Question

Evalaute $\underset{0}{\overset{\frac{\pi }{4}}{\int }}\frac{\sin x+\cos x}{9+16(1-{(\sin x-\cos x)}^2)}dx$

Answer 1

Let $t=\sin x-\cos x\Rightarrow dt=(\cos x+\sin x)dx$

${\int }_0^{\frac{\pi }{4}}\frac{(\sin x+\cos x)dx}{9+16(1-{(\sin x-\cos x)}^2)}$

$={\int }_0^{\frac{\pi }{4}}\frac{dt}{9+16(1-t^2)}$

$={\int }_0^{\frac{\pi }{4}}\frac{dt}{9+16-16t^2}$

$={\int }_0^{\frac{\pi }{4}}\frac{dt}{25-16t^2}$

$={\int }_0^{\frac{\pi }{4}}\frac{dt}{16(\frac{25}{16}-t^2)}$

$=\frac{1}{16}{\int }_0^{\frac{\pi }{4}}\frac{dt}{(\frac{25}{16}-t^2)}$

We know that $\int \frac{dx}{a^2-x^2}=\frac{1}{2a}\ln \left|\frac{a+x}{a-x}\right|+c$

$={[\frac{1}{16}\times \frac{1}{2\times \frac{5}{4}}\ln \left|\frac{\frac{5}{4}+t}{\frac{5}{4}-t}\right|]}_0^{\frac{\pi }{4}}$

$=\frac{1}{16}\times \frac{4}{2\times 5}{\left[\ln \left|\frac{\frac{5+4t}{4}}{\frac{5-4t}{4}}\right|\right]}_0^{\frac{\pi }{4}}$

$=\frac{1}{40}{\left[\ln \left|\frac{\frac{5+4(\sin x-\cos x)}{4}}{\frac{5-4(\sin x-\cos x)}{4}}\right|\right]}_0^{\frac{\pi }{4}}$

$=\frac{1}{40}{\left[\ln \left|\frac{5+4(\sin x-\cos x)}{5-4(\sin x-\cos x)}\right|\right]}_0^{\frac{\pi }{4}}$

$=\frac{1}{40}[\ln \left|\frac{5+4(\sin \frac{\pi }{4}-\cos \frac{\pi }{4})}{5-4(\sin \frac{\pi }{4}-\cos \frac{\pi }{4})}\right|-\ln \left|\frac{5+4(\sin 0-\cos 0)}{5-4(\sin 0-\cos 0)}\right|]$

$=\frac{1}{40}[\ln \left|\frac{5+0}{5-0}\right|-\ln \left|\frac{5+4(-1)}{5-4(0-1)}\right|]$

$=\frac{1}{40}[\ln 1-\ln \frac{5-4}{5+4}]$

$=\frac{1}{40}[0-\ln \frac{1}{9}]$

$=-\frac{1}{40}\ln 3^{-2}=\frac{2}{40}\ln 3=\frac{1}{20}\ln 3$