 Hi and welcome to the session. I am Purva and I will help you with the following question proof the following Integral sign raised to the path 3x dx where limit is from 0 to pi by 2 is equal to 2 by 3 Now we begin with this solution We denote this left-hand side by i so we have i is equal to integral limit from 0 to pi by 2 Sign raised to the path 3x dx now we can write this as this is equal to integral limit is from 0 to pi by 2 sign square x into sin x dx This is further equal to integral limit is from 0 to pi by 2 Now we can write sin square x as 1 minus cos square x. So we have 1 minus cos square x into sin x dx since sin square x is equal to 1 minus cos square x Now we put cos x is equal to t. So put cos x is equal to t differentiating we get minus sin x dx is equal to dt right or We can write this as sin x dx is equal to minus dt Now when x is equal to pi by 2 we have t is equal to cos pi by 2 and We know that cos pi by 2 is equal to 0 So when x is equal to pi by 2 then t is equal to 0 and when x is equal to 0 then t is equal to cos 0 and Cos 0 is 1 so t is equal to 1 Putting all these values in i we get i is equal to integral limit is from 1 to 0 minus 1 minus t square dt Or we can write this as this is equal to integral limit is from 0 to 1 1 minus t square dt And this is equal to integral limit is from 0 to 1 dt Minus integral limit is from 0 to 1 t square dt Now integrating we get this is equal to t Limit is from 0 to 1 minus Integrating t square we get t cube by 3 Limit is from 0 to 1 This is equal to now putting upper limit in place of t we get 1 Minus putting lower limit we get 0 minus Again putting upper limit we get 1 by 3 Minus putting lower limit we get 0 so we get this is equal to 1 minus 1 by 3 which is equal to 2 by 3 So we have got i is equal to 2 by 3 thus we have proved that Integral limit is from 0 to pi by 2 sine cube x dx is equal to 2 by 3 Hope you have understood the solution take care and God bless you