Question

If $\int\limits_{0}^{a}{\dfrac{1}{4+{{x}^{2}}}dx=\dfrac{\pi }{8},}$ find the value of a.

Answer 1

Hint: What we will do is firstly we will solve the integration $\int\limits_{0}^{a}{\dfrac{1}{4+{{x}^{2}}}dx}$ using various integration properties and formula, then we will put value of integration equals to\[\dfrac{\pi }{8}\], and then we will solve for the value of a.

Complete step by step answer:
To answer such a question, we must know some properties of definite integration and indefinite integration.
One of the most important property of definite integration is \[\int\limits_{a}^{b}{f(x)dx=F(a)-F(b)}\] ,where F is the integration of f ( x ) and a is lower limit and b is upper limit. That is for example we have $f(x)=x$ and we have to find integration of f ( x ) = x from lower limit 2 to upper limit 3, then
 \[\int\limits_{2}^{3}{xdx=\left[ \dfrac{{{x}^{2}}}{2} \right]}_{2}^{3}\]
\[\int\limits_{2}^{3}{xdx=}\dfrac{{{(3)}^{2}}}{2}-\dfrac{{{(2)}^{2}}}{2}=2.5\], as integration of x is $\dfrac{{{x}^{2}}}{2}$ because $\int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n}+C}$ , where C is constant which does not appear I definite integral but does appear in indefinite integral.
And, another most important formula of indefinite integration is $\int{\dfrac{1}{{{x}^{2}}+{{a}^{2}}}=\dfrac{1}{a}{{\tan }^{-1}}\left( \dfrac{x}{a} \right)}$ , where a is any constant.
Now, in question it is given that, $\int\limits_{0}^{a}{\dfrac{1}{4+{{x}^{2}}}dx=\dfrac{\pi }{8},}$and we are asked to evaluate the value of a
So, firstly we solve the $\int\limits_{0}^{a}{\dfrac{1}{4+{{x}^{2}}}dx}$.
We can re – write $\int\limits_{0}^{a}{\dfrac{1}{4+{{x}^{2}}}dx}$as ,
$\int\limits_{0}^{a}{\dfrac{1}{4+{{x}^{2}}}dx=\int\limits_{0}^{a}{\dfrac{1}{{{2}^{2}}+{{x}^{2}}}dx}}$
$\int\limits_{0}^{a}{\dfrac{1}{{{2}^{2}}+{{x}^{2}}}dx}=\left[ {{\tan }^{-1}}\left( \dfrac{x}{2} \right) \right]_{0}^{a}$, by using formula$\int{\dfrac{1}{{{x}^{2}}+{{a}^{2}}}=\dfrac{1}{a}{{\tan }^{-1}}\left( \dfrac{x}{a} \right)}$, where constant is 2.
$\left[ {{\tan }^{-1}}\left( \dfrac{x}{2} \right) \right]_{0}^{a}=\dfrac{1}{2}{{\tan }^{-1}}\left( \dfrac{a}{2} \right)-\dfrac{1}{2}{{\tan }^{-1}}\left( \dfrac{0}{2} \right)$
$\dfrac{1}{2}{{\tan }^{-1}}\left( \dfrac{a}{2} \right)-\dfrac{1}{2}{{\tan }^{-1}}\left( 0 \right)=\dfrac{1}{2}{{\tan }^{-1}}\left( \dfrac{a}{2} \right)$
Now, it given that $\int\limits_{0}^{a}{\dfrac{1}{4+{{x}^{2}}}dx=\dfrac{\pi }{8},}$
So, $\dfrac{1}{2}{{\tan }^{-1}}\left( \dfrac{a}{2} \right)=\dfrac{\pi }{8}$
On simplifying, we get
${{\tan }^{-1}}\left( \dfrac{a}{2} \right)=\dfrac{\pi }{4}$
We can re – write above equation as
$\left( \dfrac{a}{2} \right)=\tan \dfrac{\pi }{4}$
As, $\tan \dfrac{\pi }{4}=1$,
So, $\left( \dfrac{a}{2} \right)=1$
Using cross multiplication, we get
a = 2

Hence, $\int\limits_{0}^{a}{\dfrac{1}{4+{{x}^{2}}}dx=\dfrac{\pi }{8},}$ if a = 2.

Note: While integrating the definite integral always use the correct formula to evaluate the integration and always try to skip calculation error as it may change the answer of the solution or may make the solution more complex. One must know all the properties and formula of indefinite and definite integrals.