 Hello and welcome to the session. Let us discuss the following questions. Questions raised. Choose the correct answer. The value of the integral from 1 upon 3 to 1 x minus x cube whole raised to the power 1 upon 3 dx upon x raised to the power 4 is a 6 b 0 c 3 d 4. We have to choose the correct answer from a b c and d. Let us now start with the solution. Now we have to find the definite integral from 1 upon 3 to 1 x minus x cube whole raised to the power 1 upon 3 upon x raised to the power 4 dx. Now first of all let us consider indefinite integral x minus x cube whole raised to the power 1 upon 3 upon x raised to the power 4 dx. Now this integral can be written as integral of x cube multiplied by 1 upon x square minus 1 whole raised to the power 1 upon 3 upon x raised to the power 4 dx. Now this is further equal to integral of x multiplied by 1 upon x square minus 1 whole raised to the power 1 upon 3 upon x raised to the power 4 dx. Now this x will cancel 1 x in the denominator and we get this integral as integral of 1 upon x square minus 1 whole raised to the power 1 upon 3 upon x cube dx. Now in this integral let us substitute x equal to cos theta. Then differentiating both the sides with respect to x we get 1 is equal to minus sin theta d theta upon dx. This implies dx is equal to minus sin theta d theta. Now integral 1 upon x square minus 1 whole raised to the power 1 upon 3 upon x cube dx is equal to integral 1 upon cos square theta minus 1 whole raised to the power 1 upon 3 upon cos cube theta multiplied by minus sin theta d theta. Now we know 1 upon cos square theta is equal to x square theta. So we can write this integral as integral x square theta minus 1 whole raised to the power 1 upon 3 upon cos square theta multiplied by minus sin theta upon cos theta d theta cos cube theta can be written as cos square theta multiplied by cos theta. Now this integral is further equal to integral x square theta multiplied by x square theta minus 1 whole raised to the power 1 upon 3 multiplied by minus tan theta d theta. We know reciprocal of cos square theta is x square theta. This bracket is written as it is and minus sin theta upon cos theta is equal to minus tan theta. Now we know derivative of tan theta is x square theta. So we will substitute tan theta is equal to t. Then differentiating both the sides with respect to theta we get x square theta d theta is equal to dt. Now substituting these values in this integral we get integral 1 plus t square minus 1 whole raised to the power 1 upon 3 multiplied by minus t multiplied by dt. We know x square theta multiplied by d theta is equal to dt tan theta is equal to t. And here x square theta is equal to 1 plus tan square theta. So we can substitute 1 plus t square for x square theta here. x square theta is equal to 1 plus tan square theta and tan theta is equal to t. So we can write t square for tan square theta here. Now this one and this one will get cancelled and we get integral 2 square raised to the power 1 upon 3 multiplied by minus t dt. Now this integral is further equal to minus integral of t raised to the power 5 upon 3 dt. Now this integral is equal to minus t raised to the power 5 upon 3 plus 1 upon 5 upon 3 plus 1. Here we will ignore the constant of integration since we have to find the definite integral by using all these integrals. Now this is further equal to minus t raised to the power 8 upon 3 upon 8 upon 3. Now simplifying further we get minus 3 upon 8 multiplied by t raised to the power 8 upon 3. Now we know t is equal to tan theta. So we will substitute tan theta for t and we get minus 3 upon 8 multiplied by tan theta raised to the power 8 upon 3. Now we know tan theta is equal to square root of 1 minus cos square theta upon cos theta. We can write tan theta in terms of cos theta as square root of 1 minus cos square theta upon cos theta. Now substituting this value of tan theta here we get minus 3 upon 8 multiplied by square root of 1 minus cos square theta upon cos theta whole raised to the power 8 upon 3. Now we know cos theta is equal to x. So substituting x for cos theta we get minus 3 upon 8 multiplied by square root of 1 minus x square upon x whole raised to the power 8 upon 3. Now we get in definite integral x minus x cube whole raised to the power 1 upon 3 upon x raised to the power 4 dx is equal to minus 3 upon 8 multiplied by square root of 1 minus x square upon x whole raised to the power 8 upon 3. Now definite integral from 1 upon 3 to 1 x minus x cube whole raised to the power 1 upon 3 upon x raised to the power 4 dx is equal to minus 3 upon 8 multiplied by 1 minus x square upon x whole raised to the power 8 upon 3. And limits of this integral are from 1 upon 3 to 1. Substituting the upper limit we get the value of this answer as 0. And substituting the lower limit we get minus minus 3 upon 8 multiplied by square root of 1 minus 1 upon 3 square upon 1 upon 3 whole raised to the power 8 upon 3. Now this is further equal to 3 upon 8 multiplied by square root of 8 upon 9 upon 1 upon 3 whole raised to the power 8 upon 3. Now simplifying further we get 3 upon 8 multiplied by square root of 8 upon 3 multiplied by 3 upon 1 whole raised to the power 8 upon 3. Now here 3 and 3 will get cancelled and we get 3 upon 8 multiplied by 2 raised to the power 3 upon 2 whole raised to the power 8 upon 3. We know we can write root 8 as 2 raised to the power 3 upon 2. Now this is further equal to 3 upon 8 multiplied by 2 raised to the power 4. We know these two powers will get multiplied. Now after multiplication 3 and 3 will get cancelled and 2 will cancel 8 by 4. So we get here 2 raised to the power 4. Now this is further equal to 3 upon 8 multiplied by 16 2 raised to the power 4 is equal to 16. Now we will cancel common factor 8 from numerator and denominator both and we get 6. So definite integral from 1 upon 3 to 1 x minus x cube whole raised to the power 1 upon 3 upon x raised to the power 4 dx is equal to 6. So correct answer is A. So A is our required answer. This completes the session. Hope you understood the solution. Take care and have a nice day.