00:01
According to the question, we have to integrate integral 36 upon 25 minus 16 x square to the power 3 by 2 with respect to x.
00:16
This integral can also be written as 36 integral 25 minus 16 x square to the power 3 by 2 dx.
00:28
Consider this equation as equation 1.
00:30
Now put x equals to 5 upon 4 sin t.
00:41
On differentiating both sides with respect to t, we have dx by dt equals to 5 by 4 differentiation of sin t with respect to t.
00:53
This implies differentiation of x with respect to t equals to 5 by 4 cos t which can be further written as dx equals to 5 upon 4 cos t.
01:10
On putting all these values in equation 1, we have i equals to 36 integral 1 upon 25 minus 16 into 25 upon 16 sin square t to the power 3 by 2 into 5 by 4 cos t.
01:43
On solving, we have 5 upon 4 into 36 upon 125 integration 1 upon 1 minus sin square t to the power 3 by 2 into cos t dt.
02:10
On further solving, we have 9 upon 25 integration 1 upon cos square t to the power 3 by 2 into cos t dt.
02:26
This gives 9 upon 25 integration 1 upon cos cube t into cos t dt which gives 9 upon 25 integration 1 upon cos square t dt...