Integration of: #intdx/{(sinx+1)(3+sinx)}# ?

1 Answer
Feb 14, 2018

#int dx/((1+sinx)(3+sinx)) = (sinx-1)/(2cosx) - 1/(4sqrt2) arctan((2sqrt2tanx)/3)-1/(4sqrt2) arctan(cosx/(2sqrt2)) +C#

Explanation:

Excluding the points #x=pi/2+2kpi#, multiply and divide by #(1-sinx)#

#1/((1+sinx)(3+sinx)) = (1-sinx)/((1-sinx)(1+sinx)(3+sinx))#

#1/((1+sinx)(3+sinx)) = (1-sinx)/((1-sin^2x)(3+sinx))#

#1/((1+sinx)(3+sinx)) = -(sinx-1)/(cos^2x(3+sinx))#

#1/((1+sinx)(3+sinx)) = -(3+ sinx-4)/(cos^2x(3+sinx))#

#1/((1+sinx)(3+sinx)) = -1/cos^2x +4/(cos^2x(3+sinx))#

So:

#int dx/((1+sinx)(3+sinx)) = -int dx/cos^2x +4 int dx/(cos^2x(3+sinx))#

#int dx/((1+sinx)(3+sinx)) = -tanx +4 int dx/(cos^2x(3+sinx))#

Using a similar approach:

#1/(cos^2x(3+sinx)) = (3-sinx)/(cos^2x(3+sinx)(3-sinx))#

#1/(cos^2x(3+sinx)) = (3-sinx)/(cos^2x(9-sin^2x))#

#1/(cos^2x(3+sinx)) = (3-sinx)/(cos^2x(8+cos^2x))#

so:

#int dx/(cos^2x(3+sinx)) = 3int dx/(cos^2x(8+cos^2x))+ int (-sinxdx)/(cos^2x(8+cos^2x))#

Solve the first integral substituting:

#u= tanx#, #du = dx/cos^2x#

and using the trigonometric identity #sec^2x = 1+tan^2x# so that:

#1/(8+cos^2x) = sec^2x/(8sec^2x+1) = (1+tan^2x)/(9+8tan^2x)#

and then:

#int dx/(cos^2x(8+cos^2x)) = int (1+u^2)/(9+8u^2)du#

Split the fraction:

# int (1+u^2)/(9+8u^2)du = 1/8 int (8+8u^2)/(9+8u^2)du #

# int (1+u^2)/(9+8u^2)du = 1/8 int (9+8u^2-1)/(9+8u^2)du #

# int (1+u^2)/(9+8u^2)du = 1/8 int du - 1/8int 1/(9+8u^2)du #

# int (1+u^2)/(9+8u^2)du = u/8 - 1/(24sqrt8) int 1/(1+((sqrt8u)/3)^2)d((sqrt8u)/3) #

# int (1+u^2)/(9+8u^2)du = u - 1/(24sqrt8) arctan((sqrt8u)/3)+c #

and undoing the substitution:

#int dx/(cos^2x(8+cos^2x)) = tanx/8- 1/(24sqrt8) arctan((sqrt8tanx)/3)+c#

Solve the second integral substituting:

#u = cosx#, #du = -sinxdx#:

#int (-sinxdx)/(cos^2x(8+cos^2x)) = int (du)/(u^2(8+u^2))#

and using partial fractions decomposition:

#1/(u^2(8+u^2)) = A/u^2 +B/(8+u^2)#

#1 = A(8+u^2)+Bu^2#

#1 = 8A + (A+B)u^2#

#A=1/8#
#B=-1/8#

so:

#int (du)/(u^2(8+u^2)) = 1/8 int (du)/u^2-1/8 int (du)/(8+u^2)#

#int (du)/(u^2(8+u^2)) = -1/(8u)-1/(8sqrt8) int (d(u/sqrt8))/(1+(u/sqrt8)^2)#

#int (du)/(u^2(8+u^2)) = -1/(8u)-1/(8sqrt8) arctan(u/sqrt8) +c#

and undoing the substitution:

#int (-sinxdx)/(cos^2x(8+cos^2x)) = -1/(8cosx)-1/(8sqrt8) arctan(cosx/sqrt8) +c#

Putting the partial solutions together and simplifying:

#int dx/(cos^2x(3+sinx)) = (3tanx)/8- 1/(8sqrt8) arctan((sqrt8tanx)/3)-1/(8cosx)-1/(8sqrt8) arctan(cosx/sqrt8) +c#

#int dx/((1+sinx)(3+sinx)) = (sinx-1)/(2cosx) - 1/(4sqrt2) arctan((2sqrt2tanx)/3)-1/(4sqrt2) arctan(cosx/(2sqrt2)) +C#