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Question : 74 of 160
Marks:
+1,
-0
Solution:
Consider the expression,
I=∫ =
∫ =
∫| sin‌x‌d‌x |
| sin2x(1+2‌cos‌x) |
=∫| sin‌x‌d‌x |
| (1−cos‌x)(1+cos‌x)(1+2‌cos‌x) |
Let
cos‌x=t then,
−sin‌x‌d‌x−=dt So,
I=−∫ By partial fraction,
= ++ 1=A(1+t)(1+2t)+B(1−t)(1+2t) +C(1+t)(1−t) Then,
A=,B=−,C= So,
I=−‌∫+‌∫ −‌∫ =
‌log(1−cos‌x)+‌log(1+cos‌x) −‌log|1+2‌cos‌x|+c
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