 Hi and welcome to the session. Let us discuss the following question. Question says, choose the correct answer. Integral dx upon square root of 9x minus 4x square equals a 1 upon 9 sin inverse 9x minus 8 upon 8 plus c b 1 upon 2 sin inverse 8x minus 9 upon 9 plus c c 1 upon 3 sin inverse 9x minus 8 upon 8 plus c d 1 upon 2 sin inverse 9x minus 8 upon 9 plus c We have to choose correct answer from a b c and d First of all, let us understand that integral dx upon square root of a square minus x square is equal to sin inverse x upon a plus c This is the key idea to solve the given question. Let us now start with the solution Now we have to find integral dx upon square root of 9x minus 4x square First of all, let us consider 9x minus 4x square. It is equal to minus 4 multiplied by minus 9x upon 4 plus x square Now this is further equal to minus 4 multiplied by x square minus 2 multiplied by 9 upon 8x plus 9 upon 8 square minus 9 upon 8 square Clearly we can see minus 4 is as it is x square is as it is in this bracket Now we will multiply and divide this term by 2 and we get this term Now here we have added and subtracted square of 9 upon 8 Now in this bracket these three terms are a square plus b square minus 2 a b And we know a square plus b square minus 2 a b is equal to a minus b whole square So we can write it as x minus 9 upon 8 whole square And we will write this term as it is Now by multiplying this minus sign inside the bracket we get 4 multiplied by square of 9 upon 8 minus x minus 9 upon 8 whole square So we get 9x minus 4x square is equal to 4 multiplied by square of 9 upon 8 minus square of x minus 9 upon 8 Now integral dx upon square root of 9x minus 4x square is equal to integral dx upon square root of 4 multiplied by square of 9 upon 8 minus square of x minus 9 upon 8 Now this integral can be further written as integral dx upon 2 multiplied by square root of 9 upon 8 square minus x minus 9 upon 8 whole square We know square root of 4 multiplied by this bracket is equal to square root of 4 multiplied by square root of this bracket And we know square root of 4 is 2 so we can write this expression as 2 multiplied by square root of 9 upon 8 square minus x minus 9 upon 8 whole square And we know this is the denominator of the given integrand so we can write denominator of the given integrand is equal to 2 multiplied by square root of square of 9 upon 8 minus x minus 9 upon 8 whole square Now we know derivative of x minus 9 upon 8 is equal to 1 so we will substitute x minus 9 upon 8 equal to t then differentiating both the sides with respect to x we get dx is equal to dt Now we get integral dx upon 2 multiplied by square root of 9 upon 8 square minus x minus 9 upon 8 whole square is equal to integral of dt upon 2 multiplied by square root of square of 9 upon 8 minus t square we know dx is equal to dt and x minus 9 upon 8 is equal to t Now this integral can be further written as 1 upon 2 multiplied by integral dt upon square root of square of 9 upon 8 minus t square Now using the formula of integral given in key idea we get this integral is equal to 1 upon 2 multiplied by sin inverse t upon 9 upon 8 plus c Here clearly we can see the variable x has been replaced by t and a is equal to 9 upon 8 Now simplifying further this expression is equal to 1 upon 2 multiplied by sin inverse 8t upon 9 plus c Now we know t is equal to x minus 9 upon 8 so substituting x minus 9 upon 8 for t here we get 1 upon 2 multiplied by sin inverse 8 upon 9 multiplied by x minus 9 upon 8 plus c Now multiplying 8 upon 9 by this bracket we get 1 upon 2 sin inverse 8x minus 9 upon 9 plus c So we get integral dx upon square root of 9x minus 4x square is equal to 1 upon 2 sin inverse 8x minus 9 upon 9 plus c Where c is the constant of integration so the correct answer is b This is our required answer this completes the session hope you understood the solution take care and have a nice day