2 answers

Stan®x.secs x dx tanºx a) 2 tan' x tanºx + +c 5 sec b) 2 sec...

Question:

Stan®x.secs x dx tanºx a) 2 tan x tanºx + +c 5 sec b) 2 sec + secx 5 + c 9 sec? sec5 x c) + c d) None of the above
Stan®x.secs x dx tanºx a) 2 tan' x tanºx + +c 5 sec b) 2 sec + secx 5 + c 9 sec? sec5 x c) + c d) None of the above

Answers

Integration of sec^5 (x) tan^5 (x) dx = Integration of sec^4(x) (sec^2 (x) - 1)^2 sec (x) tan (x) dx [ Since, sec^2 (x) - tan^2 (x) = 1 so, tan^4 (x) = (sec^2 (x) - 1)^2 ] Now: Let us assume that: sec (x) = u => sec(x) tan(x) dx = du Substituting sec(x) = u into the integration and integrating with respect to du, we have: = Integration of u^4 (u^2 - 1)^2 du = Integration of (u^8 - 2u^6 + u^4) du = Integration of u^8 du - 2 Integration of u^6 du + Integration of u^4 du = u^9/9 - 2u^7/7 + u^5/5 + C [ Since, Integration of x^n dx = (x^{n+1})/(n+1) + C ( Where C is a constant of integration) ] Now: Undo substitution u = sec (x) , we have: = sec^9 (x) / 9 - 2sec^7 (x)/7 + sec^5 (x)/5 + C Hence: Option(b) is correct..

I Pat u = Seca. dua Seca. Tanuda S tem se see dx . I= s tantx.sectx seex. land de = /4²1².</p><p>Ut du , (ustut. 2urs du See you +

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