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Question : 71 of 160
Marks:
+1,
-0
Solution:
‌ Let ‌l‌=∫(tan3x+tan‌x)‌dx‌=∫tan‌x(tan6x+1)‌dx‌=∫tan‌x((tan2x)3+13)‌dx‌=∫tanx(tan2x+1)(tan4x+1−tan2x)‌dx‌=∫(tan3x−tan3x+tan‌x) sec3x‌dx‌‌ Let ‌tan‌x=t‌ sec2x,dx=dt‌I=∫(t5−t3+t)‌dt‌=‌−‌+‌+C=‌+C‌=‌(2tan3x−3tan2x+6)+C
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