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Question : 73 of 160
Marks:
+1,
-0
Solution:
I=∫‌| (3x−2)‌tan(√9x2−12x+1) |
| √9x2−12x+1 |
‌dxPut
9x2−12x+1=t2‌⇒‌‌(18x−12)‌dx=2t‌dt‌⇒‌‌(3x−2)‌dx=‌t‌dt‌⇒‌‌I=‌‌∫‌‌dt=‌‌∫tan‌t‌dt‌‌‌=‌‌log| sect|+C‌‌‌=‌‌log‌| sec√9x2−12x+1|+C
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