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Question : 72 of 160
Marks:
+1,
-0
Solution:
Let
I=∫e−2x(tan‌2‌x−2 sec22x‌tan‌2‌x)‌dxPut
−2x=t⇒‌‌dx=−‌∴I=‌∫et(tan(−t)−2 sec2(−t)‌tan(−t))(−‌)=‌−‌‌∫et(2 sec2t‌tan‌t−tan‌t)‌dt=‌−‌‌∫et[( sec2t−tan‌t).‌+(2 sec2t‌tan‌t− sec2t)]‌dtWe know that,
∫etf(t)+f′(t)‌dt=etf(t)+CHere,
‌‌f(t)= sec2t−tan‌tand
‌‌f′(t)=2 sec2t‌tan‌t− sec2t∴‌‌I=−‌et[ sec2t−tan‌t]+C‌=−‌e−2x⋅[ sec2(−2x)−tan(−2x)]+C‌=−‌[ sec22x+tan‌2‌x]+c
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