integration of 1/ (3sinx + 4cosx)?

Question

Please help me solve this integration problem 1/ (3sinx + 4cosx) dx

Please try not to use arc we havent got there yet

I started this problem by multiplying with the conjugate.. but i am stuck/ or made it worse lol

Please explain how you got your answer!

Thanks so much,

Answer 1

let
3sin x + 4 cos x = k cos (x - α) where k >0
= (k cos α ) cos x + (k sinα) sinx
4 = k cosα
3 = k sinα
tan α = 3/4 (1st quadrant)
α = 36.9°
16 + 9 = k²
k = 5
I = ∫ 1 / 5 cos (x - 36.9) dx
I = (1/5) ∫ sec (x - 36.9) dx
I = (1/5)[log [(sec x - 36.9) + tan (x - 36.9)] + C

Answer 2

partial fractions is the only thank you to pick for this. A / ( x + 5 ) + B / ( x + 5)^2 + C / ( x - a million ) = a million / [ ( x + 5)^2 ( x - a million) ] [ A(x + 5) + B ] / ( x + 5)^2 + C / ( x - a million ) = a million / [ (x + 5)^2 ( x - a million) ] A( x + 5 )(x - a million) + B( x - a million ) + C ( x + 5 )^2 = a million Ax^2 + 4Ax - 5A + Bx - B + Cx^2 + 10X + 25C = a million (A + C)x^2 + ( 4A + B + 10)x + ( 25C - 5A - B) = a million A + C = 0 4A + B + 10 = 0 25C - 5A - B = a million pick those.

Answer 3

Hint:
3sinx + 4cosx = 5(3/5 sinx + 4/5 cosx) = 5sin(x+α), where α = arctan(4/3)

∫1/sinx dx = ln[(tan(x/2)]

Answer 4

According to Maple the answer is:

-(1/5)*ln(tan((1/2)*x)-2)
+(1/5)*ln(2*tan((1/2)*x)+1)

Looks like a lot of work to get there by hand