+\sqrt{\ln(6-u)}}\,du\)

\(\displaystyle I=\int_{0}^{1}\frac{\sqrt{\ln(6-u)}+\sqrt{\ln(6+u)}}{\sqrt{\ln(6-u)}+\sqrt{\ln(6+u)}}\,du\)

\(\displaystyle I=\int_{0}^{1}\,du=_0^1=1-0=1\)

That's a smart approach, MarkFL! Well done!

 

[Jester]

A little more direct

Spoiler

Let $x = 6-u$ so the integral become

$I = \displaystyle - \int_4^2 \dfrac{\sqrt{\ln (3+u)}}{ \sqrt{\ln (3+u)} +\sqrt{\ln (9-u)}}\;du$

Replace $u$ with $x$ and add to the original and like Mark said $2I = \int_2^4 1dx$ gives $I = 1$