Question
Mathematics Question on integral
∫1−cosx−sinxdx is equal to
A
log1+cot2x+C
B
log1−tan2x+C
C
log1−cot2x+C
D
log1+tan2x+C
Answer
log1−cot2x+C
Explanation
Solution
LetI=∫1−cosx−sinxdx put cosx=1+tan22x1−tan22xandsinx=1+tan22x2tan2x ∴I=∫1−(1+tan22x1−tan22x)−1+tan22x2tan2xdx =∫2tan22x−2tan2xsec22xdx=∫tan22x−tan2x21sec22xdx Put tan2x=t⇒21sec22xdx=dt ∴I=∫t2−tdt=[t−11−t1]∫+C =log(t−1)−logt+C=logtt−1+C =log1−cot2x+C