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Question : 77 of 160
Marks:
+1,
-0
Solution:
∫‌| (1−cos‌x)2∕7 |
| (1+cos‌x)9∕7 |
dx =∫‌| (1−cos‌x)2∕7 |
| (1+cos‌x)(1+cos‌x)2∕7 |
dx =∫(‌)2∕7‌ =∫(‌)2∕7⋅‌ =∫(tan2‌)2∕7⋅‌sec2‌dx =‌‌∫tan4∕7‌⋅sec2‌dx Let
tan‌=t ⇒‌‌sec2‌dx‌‌=2dt ‌‌=∫t4∕7dt ‌‌=‌+C=‌t11∕7+C ‌‌=‌tan11∕7‌+C
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