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Question : 74 of 160
Marks:
+1,
-0
Solution:
Consider the given expression,
∫=∫ =∫ =∫| sinxdx |
| sin2x(1+2cosx) |
=∫| −sinxdx |
| (cos2x−1)(1+2cosx) |
Let
cosx=t then
−sinxdx=dt So ∫ By partial fraction
A=,B= and C=− Now,
∫=∫dt+∫−∫ =+ln(t+1)−ln(t+)+C =[| +ln(cosx+1) |
| −ln(cosx+)+C |
]. =[|(1+cosx)|+|ln|(1−cosx);−ln|cosx+|+C]. Further simplify the above expression
∫ =ln(1+cosx)+ln|(1−cosx)|−ln|1+2cosx|+C
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