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Prove that 

\int_{0}^{a}f(x)dx=\int_{0}^{a}f(a-x)dx

and hence evaluate 

\int_{0}^{\frac{\pi}{2}}\frac{x}{\sin x+\cos x}dx

 

 

 

 
 
 
 
 

Answers (1)

L.H.S = \int_{0}^{a}f(x)dx

Let a-x=v

-1=\frac{\mathrm{d} v}{\mathrm{d} x}\: \: for\: \: x=0,v=a

x=a,v=0

Then,

R.H.S= \int_{a}^{0}-f(a-v)dv\Rightarrow \int_{0}^{a}f(a-v)(dv)\: \: \: \left [ \because \int_{a}^{b}f(x)dx=-\int_{b}^{a}f(x)dx \right ]

Now, replacing v by x, 

 =\int_{0}^{a}f(a-x)dx

Hence, \int_{0}^{a}f(x)dx=\int_{0}^{a}f(a-x)dx  Hence Proved.

Now, Let I=\int_{0}^{\frac{\pi}{2}}\frac{x}{\sin x+\cos x}dx\: \: \: \: \: -(i)

I=\int_{0}^{\frac {\pi}{2}}\frac{\frac{\pi}{2}-x}{\cos x +\sin x}dx\: \: \: (ii)

Adding equations (i) and (ii), we get 

2I=\int_{0}^{\frac{\pi}{2}}\frac{x+\frac{\pi}{2}-x}{\sin x +\cos x}dx

2I=\int_{0}^{\frac{\pi}{2}}\frac{\frac{\pi}{2}}{\sin x +\cos x}dx

2I=\frac{\pi}{2}\int_{0}^{\frac{\pi}{2}}\frac{1}{\sin x +\cos x}dx

2I=\frac{\pi}{2}\int_{0}^{\frac{\pi}{2}}\frac{1}{\frac{2\tan \frac{x}{2}}{1+\tan^2 \frac{x}{2}}+\frac{1-\tan^2 \frac{x}{2}}{1+\tan^2 \frac{x}{2}}}dx

2I=\frac{\pi}{2}\int_{0}^{\frac{\pi}{2}}\frac{1+\tan^2 \frac{x}{2}}{2\tan \frac{x}{2}+1-\tan^2 \frac{x}{2}}dx

2I=\frac{\pi}{2}\int_{0}^{\frac{\pi}{2}}\frac{\sec^2 \frac{x}{2}}{-\tan^2 \frac{x}{2}+2\tan \frac{x}{2}+1}dx

Ley, \tan \frac{x}{2}=t

Then, \frac{\mathrm{d} }{\mathrm{d} x}(\tan \frac{x}{2})=\frac{\mathrm{d} }{\mathrm{d} t}(t)

\frac{1}{2}\sec^2\frac{x}{2}dx=dt

\sec^2\frac{x}{2}dx=2dt

Also x=0

\Rightarrow t=\tan 0 = 0 \: \: \: \: \: \: \: \: \: \:

and x=\frac{\pi}{2} , t=1

\therefore 2I=\frac{\pi}{2}\int_{0}^{1}\frac{2dt}{-t^2+2t+1} 

2I= \pi\int_{0}^{1}\frac{\mathrm{d} t}{\mathrm{}-t^2+2t+1 }

                                       \Rightarrow \pi\int_{0}^{1}\frac{\mathrm{d} t}{\mathrm{}-(t^2-2t-1) }

 \Rightarrow \pi\int_{0}^{1}\frac{\mathrm{d} t}{\mathrm{}-\left [ (t-1)^2-2 \right ] }\Rightarrow \pi\int_{0}^{1}\frac{dt}{(\sqrt2)^2-(t-1)^2}

\Rightarrow \pi \times \frac{1}{2\sqrt{2}}\left [ \log \left | \frac{\sqrt{2}+t-1}{\sqrt{2}-t+1} \right | \right ]_{_{}^{0}}^{1}\Rightarrow \frac{\pi}{2}\times \frac{1}{\sqrt{2}}\left | \log 1-\log \left ( \frac{\sqrt{2}-1}{\sqrt{2}+1} \right )\right |

\Rightarrow \frac{-\pi}{2\sqrt{2}}\log \left ( \frac{\sqrt{2}-1}{\sqrt{2}+1} \right )\Rightarrow \frac{\pi}{2\sqrt2}\log \left ( \frac{\sqrt{2}+1}{\sqrt{2}-1} \right )

\Rightarrow 2I=\frac{\pi}{2\sqrt2}\log \left \{ \frac{\left ( \sqrt2+1 \right )^2}{(\sqrt2-1)(\sqrt2+1)} \right \}

=I=\frac{\pi}{4\sqrt2}\log (\sqrt2+1)^2

\Rightarrow \frac{2\pi}{4\sqrt2}\log (\sqrt2+1)

I=\frac{\pi}{2\sqrt2}\log(\sqrt2+1)

Posted by

Ravindra Pindel

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