question_answer

Evaluate \[\int{\frac{\cos \,x}{(1-sinx)(2-\sin x)}dx.}\]

OR

Prove that \[\int_{0}^{\pi }{\frac{x}{(1+sin\,x)}}\,dx=\pi .\]

Answer:

Let        \[l=\int{\frac{\cos x}{(1-\sin x)\,\,(2-\sin x)}\,\,dx}\]

Put \[\sin x=t\Rightarrow \cos x=\frac{dt}{dx}\Rightarrow dx=\frac{dt}{\cos x}\]

\[\therefore \]\[l=\int{\frac{\cos x}{(1-t)\,\,(2-t)}\frac{dt}{\cos x}}\Rightarrow \int{\frac{1}{(1-t)\,\,(2-t)}\,\,dt}\]

\[=\int{\left[ \frac{A}{1-t}+\frac{B}{2-t} \right]}\,\,dt\]                ?(i)

\[\therefore \]\[\frac{1}{(1-t)\,\,(2-t)}=\frac{A\,\,(2-1)+B\,\,(1-t)}{(1-t)\,\,(2-t)}\]

\[\Rightarrow \]   \[1=2A-tA+B-Bt\]

\[\Rightarrow \]   \[1=1\,(2A+B)+t(-A-B)\]

On comparing the coefficients of t and constant term both sides, we get

\[2A+B=1\]and \[-\,A-B=0\]

On adding above equations, we get

\[A=1\]and then \[B=-\,1\]

\[\therefore \]      \[l=\int{\left( \frac{1}{1-t}-\frac{1}{2-t} \right)\,\,dt}\]   [from Eq. (i)]

\[=\int{\frac{1}{(1-t)}\,dt-\int{\frac{1}{(2-t)}\,\,dt}}\]

\[=\frac{\log |1-t|}{(-\,1)}-\frac{\log |2-t|}{(-\,1)}+C\]

\[=\log \left| \frac{2-t}{1-t} \right|+C=\log \left| \frac{2-\sin x}{1-\sin x} \right|+C\]

\[[put\,\,t=\sin x]\]

OR

Let \[l=\int_{0}^{\pi }{\frac{x}{1+\sin x}\,dx}\]                       ?(i)

\[=\int_{0}^{\pi }{\frac{\pi -x}{1+\sin (\pi -x)}\,dx\left[ \because \int_{0}^{a}{f\,(x)\,\,dx}=\int_{0}^{a}{f\,(a-x)\,dx} \right]}\] \[=\int_{0}^{\pi }{\frac{\pi -x}{1+\sin x}\,dx}\]                       ?(ii)

On adding Eqs. (i) and (ii), we get

\[2l=\int_{0}^{\pi }{\frac{\pi }{(1+\sin x)}\,d}x=\pi \int_{0}^{\pi }{\frac{1}{1+\sin x}\,dx}\]

\[=\pi \int_{0}^{\pi }{\frac{1-\sin x}{(1+\sin x)\,\,(1-\sin x)}\,dx}\]

[multiply numerator and denominator by \[(1-\sin x)\]]

\[2l=\pi \int_{0}^{\pi }{\frac{1-\sin x}{1-{{\sin }^{2}}x}\,dx}=\pi \int_{0}^{\pi }{\frac{1}{{{\cos }^{2}}x}\,dx-\pi \int_{0}^{\pi }{\frac{\sin x}{{{\cos }^{2}}x}\,dx}}\] \[[\because {{\sin }^{2}}x+{{\cos }^{2}}x=1]\]

\[\Rightarrow \]   \[2l=\pi \int_{0}^{\pi }{{{\sec }^{2}}x\,dx-\pi \int_{0}^{\pi }{\sec x\tan x\,dx}}\]

\[\Rightarrow \]   \[2l=\pi \,[\tan x-\sec x]_{0}^{\pi }\]

\[=\pi \,[\tan \pi -\sec \pi -(\tan 0-\sec 0)]\]

\[=\pi \,[0+1-0+1]=2\pi \]

\[\Rightarrow \]   \[l=\pi \]

\[\therefore \]      \[LHS=RHS\]         Hence proved.