##-1\/2(ln(abs(sin(x)+cos(x)))-x)+C##<\/p>\n
Another method is to write this using all tangents:<\/p>\n
##I=int1\/(1+cot(x))dx=inttan(x)\/(tan(x)(1+cot(x)))=inttan(x)\/(tan(x)+1)dx##<\/p>\n
Now, since all we have are tangents, we need a ##sec^2(x)## in order to substitute. We can multiply the fraction by ##sec^2(x)## in the numerator, but express it as ##tan^2(x)+1## in the denominator (they’re equal through the Pythagorean identity).<\/p>\n
##I=int(tan(x)sec^2(x))\/((tan(x)+1)(tan^2(x)+1))dx##<\/p>\n
Letting ##u=tan(x)## so that ##du=sec^2(x)dx##, we see that:<\/p>\n
##I=intu\/((u+1)(u^2+1))du##<\/p>\n
Now, we have to perform partial fraction decomposition:<\/p>\n
##u\/((u+1)(u^2+1))=A\/(u+1)+(Bu+C)\/(u^2+1)##<\/p>\n
Multiplying through:<\/p>\n
##u=A(u^2+1)+(Bu+C)(u+1)##<\/p>\n
##u=Au^2+A+Bu^2+Bu+Cu+C##<\/p>\n
Factor in three groups: those with ##u^2##, those with ##u##, and constants.<\/p>\n
##u=u^2(A+B)+u(B+C)+(A+C)##<\/p>\n
##color(purple)0u^2+color(red)1u+color(brown)0=u^2color(purple)((A+B))+ucolor(red)((B+C))+color(brown)((A+C))##<\/p>\n
Comparing the two sides, we see that:<\/p>\n
##{(A+B=0),(B+C=1),(A+C=0):}##<\/p>\n
Subtracting the second equation from the third, we see that ##A-B=-1##. Adding this to the first equation shows that ##2A=-1## and ##A=-1\/2##. It then follows that:<\/p>\n
##{(A=-1\/2),(B=1\/2),(C=1\/2):}##<\/p>\n
So:<\/p>\n
##u\/((u+1)(u^2+1))=1\/2(1\/(u+1))+1\/2((u+1)\/(u^2+1))##<\/p>\n
Returning to the integral now:<\/p>\n
##I=-1\/2int1\/(u+1)du+1\/2int(u+1)\/(u^2+1)du##<\/p>\n
##I=-1\/2int1\/(u+1)du+1\/2intu\/(u^2+1)du+1\/2int1\/(u^2+1)du##<\/p>\n
Modifying the second integral slightly:<\/p>\n
##I=-1\/2int1\/(u+1)du+1\/4int(2u)\/(u^2+1)du+1\/2int1\/(u^2+1)du##<\/p>\n
Now all three integrals can be integrated rather painlessly:<\/p>\n
##I=-1\/2ln(abs(u+1))+1\/4ln(abs(u^2+1))+1\/2arctan(u)##<\/p>\n
##I=-1\/2ln(abs(tan(x)+1))+1\/4ln(tan^2(x)+1)+1\/2arctan(tan(x))##<\/p>\n
##color(blue)(I=-1\/2ln(abs(tan(x)+1))+1\/4ln(sec^2(x))+1\/2x##<\/p>\n
This is a fine final answer, once the constant of integration is added, but we can fiddle around a little more to achieve some fun simplification.<\/p>\n
##I=-1\/2ln(abs((sin(x)+cos(x))\/cos(x)))+1\/2(1\/2ln(sec^2(x)))+1\/2x##<\/p>\n
Rather sneakily, bring one of the ##1\/2##s outside the ##ln(sec^2(x))## in, effectively using the ##log(a^b)=blog(a)## rule in reverse. (Absolute value bars will be added since we’ve just taken the square root:)<\/p>\n
##I=-1\/2ln(abs((sin(x)+cos(x))\/cos(x)))+1\/2ln(abs(sec(x)))+1\/2x##<\/p>\n
Now we can bring a ##-1## in as a ##-1## power to make ##sec(x)## into ##cos(x)##:<\/p>\n
##I=-1\/2ln(abs((sin(x)+cos(x))\/cos(x)))-1\/2ln(abs(cos(x)))+1\/2x##<\/p>\n
Factor ##-1\/2## and use the rule that ##log(A)+log(B)=log(AB)##:<\/p>\n
##I=-1\/2(ln(abs((sin(x)+cos(x))\/cos(x)))+ln(abs(cos(x)))-x)##<\/p>\n
##color(green)(I=-1\/2(ln(abs(sin(x)+cos(x)))-x)+C##<\/p>\n","protected":false},"excerpt":{"rendered":"
##-1\/2(ln(abs(sin(x)+cos(x)))-x)+C## Another method is to write this using all tangents: ##I=int1\/(1+cot(x))dx=inttan(x)\/(tan(x)(1+cot(x)))=inttan(x)\/(tan(x)+1)dx## Now, since all we have are tangents, we need a ##sec^2(x)## in order to substitute. We can multiply the fraction by ##sec^2(x)## in the numerator, but express it as Read More …<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[],"tags":[],"class_list":["post-28199","post","type-post","status-publish","format-standard","hentry"],"_links":{"self":[{"href":"https:\/\/www.benedictsol.com\/blogs\/wp-json\/wp\/v2\/posts\/28199","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.benedictsol.com\/blogs\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.benedictsol.com\/blogs\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.benedictsol.com\/blogs\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.benedictsol.com\/blogs\/wp-json\/wp\/v2\/comments?post=28199"}],"version-history":[{"count":0,"href":"https:\/\/www.benedictsol.com\/blogs\/wp-json\/wp\/v2\/posts\/28199\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.benedictsol.com\/blogs\/wp-json\/wp\/v2\/media?parent=28199"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.benedictsol.com\/blogs\/wp-json\/wp\/v2\/categories?post=28199"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.benedictsol.com\/blogs\/wp-json\/wp\/v2\/tags?post=28199"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}