integration 1-cosx/cosx(1+cosx)dx
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Given, ∫
cosx(1+cosx)
(1−cosx)dx
⇒ ∫
cosx(1+cosx)
(1+cosx)
−
cosx(1+cosx)
2cosx
dx
⇒ ∫secx−
1+cosx
2
dx
⇒ ∫secx−
2cos
2
2
x
2
dx since [cosx=2cos
2
2
x
−1]
⇒ ∫secx−sec
2
2
x
dx
⇒ log(secx+tanx)−2tan
2
x
+c is our answer.
Step-by-step explanation:
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