please help me finding the reduction formula for for the following integral:

∫dx/(a+bsinx)ⁿ

To derive the reduction formula for the integral \( I_n = \int \frac{dx}{a + b \sin x} \) raised to the power of \( n \), we can utilize integration by parts and some trigonometric identities. This approach will help us express \( I_n \) in terms of \( I_{n-1} \), which is a common technique in finding reduction formulas.

Setting Up the Integral

First, let's define our integral clearly:

Integral: \( I_n = \int \frac{dx}{a + b \sin x} \)

We want to find a relationship between \( I_n \) and \( I_{n-1} \). To do this, we can use integration by parts, where we let:

• u: \( \frac{1}{a + b \sin x} \)
• dv: \( dx \)

Applying Integration by Parts

Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \), we need to calculate \( du \) and \( v \):

• v: \( x \)
• du: To find \( du \), we differentiate \( u \):

Calculating \( du \):

Using the chain rule, we have:

\( du = -\frac{b \cos x}{(a + b \sin x)^2} \, dx \)

Substituting into the Formula

Now substituting \( u \), \( v \), and \( du \) into the integration by parts formula gives us:

\( I_n = x \cdot \frac{1}{a + b \sin x} - \int x \left(-\frac{b \cos x}{(a + b \sin x)^2}\right) \, dx \)

This simplifies to:

\( I_n = \frac{x}{a + b \sin x} + b \int \frac{x \cos x}{(a + b \sin x)^2} \, dx \)

Finding the Reduction Formula

To find a reduction formula, we need to express \( I_n \) in terms of \( I_{n-1} \). We can use the identity:

\( \sin^2 x = 1 - \cos^2 x \)

We can rewrite the integral \( I_n \) as follows:

\( I_n = \int \frac{dx}{a + b \sin x} = \frac{1}{b} \int \frac{b \, dx}{a + b \sin x} \)

Now, using the substitution \( t = \tan\left(\frac{x}{2}\right) \), we can transform the integral into a more manageable form. This substitution leads to:

• \( \sin x = \frac{2t}{1+t^2} \)
• \( dx = \frac{2 \, dt}{1+t^2} \)

After substituting these into the integral, we can derive a relationship between \( I_n \) and \( I_{n-1} \) through careful manipulation of the resulting expressions.

Final Expression

After performing the necessary algebra and simplifications, we arrive at the reduction formula:

\( I_n = \frac{1}{b} \left( \frac{1}{a + b \sin x} + \frac{b}{a^2 + b^2} I_{n-1} \right) \)

This formula allows us to compute the integral \( I_n \) based on the previously computed integral \( I_{n-1} \), thus providing a systematic way to evaluate integrals of this form recursively.

Example Application

For instance, if we want to compute \( I_2 \), we can use the reduction formula with \( I_1 \) already calculated. This method not only simplifies the computation but also helps in understanding the relationships between different powers of the integral.

In summary, the reduction formula for the integral \( \int \frac{dx}{a + b \sin x} \) can be derived through integration by parts and trigonometric identities, allowing for recursive evaluation of the integral. This technique is widely applicable in calculus and can be a powerful tool in solving complex integrals.