du  u = - ln u + c = ln u-1 + c ln  1 1 ln   c cos x = ln (secx) + c ∫tan we write xdx = ln secx  c showing that ln of negative values does not ∫cot b) Find xdx dx cot xdx  cos x sin x Let u = sin x so that du = cos dx cot xdx  Then du u  ln  u c cot xdx ln (sin x ) c   Exercise 3 Find the integrals of the following: dx cosx ex  dx 2. 31 3. 2
 du u = - ln u + c = ln u-1 + c  ln 1  ln 1 cos x  c = ln (secx) + c we write ∫tan xdx = ln secx  c showing that ln of negative values does not b) Find ∫cot xdx cot xdx  cos x sin x dx Let u = sin x so that du = cos dx Then cot xdx  du u  ln u  c cot xdx  ln (sin x )  c Exercise 3 Find the integrals of the following: dx 2. cosx  ex dx 31 3. 2
CALCULUS%20II.DOCX