00:01
Problem 14 says that we have to integrate 4x, 6 square 2x dx using integration by parts.
00:11
Let's take this four outside since it is a constant and we have to integrate just x, x, x, square, 2x, dx.
00:19
This is algebraic, so this is first term, trigonometric will be second term.
00:23
So the four remains outside.
00:25
Now the first term is as it is we integrate sex square 2x, which will be tan 2x over 2, minus integration of differentiation of x is one integration of x squared 2x is tan 2x over 2 and this complete expression is once again integrated so 4 remains outside this becomes x tan x tan 2x over 2 so this bracket will close here x tan 2x over 2 minus 1 over 2 comes outside and then we are left with tan 2x integration the integration of 10 to x will be minus ln cost 2x over 2.
01:12
And then we have a constant of integral as well...