MHB How Can You Effectively Evaluate a Challenging Definite Integral?

[anemone]

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

 

[MarkFL]

My solution:

Spoiler

We are given to evaluate:

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

Let:

$$u=x-3\,\therefore\,du=dx$$

and we have:

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

Using the property of definite integrals:

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

we may write:

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

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

$$I=\int_{0}^{1}\,du=_0^1=1-0=1$$

 

[anemone]

My solution:

Spoiler

We are given to evaluate:

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

Let:

$$u=x-3\,\therefore\,du=dx$$

and we have:

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

Using the property of definite integrals:

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

we may write:

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

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

$$I=\int_{0}^{1}\,du=_0^1=1-0=1$$

That's a smart approach, MarkFL! Well done!(Clapping)

 

[Mathwebster]

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 $I = 1$