Ideas required to solve the problems:

* 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 =

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

Then substitute numerator as -

Where A, B and C are constants

We have, I =

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:

I =

I =

∴ Let I₁ = and I₂ =

⇒ I = I₁ + I₂ ….equation 1

I₁ =

So, I₁ reduces to:

I₁ = …..equation 2

As, I₂ =

To solve the integrals of the form

To apply substitution method we take following procedure.

We substitute:

∴ I₂ =

⇒ I₂ =

⇒ I₂ =

⇒ I₂ =

Let, t = ⇒

∴ I₂ =

As, the denominator is polynomial without any square root term. So one of the special integral will be used to solve I₂.

I₂ =

⇒ I₂ =

∴ I₂ = {∵ a² + 2ab + b² = (a+b)²}

As, I₂ matches with the special integral form

I₂ =

Putting value of t we have:

∴ I₂ = + C₂ ……equation 3

From equation 1,2 and 3:

I = + C₂

∴ I = + C ….ans