Question

If an integral expression is given as $\int {\dfrac{{dx}}{{\sqrt {16 - 9{x^2}} }}} = A{\sin ^{ - 1}}\left( {Bx} \right) + C$ then A + B =
$\left( a \right)\dfrac{9}{4}$
$\left( b \right)\dfrac{{19}}{4}$
$\left( c \right)\dfrac{3}{4}$
$\left( d \right)\dfrac{{13}}{{12}}$

Answer 1

In this particular question use the direct integration formula which is given as $\int {\dfrac{{dx}}{{\sqrt {{a^2} - {x^2}} }} = } {\sin ^{ - 1}}\left( {\dfrac{x}{a}} \right) + C$ where C is some arbitrary integration constant, by converting the given integration into standard form so use these concepts to reach the solution of the question.

Complete step-by-step solution:
Given integral:
$\int {\dfrac{{dx}}{{\sqrt {16 - 9{x^2}} }}} = A{\sin ^{ - 1}}\left( {Bx} \right) + C$
Now we have to find out the value of A + B.
Consider the LHS of the given integral we have,
$\Rightarrow \int {\dfrac{{dx}}{{\sqrt {16 - 9{x^2}} }}}$
So first convert the integral into standard form by taking square root of 9 common from the denominator of the integral so we have,
$\Rightarrow \int {\dfrac{{dx}}{{\sqrt 9 \sqrt {\dfrac{{16 - 9{x^2}}}{9}} }}}$
Now simplify this we have,
$\Rightarrow \int {\dfrac{{dx}}{{\sqrt 9 \sqrt {\dfrac{{16}}{9} - {x^2}} }}}$
$\Rightarrow \int {\dfrac{{dx}}{{3\sqrt {{{\left( {\dfrac{4}{3}} \right)}^2} - {x^2}} }}}$
$\Rightarrow \dfrac{1}{3}\int {\dfrac{{dx}}{{\sqrt {{{\left( {\dfrac{4}{3}} \right)}^2} - {x^2}} }}}$
Now as we know that $\int {\dfrac{{dx}}{{\sqrt {{a^2} - {x^2}} }} = } {\sin ^{ - 1}}\left( {\dfrac{x}{a}} \right) + C$ where C is some arbitrary integration, so use this property in the above equation we have,
Where $a = \dfrac{4}{3}$
$\Rightarrow \dfrac{1}{3}{\sin ^{ - 1}}\left( {\dfrac{x}{{\dfrac{4}{3}}}} \right) + C$
$\Rightarrow \dfrac{1}{3}{\sin ^{ - 1}}\left( {\dfrac{{3x}}{4}} \right) + C$
Now compare this with given value of the integral,
$\Rightarrow \dfrac{1}{3}{\sin ^{ - 1}}\left( {\dfrac{{3x}}{4}} \right) + C = A{\sin ^{ - 1}}\left( {Bx} \right) + C$
Now on comparing we have,
$\Rightarrow A = \dfrac{1}{3},B = \dfrac{3}{4}$
So, $A + B = \dfrac{1}{3} + \dfrac{3}{4} = \dfrac{{13}}{{12}}$
So this is the required answer.
Hence option (d) is the correct answer.

Note: Whenever we face such types of questions the key concept we have to remember is that always recall the basic integration formula which is the key to solve this problem otherwise we cannot, so first convert the given integral into standard form as above then apply the standard integral formula as above then compare the result with the given value of the integral we will get the values of A and B, then calculate the sum of A and B we will get the required answer.