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Question : 18 of 89
Marks:
+1,
-0
Solution:
Let
I=∫‌‌dxPut
secx+tan‌x=t∴ secx−tan‌x=‌∴ secx=‌(t+‌) and
tan‌x=‌(t−‌)Differentiate w.r.t.
x, we get
‌⇒‌‌( secx‌tan‌x+ sec2x)‌dx=dt‌⇒‌‌dx=‌‌I=∫‌×‌‌=∫‌×‌‌=∫(‌(t+‌))⋅‌×‌‌dt‌=‌‌∫(t+‌)t‌‌dt‌=‌[∫t‌‌dt+∫t‌‌dt]‌=‌[‌+‌]+C‌=‌[‌×‌−‌‌]+C‌=‌( secx−tan‌x)‌−‌( secx−tan‌x)‌+C
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