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Question : 77 of 160
Marks:
+1,
-0
Solution:
∫dx Let‌x+=t ⇒x=t− dx=dt ∫dt =∫√2.| (sin‌t‌cos‌−cos‌t‌sin‌) |
| sin‌t |
dt =∫√2.dt =∫dt=∫(1−cot‌t)dt =t−ln|sin‌t|+c =x+−ln|sin(x+)|+c =x−log|sin(x+)|+c
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