Prove that
\int_{a}^{b}f\left ( x \right )dx= \int_{a}^{b}f\left ( a+b-x \right )dx  and hence evaluate
\int_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{dx}{1+\sqrt{\tan x}}

 

 

 

 
 
 
 
 

Answers (1)

Let a+b-x= t                  given: \int_{a}^{b}f\left ( x \right )dx= \int_{a}^{b}f\left ( a+b-x \right )dx
\Rightarrow dx= -dt
when x= a,t= b\: and\: \: x= b,t= a
\int_{a}^{b}f\left ( x \right )dx= \int_{a}^{b}f\left ( a+b-x \right )dt
= \int_{a}^{b}f\left ( a+b-t \right )dt\: \: \left [ \because \int_{a}^{b}f\left ( x \right )dx= -\int_{b}^{a}f\left ( x \right )dx \right ]
= \int_{a}^{b}f\left ( a+b-x \right )dx
Let I= \int_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{dx}{1+\sqrt{\tan x}}=\int_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{\sqrt{\cos xdx}}{\sqrt{\cos x}+\sqrt{\sin x}}
I= \int_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{\sqrt{\cos \left ( \frac{\pi }{3}+\frac{\pi }{6} -x\right )dx}}{\sqrt{\cos \left ( \frac{\pi }{3}+\frac{\pi }{6} -x\right )+\sqrt{\sin \left ( \frac{\pi }{3}+\frac{\pi }{6}-x \right )}}}
I= \int_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{\sqrt{\sin x\, dx}}{\sqrt{\sin x}+\sqrt{\cos x}}---\left ( iii \right )
2I= \int_{\frac{\pi }{6}}^{\frac{\pi }{3}}dx= \left [ x \right ]_{\frac{\pi }{6}}^{\frac{\pi }{3}}= \frac{\pi }{3}-\frac{\pi }{6}= \frac{\pi }{6}
    \Rightarrow = \frac{\pi }{12}

 

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