 That's what I did, 1 by z, so when you subtract it you get 2 y sizes, okay, into sine to the power 4, if I am not mistaken it will be z plus 1 by z to the power 4 and again 1 by 16 z minus 1 by z to the power 4, please note i to the power 4 becomes 1, so you realize integration, I will be already, so it will be either 1, so 16 into 16 to 56, can I write this as z square minus 1 by z square to the power 4, I am just writing the final terms, will you get 6 plus 1 by z to the power 6, correct, then plus 6 z to the power 4 plus 1 by z to the power 4, z square plus, this term, if I raise it to 16 i to the power 4 which is 1 by 16, correct, z minus 1 by z to the power 2, now expanding this multiplying z minus 1 by z and z plus 1 by z I get z square minus 1 by z square and both of this, so that you will understand it easily, so z to the power 8, next term would be what 2, minus 4 z to the power 6 into 1 by z, next term would be what 6, 6 z to the power 4, next term would be plus z to the power 4 and this is minus 4 minus 8 cos 4 x and 6, so instead of sum which is much easier to integrate, so this will become 1 by 256, this will become 2 sin 8 x by 8, this will become, this will become 6 x plus c, don't start up by, by how we are going to do this, which is case number 4 minus 1 by 3, you get minus 12, which is a, so in this case, by sin to the power 11 by 3, when you do that you will get x, we get tan, put tan x not k, I already use k, let's put it as t, so what will happen, 6 square x d x square which is 1 plus t square, which you can further write it as t to the power minus 8 by 3 into minus 3 by 8 and minus 3 by 2 tan x to the power, it becomes very easy to separate a c square from it and the beginning that's the beauty of a c term, so c square should be separated out to facilitate becoming d t part, this part becomes, so we can take sin to the power k, so we get cos c can do that also, cos c part also is equal to, it will give you this, so first of all if you divide this by cos square, so ultimately you are, so c can square x d x will become d t, so this integral converts to d t by t to the power 7 by 2, which is t to the power 2 into minus 2 by 2 minus c, which is minus 2 by 5 tan x to the power of minus 5 by 2 plus, does it help? Sir, the special