Solve the trigonometric integral $\int\frac{3\sin\left(x\right)+4\cos\left(x\right)}{4\sin\left(x\right)-3\cos\left(x\right)}dx$

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Final answer to the problem

$\ln\left|4\sin\left(x\right)-3\cos\left(x\right)\right|+C_0$
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Step-by-step Solution

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  • Integrate by partial fractions
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1

We can solve the integral $\int\frac{3\sin\left(x\right)+4\cos\left(x\right)}{4\sin\left(x\right)-3\cos\left(x\right)}dx$ by applying integration by substitution method (also called U-Substitution). First, we must identify a section within the integral with a new variable (let's call it $u$), which when substituted makes the integral easier. We see that $4\sin\left(x\right)-3\cos\left(x\right)$ it's a good candidate for substitution. Let's define a variable $u$ and assign it to the choosen part

$u=4\sin\left(x\right)-3\cos\left(x\right)$

Differentiate both sides of the equation $u=4\sin\left(x\right)-3\cos\left(x\right)$

$du=\frac{d}{dx}\left(4\sin\left(x\right)-3\cos\left(x\right)\right)$

Find the derivative

$\frac{d}{dx}\left(4\sin\left(x\right)-3\cos\left(x\right)\right)$

The derivative of a sum of two or more functions is the sum of the derivatives of each function

$\frac{d}{dx}\left(4\sin\left(x\right)\right)+\frac{d}{dx}\left(-3\cos\left(x\right)\right)$

The derivative of a function multiplied by a constant is equal to the constant times the derivative of the function

$4\frac{d}{dx}\left(\sin\left(x\right)\right)-3\frac{d}{dx}\left(\cos\left(x\right)\right)$

The derivative of the sine of a function is equal to the cosine of that function times the derivative of that function, in other words, if ${f(x) = \sin(x)}$, then ${f'(x) = \cos(x)\cdot D_x(x)}$

$4\cos\left(x\right)-3\frac{d}{dx}\left(\cos\left(x\right)\right)$

The derivative of the cosine of a function is equal to minus the sine of the function times the derivative of the function, in other words, if $f(x) = \cos(x)$, then $f'(x) = -\sin(x)\cdot D_x(x)$

$4\cos\left(x\right)+3\sin\left(x\right)$
2

Now, in order to rewrite $dx$ in terms of $du$, we need to find the derivative of $u$. We need to calculate $du$, we can do that by deriving the equation above

$du=\left(4\cos\left(x\right)+3\sin\left(x\right)\right)dx$
3

Isolate $dx$ in the previous equation

$\frac{du}{\left(4\cos\left(x\right)+3\sin\left(x\right)\right)}=dx$

Simplify the fraction $\frac{\frac{3\sin\left(x\right)+4\cos\left(x\right)}{u}}{4\cos\left(x\right)+3\sin\left(x\right)}$ by $4\cos\left(x\right)+3\sin\left(x\right)$

$\int\frac{1}{u}du$
4

Substituting $u$ and $dx$ in the integral and simplify

$\int\frac{1}{u}du$
5

The integral of the inverse of the lineal function is given by the following formula, $\displaystyle\int\frac{1}{x}dx=\ln(x)$

$\ln\left|u\right|$

Replace $u$ with the value that we assigned to it in the beginning: $4\sin\left(x\right)-3\cos\left(x\right)$

$\ln\left|4\sin\left(x\right)-3\cos\left(x\right)\right|$
6

Replace $u$ with the value that we assigned to it in the beginning: $4\sin\left(x\right)-3\cos\left(x\right)$

$\ln\left|4\sin\left(x\right)-3\cos\left(x\right)\right|$
7

As the integral that we are solving is an indefinite integral, when we finish integrating we must add the constant of integration $C$

$\ln\left|4\sin\left(x\right)-3\cos\left(x\right)\right|+C_0$

Final answer to the problem

$\ln\left|4\sin\left(x\right)-3\cos\left(x\right)\right|+C_0$

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Function Plot

Plotting: $\ln\left(4\sin\left(x\right)-3\cos\left(x\right)\right)+C_0$

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a
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m
n
u
v
w
x
y
z
.
(◻)
+
-
×
◻/◻
/
÷
2

e
π
ln
log
log
lim
d/dx
Dx
|◻|
θ
=
>
<
>=
<=
sin
cos
tan
cot
sec
csc

asin
acos
atan
acot
asec
acsc

sinh
cosh
tanh
coth
sech
csch

asinh
acosh
atanh
acoth
asech
acsch

How to improve your answer:

Main Topic: Trigonometric Integrals

Integrals that contain trigonometric functions and their powers.

Used Formulas

See formulas (1)

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