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Question : 72 of 160
Marks:
+1,
-0
Solution:
Let
Let
tan‌x=t2⇒ sec2x‌dx=2t‌dt I‌=∫‌‌=2‌∫‌‌dt−2‌∫‌‌dt‌=2‌∫‌‌dt−2‌∫‌‌dt‌=−‌+‌×‌+C‌=‌+C‌=‌| 2−3(1+√tan‌x) |
| 3(1+√tan‌x)3 |
+C‌=‌| −(1+3√tan‌x) |
| 3(1+√tan‌x)3 |
+C
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