I just learned calculus, so I want to be able to figure out how to solve each step. If it is a paper answer hope to be able to clear the word, I am afraid that I will not understand the font, grateful.

Integration of trigonometric forms sin udu = -cosu + C [sec u du = tanu + C cosu du = sinu + C cschu du = -cotu + C tanu du = -In cosu + C secu tanu du = secu + C cotudu = In sinu + C cscu cotudu = -cscu + C secu du = In secu + tanu + C cscudu = -In cscu + cotu + C = In cscu -cotu + C19.1 Homework Practice Answers Exercise 19.1: a) cos 6x + C 6 b) - sin (2x -3) +C 2 C) - tan (x2 -3) + C d) - csc 4x + C 4 e) 2.44 f) In csce -cote +C 2 g) -_ In cost + C h) IT + 2 4 i) - tan 4t - In cos 4t| + 2t + C 4 j) 1.70 k) y = sin 3x + cosx+-Exercise 19.1: Integrate the following: a) sin 6xdx b) [cos (2x -3) dx c) xsec? (x2 -3) dx d) csc 4x cot 4xdx e ) [x csc' x'dx ( Vcot2 0 + 1de g) ( sin t at sin 2t h) ( sin 2r + cos 2r) dr i) [ ( tan 41 + 1) (tan 4t + 3) dt j) Find the area under the curve of y = sec - tan ~ between x= 0 and x=2. k) Find y given that y' = 3cos 3x -sin x and it passes through the point ,1