Given,sec tcos t dt=1cos tcos t dt=1cos t·1cos t dt=sec t · sec t dt=sec2t dt= tan t + C
Problem 9 :
1sin2t dt
Solution :
Given,1sin2t dt=csc2 t dt=-cot t + C
Problem 10 :
sec w sin wcos w dw
Solution :
Given,sec w sin wcos w dw=1cos w·sin w cos w dw=sin wcos wcos w dw=sin wcos w·1cos w dw=sin wcos2 w dwu = cos w, du =-sin w dwdw =1-sin w du=sin wu·1u·1-sin w du=-1u·1u du=-1u2 du=-u-2 du=-u-2 + 1-2 + 1=-u-1-1=--u-1=--cos-1 w= sec w + C
Problem 11 :
csc w cos wsin w dw
Solution :
Given,csc w cos wsin w dw=1sin w·cos w sin w dw=cos wsin wsin w dw=cos wsin w·1sin w dw=cos wsin2w dwu = sin w, du = cos w dwdw =1cos w du=cos wu2·1cos w du=1u2 du=u-2 du=u-2 + 1-2 + 1=u-1-1=-u-1=-csc w + C
Problem 12 :
1 + cot2z cot zcsc z dz
Solution :
Given,1 + cot2z cot zcsc z dz=csc2z cot zcsc z dz=csc z cot z dz=-csc z + C
Problem 13 :
tan zcos z dz
Solution :
Given,tan zcos z dz=sin zcos zcos z dz=sin zcos z·1cos z dzu=cos zdu =-sin z dzdz =1-sin z du =sin zu·1u·1-sin z du=-1u·1u du=-1u2 du=-u-2 du=-u-2 + 1-2 + 1=-u-1-1=--u-1=--cos-1 z= sec z + C