 Hi and welcome to the session. Let us discuss the following question which says evaluate integral of square root of a-x upon a-x dx from a-a to a. Before moving on to the solution let's recall some formulas which will be helpful in solving this question. So first upon we have integral of 1 upon square root of a square minus x square dx is equal to sin inverse x upon a plus c and integral of f of x dx from a to a is equal to 0. So this is the key idea for this question. Now let's move on to its solution. We need to evaluate integral of square root of a minus x upon a plus x dx from minus a to a. So let us assume that this is equal to i. Now we need to find out the value of i. So first of all in this let us multiply the numerator and denominator by square root of a minus x. So this will be equal to integral of square root of a minus x upon a plus x into square root of a minus x upon square root of a minus x dx from minus a to a. So we got this by multiplying numerator and denominator by square root of a minus x. So this will be equal to integral of a minus x upon square root of a square minus x square dx from minus a to a as in the numerator square root of a minus x into square root of a minus x will be a minus x and in denominator square root of a plus x into square root of a minus x will be square root of a square minus x square because a plus b into a minus b is a square minus b square and this will be equal to integral of a upon square root of a square minus x square dx from minus a to a plus integral of minus x upon square root of a square minus x square dx from minus a to a and this can also be written as a into integral of 1 upon a square minus x square dx from minus a to a plus 1 by 2 into integral of minus 2x upon square root of a square minus x square dx from minus a to a. Now from the key idea we already know that integral of 1 upon square root of a square minus x square dx is equal to sin inverse x upon a plus c. So here we will get a into sin inverse x upon a from minus a to a plus 1 by 2 here we will assume this integral as i1. So we get 1 by 2 into i1 where i1 is integral of minus 2x upon square root of a square minus x square dx from minus a to a. Now let us solve i1 first so i1 is equal to integral of minus 2x upon square root of a square minus x square dx from minus a to a. Now here let us substitute a square minus x square equal to t so this implies that minus 2x dx is equal to dt. Now when x is equal to minus a then t will be equal to a square minus minus a square that is a square minus a square which is equal to 0 and when x is equal to a then t will be equal to a square minus a square that is 0. So integral i1 will become integral of dt upon square root of t from 0 to 0 and we know that integral of f of x dx from a to a is 0. So this will be equal to 0. Now here we have i1 equal to 0 so let us find i. Now we have i equal to a into sin inverse x upon a from minus a to a plus 1 by 2 into i1. So this will be equal to this will be equal to a into sin inverse a upon a minus sin inverse minus a upon a plus 1 by 2 into 0. So this gives a into sin inverse of 1 minus sin inverse of minus 1 plus 0 that is a into sin inverse of 1 plus sin inverse of 1 because sin inverse of minus theta is equal to minus of sin inverse theta. Now sin inverse 1 is equal to pi by 2 so i will be equal to a into pi by 2 plus pi by 2 because sin inverse 1 is equal to pi by 2. So this is equal to a into pi that is pi a thus integral of a square root of a minus x upon a plus x dx from minus a to a is equal to pi into a. So this is the required answer to this question. With this we finish this session. Hope you must have understood the question. Goodbye, take care and have a nice day.