Integral of Trig. Function, PowerDate: 7/19/96 at 11:25:22 From: Suryadie Gemilang Subject: Integral of Trig. Function, Power Can you solve this one? Integral of dx/cos^4 x Thanks Sg Date: 7/19/96 at 17:12:26 From: Doctor Anthony Subject: Re: Integral of Trig. Function, Power I = INT[dx/cos^4(x)] = INT[sec^4(x)*dx] = INT[(1+tan^2(x))*sec^2(x) dx] Integrate by parts I = (1+tan^2(x))*tan(x) - INT[tan(x)*2tan(x)*sec^2(x)*dx] = (1+tan^2(x))*tan(x) - 2*INT[tan^2(x)*sec^2(x)*dx] = (1+tan^2(x))*tan(x) - 2*Int[(sec^2(x)-1)*sec^2(x)*dx] = tan(x)+tan^3(x) - 2*INT[{sec^4(x) - sec^2(x)}*dx] I = tan(x)+tan^3(x) - 2*I + 2*INT[sec^2(x)*dx] 3I = tan(x)+tan^3(x) + 2*tan(x) + Const I = (1/3)(tan(x)+tan^3(x)) + (2/3)tan(x) + C = tan(x) + (1/3)tan^3(x) + C -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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