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.
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