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Question : 151 of 200
Marks:
+1,
-0
Solution:
We have,
‌I=∫‌‌=∫‌‌=∫‌| sec2‌dx |
| 1+tan2‌−1+tan2‌−2‌tan‌ |
=‌‌∫‌| sec2‌dx |
| tan‌(tan‌−1) |
Let
tan‌=z⇒‌ sec2‌‌dx=dz⇒ sec2‌‌dx=2dz∴I‌=‌‌∫‌=∫(‌−‌)dz‌=log(z−1)−log‌z+C‌=log(‌)+C=log[‌]+C‌=log[1−cot‌]+C
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