Question

If $I_a={\int }_0^{\pi /2}\frac{dx}{2\cos x+\sin x+a}$ then the value of $I_a$ for

Answer 1

A) For $I_1={\int }_0^{\frac{\pi }{2}}\frac{dx}{2cosx+sinx+1}$
Let $I=\int \frac{dx}{2cosx+sinx+1}$
Substituting $\tan \frac{x}{2}=t\Rightarrow \frac{1}{2}{\sec }^2\frac{x}{2}dx=dt$
we get $\sin x=\frac{2t}{t^2+1},\cos x=\frac{1-t^2}{1+t^2},dx=\frac{2dt}{1+t^2}$
Then
$I=2\int \frac{dt}{-t^2+2t+3}=2\int \frac{dt}{4-{(t-1)}^2}=\frac{1}{2}\log \left|\frac{t-1+2}{t-1-2}\right|=\frac{1}{2}\log \left|\frac{t+1}{t-3}\right|=\frac{1}{2}\log \left|\frac{\tan \frac{x}{2}+1}{\tan \frac{x}{2}-3}\right|$
Hence
$I_1={\left[\frac{1}{2}\log \left|\frac{\tan \frac{x}{2}+1}{\tan \frac{x}{2}-3}\right|\right]}_0^{\frac{\pi }{2}}=\frac{1}{2}\log \left|\frac{1+1}{1-3}\right|-\frac{1}{2}\log \left|\frac{0+1}{0-3}\right|=\frac{1}{2}\log 3$
B) For $I_3={\int }_0^{\frac{\pi }{2}}\frac{dx}{2cosx+sinx+3}$
Let $I=\int \frac{dx}{2cosx+sinx+3}$
Substituting $\tan \frac{x}{2}=t\Rightarrow \frac{1}{2}{\sec }^2\frac{x}{2}dx=dt$
We get $\sin x=\frac{2t}{t^2+1},\cos x=\frac{1-t^2}{1+t^2},dx=\frac{2dt}{1+t^2}$
Then
$I=2\int \frac{dt}{t^2+2t+5}=2\int \frac{dt}{{(t+1)}^2+4}={\tan }^{-1}\frac{t+1}{2}={\tan }^{-1}\frac{\tan \frac{x}{2}+1}{2}$
Hence
$I_3={\left[{\tan }^{-1}\frac{\tan \frac{x}{2}+1}{2}\right]}_0^{\frac{\pi }{2}}={\tan }^{-1}1-{\tan }^{-1}\frac{1}{2}={\tan }^{-1}\left(\frac{1-\frac{1}{2}}{1+\frac{1}{2}}\right)={\tan }^{-1}\frac{1}{3}$
C) For $I_2={\int }_0^{\frac{\pi }{2}}\frac{dx}{2cosx+sinx+2}$
Let $I=\int \frac{dx}{2cosx+sinx+2}$
Taking $\tan \frac{x}{2}=t\Rightarrow \frac{1}{2}{\sec }^2\frac{x}{2}dx=dt$
we get
$\sin x=\frac{2t}{t^2+1},\cos x=\frac{1-t^2}{1+t^2},dx=\frac{2dt}{1+t^2}$
Therefore,
$I=\int \frac{dt}{t+2}=\log (t+2)=\log (\tan \frac{x}{2}+2)$
Hence
$I_2={\left[\log (\tan \frac{x}{2}+2)\right]}_0^{\frac{\pi }{2}}=\log 3-\log 2=\log \frac{3}{2}$
D) $I_4={\int }_0^{\frac{\pi }{2}}\frac{dx}{2cosx+sinx+4}$
Let $I=\int \frac{dx}{2cosx+sinx+4}$
Taking $\tan \frac{x}{2}=t\Rightarrow \frac{1}{2}{\sec }^2\frac{x}{2}dx=dt$, we get
$\sin x=\frac{2t}{t^2+1},\cos x=\frac{1-t^2}{1+t^2},dx=\frac{2dt}{1+t^2}$
Therefore,
$I=\int \frac{dt}{t^2+t+3}=\int \frac{dt}{{(t+\frac{1}{2})}^2+\frac{11}{4}}=\frac{2}{\sqrt{11}}{\tan }^{-1}\left(\frac{2t+1}{\sqrt{11}}\right)$
Hence
$I_2=\frac{2}{\sqrt{11}}{\left[{\tan }^{-1}\left(\frac{2\tan \frac{x}{2}+1}{\sqrt{11}}\right)\right]}_0^{\frac{\pi }{2}}=\frac{2}{\sqrt{11}}({\tan }^{-1}\frac{3}{\sqrt{11}}-{\tan }^{-1}\frac{1}{\sqrt{11}})$