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Question : 76 of 160
Marks:
+1,
-0
Solution:
I‌=‌| dx |
| 5cos2x+16sin‌2x+8sin‌x‌cos‌x |
‌=‌| sec2x‌dx |
| 5+16tan2x+8‌tan‌x |
Put,
tan‌x=t⇒ sec2x‌dx=dtI‌=‌‌=‌‌‌‌=‌‌‌‌=‌‌‌‌=‌×‌tan−1(‌)01‌=‌[tan−1(‌)−tan−1(‌)]‌=‌tan−1(‌)=‌tan−1(‌)
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