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Question : 53 of 160
Marks:
+1,
-0
Solution:
∫| sin((x+4)−(x+2)) |
| sin‌2‌cos(x+4)‌cos(x+2) |
dx=∫| sin(x+4)‌cos(x+2)−cos(x+4)‌sin(x+2) |
| sin‌2‌cos(x+4)‌cos(x+2) |
=‌∫tan(x+4)‌d‌x −‌∫tan(x+2)‌d‌x=={log|sec(x+4)| −log|sec(x+2)|}=‌log||+c
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