 Hi and welcome to the session. Let us discuss the following question. Question says evaluate the integrals using substitution. Given definite integral is from 0 to pi upon 2 square root of sin phi multiplied by cos phi raised to the power 5 d phi. First of all let us understand that integral x raised to the power n dx is equal to x raised to the power n plus 1 upon n plus 1 plus c. This formula of integration is the key idea to solve the given question. Let us now start with the solution. We have to evaluate definite integral from 0 to pi upon 2 square root of sin phi multiplied by cos phi raised to the power 5 d phi. Clearly we can see derivative of sin phi is cos phi. So we will solve this integral by substitution method. Now let sin phi is equal to t. Now differentiating both the sides with respect to phi we get cos phi is equal to dt upon d phi. Now this further implies cos phi d phi is equal to dt. Now we will find out new limits. Clearly we can see when phi is equal to 0 t is equal to 0 and when phi is equal to pi upon 2 t is equal to 1 so we get as phi varies from 0 to pi upon 2 t varies from 0 to 1. Now this given definite integral is equal to definite integral from 0 to pi upon 2 square root of sin phi multiplied by square of cos square phi multiplied by cos phi d phi. We know cos phi raised to the power 5 is equal to square of cos square phi multiplied by cos phi. Now cos square phi can be written as 1 minus sin square phi. We know cos square theta is equal to 1 minus sin square theta. So cos square phi is equal to 1 minus sin square phi. So we will write definite integral from 0 to pi upon 2 square root of sin phi multiplied by 1 minus sin square phi whole square multiplied by cos phi d phi. Now substituting t for sin phi and dt for cos phi d phi in this integral and we know limits of t are from 0 to 1. So we can write this integral is equal to definite integral from 0 to 1 square root of t multiplied by 1 minus t square whole square multiplied by dt. Now this is further equal to definite integral from 0 to 1 square root of t multiplied by 1 plus t raised to the power 4 minus 2t square dt below a minus b whole square is equal to a square plus b square minus 2ab. Now this integral is further equal to definite integral from 0 to 1 t raised to the power 1 upon 2 plus t raised to the power 4 plus 1 upon 2 minus 2 multiplied by t raised to the power 2 plus 1 upon 2 dt. We know square root of t is equal to t raised to the power 1 upon 2. Now multiplying this square root of t with every term of this bracket we get these terms. Now this is further equal to definite integral from 0 to 1 t raised to the power 1 upon 2 plus t raised to the power 9 upon 2 minus 2 multiplied by t raised to the power 5 upon 2 dt. Now this definite integral can be further written as definite integral from 0 to 1 t raised to the power 1 upon 2 dt plus definite integral from 0 to 1 t raised to the power 9 upon 2 dt minus 2 multiplied by integral from 0 to 1 t raised to the power 5 upon 2 dt. Now using the formula of integration given in t idea this integral is equal to 2 upon 3 multiplied by t raised to the power 3 upon 2 and limit of the integral varies from 0 to 1 plus again here we will apply the formula given in t idea. Now this integral is equal to 2 upon 11 multiplied by t raised to the power 11 upon 2. Limit of the integral varies from 0 to 1 and this integral is equal to 4 upon 7 multiplied by t raised to the power 7 upon 2 and limit of the integral varies from 0 to 1. For finding these three integrals we have used formula of integration given in t idea. Now substituting the limits in this function we get 2 upon 3 multiplied by 1 raised to the power 3 upon 2 minus 0. Now we will write plus sign as it is now substituting limits in this function we get 2 upon 11 multiplied by 1 raised to the power 11 upon 2 minus 0. Now we will write this minus sign as it is and we will substitute these limits in this function we get 4 upon 7 multiplied by 1 raised to the power 7 upon 2 minus 0. Now we know 1 raised to any power is equal to 1 only so this expression is further equal to 2 upon 3 plus 2 upon 11 minus 4 upon 7 now this is further equal to 154 plus 42 minus 132 upon 231. Now simplifying further this expression is equal to 64 upon 231. So definite integral from 0 to pi upon 2 square root of sin phi multiplied by cos phi raised to the power 5 d phi is equal to 64 upon 231. So this is our required answer this completes the session hope you understood the solution take care and keep smiling.