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How do you integrate #sec(x)/(4-3tan(x)) dx#?

Answer 1

#1/5ln(tan(x/2)+2)-1/5Ln(2tan(x/2)-1)+C#

#int secx/(4-3tanx)*dx#

=#int dx/(4cosx-3sinx)#

After using #y=tan(x/2)#, #dx=(2dy)/(y^2+1)#, #cosx=(1-y^2)/(y^2+1)# and #sinx=(2y)/(y^2+1)# transforms, this integral became

#int ((2dy)/(y^2+1))/(4*(1-y^2)/(y^2+1)-3*(2y)/(y^2+1))#

=#int ((2dy)/(y^2+1))/((4-6y-4y^2)/(y^2^1))#

=-#int dy/(2y^2+3y-2)#

=-#int dy/((y+2)*(2y-1))#

=-#1/5int (5dy)/((y+2)*(2y-1))#

=-#1/5int ((2y+4)*dy)/((y+2)(2y-1))#+#1/5int ((2y-1)*dy)/((y+2)(2y-1))#

=#1/5int dy/(y+2)-2/5 int dy/(2y-1)#

=#1/5ln(y+2)-1/5Ln(2y-1)+C#

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

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Answer 2

Adam Adkins

To integrate sec(x)/(4-3tan(x)) dx, you can use the substitution method. Let u = tan(x), then du = sec^2(x) dx. After substitution, the integral becomes 1/(4-3u) du, which can be solved using partial fraction decomposition. Once decomposed, integrate each term separately and then resubstitute u = tan(x) back to obtain the final result.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.