Question

If ${\int }_0^{\pi /2}\frac{dx}{3+4\sin x}dx=\frac{1}{\sqrt{\left(k\right)}}\log \left(\frac{4+\sqrt{7}}{3}\right)$ then find the value of $k$.

Answer 1

Let $I={\int }_0^{\pi /2}\frac{dx}{3+4\sin x}={\int }_0^{\pi /2}\frac{dx}{3+8\sin \left(x/2\right)\cos \left(x/2\right)}$

$={\int }_0^{\pi /2}\frac{{\sec }^2\left(x/2\right)dx}{3(1+{\tan }^2\left(x/2\right))+8\tan \left(x/2\right)}={\int }_0^{\pi /2}\frac{{\sec }^2\left(x/2\right)dx}{3{\tan }^2\left(x/2\right)+8\tan \left(x/2\right)+3}$

Substitute $\tan \left(x/2\right)=t\Rightarrow \left(1/2\right){\sec }^2\left(x/2\right)dx=dt$

Therefore
$I={\int }_0^1\frac{2dt}{3t^2+8t+3}=\frac{2}{3}{\int }_0^1\frac{dt}{t^2+\frac{8}{3}t+1}=\frac{2}{3}{\int }_0^1\frac{dt}{{(t+\frac{4}{3})}^2-{\left[\sqrt{\left(7\right)/3}\right]}^2}$
$=\frac{2}{3}.\frac{1}{2.\left[\sqrt{\left(7\right)/3}\right]}\left[\log \frac{t+\frac{4}{3-\left(\sqrt{7/3}\right)}}{t+\frac{4}{3}+\left[\sqrt{\left(7\right)/3}\right]}\right]$
$=\frac{1}{\sqrt{\left(7\right)}}[\log \frac{1+\frac{4}{3}-\left(\sqrt{7/3}\right)}{1+\frac{4}{3}+\left[\sqrt{\left(7\right)/3}\right]}-\log \frac{\frac{4}{3}-\left(\sqrt{7/3}\right)}{\frac{4}{3}+\left[\sqrt{\left(7\right)/3}\right]}]$
$=\frac{1}{\sqrt{\left(7\right)}}\log \frac{7-\sqrt{7}}{7+\sqrt{\left(7\right)}}-\log \frac{-4\sqrt{7}}{4+\sqrt{\left(7\right)}}=\frac{1}{\sqrt{\left(7\right)}}\log \left(\frac{\sqrt{7}-1}{\sqrt{\left(7\right)}+1}\right).\left(\frac{4+\sqrt{7}}{4-\sqrt{\left(7\right)}}\right)$
$=\frac{1}{\sqrt{\left(7\right)}}\log [\frac{{(\sqrt{7}-1)}^2}{7-1}\times \frac{{(4+\sqrt{7})}^2}{16-7}]=\frac{1}{\sqrt{\left(7\right)}}\log [\frac{4-\sqrt{7}}{3}\times \frac{23+8\sqrt{7}}{9}]$
$=\frac{1}{\sqrt{\left(7\right)}}\log \left(\frac{36+9\sqrt{7}}{27}\right)=\frac{1}{\sqrt{\left(7\right)}}\log \left(\frac{4+\sqrt{7}}{3}\right)$