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Question : 75 of 160
Marks:
+1,
-0
Solution:
∫‌=∫‌| dx |
| 2‌tan‌x∕2 | | 1+tan2x∕2 | +‌ |
[| sin‌2‌x=‌ |
| ∵‌‌cos‌2‌x=‌ |
]]. =∫‌| (1+tan2x∕2)dx |
| 2‌tan‌+1−tan2‌ |
=∫‌| sec2x∕2dx |
| 2‌tan‌+1−tan2‌ |
Put
‌‌tan‌=t⇒sec2‌dx=2dt ‌‌=2‌∫‌=−2‌∫‌ ‌‌=−2‌∫‌ ‌‌=2‌∫‌ ‌‌=2⋅‌‌log|‌| √2+(t−1) |
| √2−(t−1) |
| ‌‌=‌‌log|‌| √2+tan‌−1 |
| √2−tan‌+1 |
|+C ‌‌=‌‌log|‌| tan‌+(√2−1) |
| 1−tan‌+√2 |
|+C tan‌‌‌=√2−1,cot‌=√2+1 ‌‌=‌‌log|tan(‌+‌)|+C
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