Question

Evaluate the given integral
$\Rightarrow \int_{ - 1}^0 {\dfrac{{dt}}{{{5^2} - {{\left( {4t} \right)}^2}}}}$

Answer 1

In this particular type of question use the concept that first simplify the integration using substitution method then according to this substitution change the variable then apply the standard integral formula so use these concepts to reach the solution of the question.

Complete step-by-step answer:
Given integral
$\int_{ - 1}^0 {\dfrac{{dt}}{{{5^2} - {{\left( {4t} \right)}^2}}}}$.............. (1)
Now substitute 4t = v........... (2)
Now differentiate equation (2) w.r.t t we have,
$\Rightarrow 4dt = dv$.................. (3)
Now change the integral limits we have,
When, t = -1
So, v = 4t = 4(-1) = -4
When t = 0
So, v = 4t = 4(0) = 0
So the integration limits become (-4 to 0)
Now substitute the values from equation (2) and (3) in equation (1) and change the integration limits we have,
$\Rightarrow \int_{ - 4}^0 {\dfrac{{\dfrac{{dv}}{4}}}{{{5^2} - {v^2}}}}$
$\Rightarrow \dfrac{1}{4}\int_{ - 4}^0 {\dfrac{{dv}}{{{5^2} - {v^2}}}}$
Now as we know the direct integral formula of ($\int {\dfrac{{dx}}{{{a^2} - {x^2}}}} = \dfrac{1}{{2a}}\log \left| {\dfrac{{a + x}}{{a - x}}} \right| + c$), where c is some arbitrary integration constant.
So use this formula in the above integral we have,
$\Rightarrow \dfrac{1}{4}\left[ {\dfrac{1}{{2\left( 5 \right)}}\log \left| {\dfrac{{5 + v}}{{5 - v}}} \right|} \right]_{ - 4}^0$
Now apply integral limit we have,
$\Rightarrow \dfrac{1}{4}\left[ {\dfrac{1}{{10}}\log \left| {\dfrac{{5 + 0}}{{5 - 0}}} \right| - \dfrac{1}{{10}}\log \left| {\dfrac{{5 + \left( { - 4} \right)}}{{5 - \left( { - 4} \right)}}} \right|} \right]$
$\Rightarrow \dfrac{1}{4}\left[ {\dfrac{1}{{10}}\log \left( 1 \right) - \dfrac{1}{{10}}\log \left( {\dfrac{1}{9}} \right)} \right]$
Now as we know that the value of log1 = 0 and use that $\log \left( {\dfrac{1}{a}} \right) = - \log a$ so we have
$\Rightarrow \dfrac{1}{4}\left[ {\dfrac{1}{{10}}\left( 0 \right) + \dfrac{1}{{10}}\log 9} \right]$
$\Rightarrow \dfrac{1}{{40}}\log 9 = \dfrac{1}{{40}}\log {3^2}$
Now as we know that $\log {a^b} = b\log a$ so we have,
$\Rightarrow \dfrac{1}{{40}}\log 9 = \dfrac{1}{{40}}\log {3^2} = \dfrac{2}{{40}}\log 3 = \dfrac{1}{{20}}\log 3$
So this is the required value of the integral.

Note: Whenever we face such types of questions the key concept we have to remember is that always recall the standard integration formulas, the formula which is used above is given as, $\int {\dfrac{{dx}}{{{a^2} - {x^2}}}} = \dfrac{1}{{2a}}\log \left| {\dfrac{{a + x}}{{a - x}}} \right| + c$, so first simplify the integration as above then apply this formula then apply integration limits as above and simplify we will get the required answer.