- Thread starter
- Admin
- #1
- Feb 14, 2012
- 3,967
Evaluate \(\displaystyle \int_2^4 \frac{\sqrt{\ln (9-x)}\,dx}{\sqrt{\ln (9-x)}+\sqrt{\ln (x+3)}}\)
My solution:
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:
\(\displaystyle I=\int_{0}^{1}\frac{\sqrt{\ln(6-u)}}{\sqrt{\ln(6-u)}+\sqrt{\ln(6+u)}}+\frac{\sqrt{\ln(6+u)}}{\sqrt{\ln(6+u)}+\sqrt{\ln(6-u)}}\,du\)
\(\displaystyle I=\int_{0}^{1}\frac{\sqrt{\ln(6-u)}+\sqrt{\ln(6+u)}}{\sqrt{\ln(6-u)}+\sqrt{\ln(6+u)}}\,du\)
\(\displaystyle I=\int_{0}^{1}\,du=_0^1=1-0=1\)