# How do solve $\int\frac{2\sin(x)+3\cos(x)}{3\sin(x)+2\cos(x)}dx$? [duplicate]

How do I solve this integral? Should I use some kind of an integral substitution?

• $u = \tan \frac x2$. Nov 20, 2016 at 17:00
• I'm tempted to edit in a minus sign. Nov 20, 2016 at 17:01
• Are you sure the denominator has no negative sign for the sin x. If that is the case see my hint below otherwise just substitute the numerator as the denominator is its derivative.
– user371838
Nov 20, 2016 at 17:02
• math.stackexchange.com/questions/1219016/… Nov 20, 2016 at 17:05
• this is close to a duplicate... Nov 20, 2016 at 17:06

Hint. By setting $$I=\int\frac{\cos(x)\:dx}{2\cos(x)+3 \sin(x)}\quad J=\int\frac{\sin(x)\:dx}{2\cos(x)+3 \sin(x)}$$ One may observe that

$$\begin{cases} 2 I+3J=\displaystyle\int 1\:dx \\ 3 I-2J=\displaystyle \int\frac{(2\cos(x)+3 \sin(x))'}{2\cos(x)+3 \sin(x)}\:dx \end{cases}$$ Can you take it from here?

• You should give credit to the person who had this idea. Nov 20, 2016 at 17:04
• @Git Gud Please let me know the name. Nov 20, 2016 at 17:06
• As if you didn't know Tao did it first... Nov 20, 2016 at 17:09
• This is actually quite an old chestnut. See question 3 in pmt.physicsandmathstutor.com/download/Maths/STEP/…. But it probably dates from before then. Nov 20, 2016 at 17:16
• I linked to a very specific comment by Wong which reads: "In this math education article the author describes giving the same problem to a young Terence Tao, aged 8; he gave essentially the same beautiful solution". And I don't believe for one minute that you didn't already know the trick. Nov 20, 2016 at 17:16

If there is no minus sign, multiply the numerator and the denominator by $sec (x)^{3}$. Then substitute $u= tan (x)$ and then the integral becomes easy. Hope it helps.

Hint:

Let $$2\sin x+3\cos x=A(3\sin x+2\cos x)+B\cdot\dfrac{d(3\sin x+2\cos x)}{dx}$$