Evaluate the integral : ∫1/(4+3tanx)dx ​

Question

Evaluate the integral:

\(\int\frac{1}{4+3tanx}\) 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 = \(\int\frac{1}{4+3tanx}\)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 = \(\int\frac{1}{4+3tanx}\)dx

= \(\int\frac{1}{4+3{\frac{sinx}{cosx}}}\)dx

= \(\int\frac{cosx}{3sinx+4cosx}\)dx

As I matches with the form described above, 

So we will take the steps as described.

Comparing both sides we have : 

C = 0 3B - 4A = 0 4B + 3A = 1 

On solving for A ,B and C we have:

 A = 3/25 , B = 4/25 and C = 0 

Thus I can be expressed as:

Let, 4 cos x + 3sin x = u 

⇒ (-4sin x + 3cos x)dx = du 

So, I₁ reduces to: