00:01
Hi, in the given problem we are given with the in the first part we are given with the double integral from 0 to 1 from 0 to root y x cube secant y cube dx dy.
00:18
So, we have to integrate and solve it.
00:20
So, since we have to integrate with respect to x and this is not a x variable term.
00:25
So, we are just going to have this as a variable and this will be considered as a constant.
00:29
So, integrating the innermost part.
00:32
So, this would be x to the power 4 over secant y cube and this goes from 0 to root over y this is with respect to y.
00:44
So, putting that we have y to the power y square over 4 secant y cube dy and integral from 0 to 1.
00:55
Now, if we take y cube as t.
01:01
So, y cube as t.
01:07
So, 3 y square dy would become dt and when y is equal to 1 t will be 1 and when y is 0 t is equal to.
01:17
So, this will integral can also be written as 0 to 1.
01:20
So, this is y square dy would be dt y square dy.
01:24
So, that would be dt over 3 and or we can have this secant.
01:31
So, write this secant y cube is t and this is dt over 3...