How do I integrate #f(x)=sec(x)/(4-3tan(x))#?

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Answer 1 · 2015-04-22T19:51:11

#int (sec(x))/(4-3tan(x))dx = int (1/cos(x))/(4-3tan(x)) dx#

Let's start with #t =tan(x/2)#

#sec(x) = (1+t^2)/(1-t^2)#

#tan(x) = (2t)/(1-t^2)#

#dx = 2/(1+t^2) dt#

So, the integral become :

#=> int((1+t^2)/(1-t^2))/(4-(6t)/(1-t^2))*2/(1+t^2)dt#

#=>int((1+t^2)/(1-t^2))/((4(1-t^2))/(1-t^2)-(6t)/(1-t^2))*2/(1+t^2)dt#

#=>int((1+t^2)/(1-t^2))/((-4t^2-6t+4)/(1-t^2))*2/(1+t^2)dt#

#=>int(1+t^2)/((-4t^2-6t+4))*2/(1+t^2)dt#

#=>int2/(2(-2t^2-3t+2))dt#

#=>int1/(-2t^2-3t+2)dt#

#Delta = b^2-4ac = 25#

#x_1 = 1/2#

#x_2 = -2#

Factorize :

#=>int-1/((2t-1)(t+2))dt#

Partial fraction :

#=>1/5int 1/(t+2)dt-1/5int2/(2t-1)dt#

#=>1/5[ln(|t+2|)]-1/5[ln(|2t-1|)]+C#

#=> 1/5((ln(|t+2|)-ln(|2t-1|))+C#

Substitute back for #t = tan(1/2x)#

#=> 1/5(ln(|tan(1/2x)+2|)-ln(|2tan(1/2x)-1|))+C#