 Hello and welcome to the session, let us discuss the following question. It says integrate the following function. The given function is x cube into sin tan inverse x to the power 4 upon 1 plus x to the power 8. Let us now proceed on with the solution. And with i be the integral x cube into sin tan inverse x to the power 4 upon 1 plus x to the power 8 dx. There we see that the derivative of tan inverse x to the power 4 is 1 upon 1 plus x to the power 8 into derivative of x to the power 4 which is 4 x cube. So here we put x to the power 4. So dt by dx is equal to 1 upon 1 plus x to the power 4 to the power 2 into dx by dx of x to the power 4. So this is equal to 1 upon 1 plus x to the power 8 into 4x cube. And this implies dt is equal to 4x cube upon 1 plus x to the power 8 dx. That is this term. But here we need to divide both sides by 4 to get this. So dividing both sides by 4 we get x cube upon 1 plus x to the power 8 dx is equal to dt by 4. So substituting x cube upon 1 plus x to the power 8 as dt by 4 and t equal to tan inverse x to the power 4 in the given integral the integral becomes sin dt by 4 which is again equal to 1 by 4 integral cos t is sin t. Now we know that the integral of sin t is minus cos t plus c. So this becomes minus 1 by 4 cos t plus c where c is the constant of integration. So again this is equal to minus 1 by 4 cos t is tan inverse x to the power 4. So substitute it plus c. Hence the integral of the given function is minus 1 by 4 cos tan inverse x to the power 4 plus c. That's all for this session. Keep practicing such questions. Bye for now. Take care and have a good day.