Evaluate the integral.
$ \displaystyle \int x \sec x \tan x dx $
\int x \sec x \tan x d x=x \sec x-\ln |\sec x+\tan x|+C
this problem is from Chapter seven, a section to problem number thirty three and the book Calculus Early Transcendental lt's a condition. My James Door. We have an indefinite integral of X times Seeking of X times Stange Innovex Let's try integration my part here. Let's take you two be x so that do you is the ex And let's take Devi to be the remaining part of the immigrant. So seek Innovex time Sandra Becks DX and we know the inner rule of that be It's just seeking the books. So playing integration by parts here this integral is UV Linus and a girl be do you So we have you times be so excited Can't minus in a girl vee do which is just seeking it Time's theatrics So we still have to evaluate this integral ves He can't what you may have memorised already which you could just apply if you memorized But if you haven't memorized this one well, let's go over a method of how to actually do this one So x again, Biggs. One way to proceed is to multiply topping the nominator Bye. See campus tangent separate us from our scratch work. We haven't changed. The integral because we had the thing in the parentheses is just one. So this is still being a roll of seeking. But doing this will make it easier to evaluate Scott and pushed the Sikh and outside the princess's multiply into the numerator. And we did seek and square plus seeking time. Sandra Denominator remains the same, and here we could see that we can apply u substitution. Let's think you to be the denominator. The reason is because then we'LL see that do you will be become a numerator. So do you become seek and time, Stan Gin plus the derivative of tangent, seek and square. So after we play early substitution we have do you on top and then you and the denominator and we have X times he can't minus natural log absolute value of you and our constant sea of integration. And let's not forget that this, you hear is not from integration. My parts This was from our U substitution. So let's be careful with the with the back substitution, we have X time seeking a Vicks minus natural log of you, which was seek and plus tangent and then plus E. And there's our answer