Evaluate the integral : ∫(3+2cosx+4sinx)/(2sinx+cosx+3)dx ​

Question

Evaluate the integral :

\(\int\frac{3+2cosx+4sinx}{2sinx+cosx+3}\)dx

Answer 1

* Integration by substitution : A change in the variable of integration often reduces an integral to one of the fundamental integration. If derivative of a function is present in an integration or if chances of its presence after few modification is possible then we apply integration by substitution method.

* Knowledge of integration of fundamental functions like sin, cos ,polynomial, log etc and formula for some special functions.

Let, I = \(\int\frac{3+2cosx+4sinx}{2sinx+cosx+3}\)dx

To solve such integrals involving trigonometric terms in numerator and denominators. 

We use the basic substitution method and to apply this simply we follow the undermentioned procedure.

If I has the form \(\int\frac{asinx+bcosx+c}{dsinx+ecosx+f}\)dx

Then substitute numerator as -

asinx + bcosx + c = A\(\frac{d}{dx}\)(dsinx + ecosx +f)+ B(dsinx + ecosx +c) + c

Where A, B and C are constants 

We have,

I = \(\int\frac{3+2cosx+4sinx}{2sinx+cosx+3}\)dx

As I matches with the form described above, 

So we will take the steps as described.

Comparing both sides we have : 

3B+ C = 3 

B + 2A = 2 

2B - A = 4 

On solving for A ,B and C we have:

 A = 0, B = 2 and C = -3 

Thus I can be expressed as

So, I₁ reduces to : 

I₁ = 2∫ dx = 2x + C₁ …..equation 2 

As, I₂ = - 3\(\int\frac{1}{2sinx+cosx+3}\) dx

To solve the integrals of the form \(\int\frac{1}{asinx+bcosx+c}\)dx

To apply substitution method we take following procedure. 

We substitute :

As, the denominator is polynomial without any square root term. 

So one of the special integral will be used to solve I₂.