I=(tan)^1/3 then,intergral this sum.please solve them.

Hello,

It is quite lengthy sum. The sum is a mixture of standard integral, substitution method, and partial fraction.

The given question by you is  a bit incorrect as it is missing one constant after tan. The question should be I = (tan x )^1/3 .

So Given, I = (tan x )^1/3

This can be written as I = 3√tan(x) dx (cube root of tan x)

on integration and applying standard integral, substitution and partial fraction to 3 √tan(x) dx , we get

= [ ln (tan 43(x )− tan 23 (x) + 1) ] divided by 4 + [ √3arctan (2 tan 23 (x)− 1√3) ] divided by 2 − [ ln (tan 23 (x) + 1) ] divided by 2 + C

or in simpler form can be written as

[ ln (tan 43 (x) − tan 23 (x)+1) + 2(√3 arctan (2 tan 23 (x) − 1√3 )− ln (tan 23 (x) + 1 )) ] whole divided by 4 + C

Therefore, Integration of I = (tan x )^1/3 is [ ln (tan 43 (x) − tan 23 (x)+1) + 2(√3 arctan (2 tan 23 (x) − 1√3 )− ln (tan 23 (x) + 1 )) ] whole divided by 4 + C

Hope this was helpful.