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#
