If $\int_{0}^{a} 1/(1+4x^{2}) dx=\pi /8$ ,then a=?

We have  \[\int_{0}^{a} 1/(1+4x^{2}) dx=\pi /8\] and our aim is to find the value of a .

So in order find a ,at first we have solve the integral or to find the value of given integral.For that consider the formulae $\mathbf{\int \frac{1}{a^{2}+(bx)^{2}} dx= \frac{1}{ab}\tan ^{-1}\frac{bx}{a}} + C$ 

Now consider the integral 

\[\int_{0}^{a} 1/(1+4x^{2}) dx=\int_{0}^{a} 1/(1^2+(4x)^{2}) dx\]

Here a=1,b=2  now on integrating we will get 

\[\int_{0}^{a} \frac{1}{(1+4x^{2})} dx=\left [\frac{1}{1.2} \tan ^{-1} \frac{2x}{1} \right ]_{0}^{a} =\frac{1}{2}[\tan ^{-1} 2a - \tan ^{-1} 0] =\frac{1}{2}\tan ^{-1} 2a \; \: (\because \tan ^{-1} 0=0)\]

\[\therefore \int_{0}^{a} \frac{1}{(1+4x^{2})} dx=\frac{1}{2}\tan ^{-1} 2a \;\] 

Also we have  $\int_{0}^{a} 1/(1+4x^{2}) dx=\pi /8 $.

Now by using these two equations we can write

\[\begin{align*} \frac{1}{2}\tan ^{-1} 2a \; = \; \pi /8 &\Rightarrow \tan ^{-1} 2a= 2\frac{\pi }{8} =\frac{\pi }{4}\\ &\Rightarrow 2a=\tan \frac{\pi }{4} \\ &\Rightarrow 2a=1 \\ &\Rightarrow a=1/2 \end{align*}\]

Hence value of a=1/2.

 

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