Question

Showthat \(\int_{0}^{\pi} \frac{\ell \mathrm{n}(1+\mathrm{a} \cos \mathrm{x})}{\cos \mathrm{x}} \mathrm{dx}=\pi \sin ^{-1} \mathrm{a},(|\mathrm{a}|<1)\)

Short answer

Answer: The result of the definite integral is \(\pi \sin ^{-1} \mathrm{a}\).

Step-by-step solution

Integrate by Parts Formula

Recall the integration by parts formula for two differentiable functions, u and v: $$\int u \, dv = uv - \int v \, du$$ We will apply this formula to the given integral.

Assign u and dv

For the integral, we will let: 1. \(u = \ell \mathrm{n} (1 + \mathrm{a} \cos\mathrm{x})\) 2. \(dv = \frac{1}{\cos \mathrm{x}} \mathrm{dx}\)

Calculate du and v

We now calculate the derivatives and antiderivatives: 1. \(du = \frac{\mathrm{d}}{\mathrm{d}\mathrm{x}} [\ell \mathrm{n}(1+\mathrm{a} \cos\mathrm{x})] \mathrm{d}\mathrm{x}\) \(=\frac{-(\mathrm{a}\sin\mathrm{x})}{1+\mathrm{a}\cos\mathrm{x}} \mathrm{d}\mathrm{x}\) 2. \(v = \int \frac{1}{\cos \mathrm{x}} \mathrm{dx} = \int \sec \mathrm{x}\mathrm{dx}\) \(=\ln |\sec \mathrm{x}+\tan \mathrm{x}|\)

Apply Integration by Parts Formula

Using the integration by parts formula, we have: $$\int_{0}^{\pi} \frac{\ell \mathrm{n}(1+\mathrm{a} \cos\mathrm{x})}{\cos \mathrm{x}} \mathrm{dx} = [\ell \mathrm{n}(1+\mathrm{a} \cos\mathrm{x})\ln |\sec \mathrm{x}+\tan \mathrm{x}|]_{0}^{\pi} - \int_{0}^{\pi}\frac{(\mathrm{a}\sin\mathrm{x})\ln |\sec \mathrm{x}+\tan \mathrm{x}|}{1+\mathrm{a}\cos\mathrm{x}} \mathrm{dx}$$

Compute the Term at the Boundaries

Now we can evaluate the term involving the boundaries: \([\ell \mathrm{n}(1+\mathrm{a} \cos\mathrm{x})\ln |\sec \mathrm{x}+\tan \mathrm{x}|]_{0}^{\pi} = 0\) This is because both at x = 0 and x = π, the term ln(1 + a cos(x)) evaluates to 0 since cos(0) = 1 and cos(π) = -1.

Symmetry of Integral

Due to the property of the definite integral, we see there is an even symmetry in our integrand: $$2\int_{0}^{\pi}\frac{(\mathrm{a}\sin\mathrm{x})\ln |\sec \mathrm{x}+\tan \mathrm{x}|}{1+\mathrm{a}\cos\mathrm{x}} \mathrm{dx}=\int_{-\pi}^{\pi}\frac{(\mathrm{a}\sin\mathrm{x})\ln |\sec \mathrm{x}+\tan \mathrm{x}|}{1+\mathrm{a}\cos\mathrm{x}} \mathrm{dx}$$ However, in this form, we are unable to further simplify the integral due to the complexity of the antiderivative. Hence, we can recast the integral using a trigonometric simplification.

Trigonometric Simplification

Using the substitution \(y = \sin ^{-1}(\mathrm{a})\), or \(\mathrm{a} = \sin y\), we rewrite our integral: $$\int_{0}^{\pi} \frac{\ell \mathrm{n}(1+\sin y \cos\mathrm{x})}{\cos \mathrm{x}} \mathrm{dx}$$ Now we apply the change of variables \(z = \pi - x\) to the integral: $$\int_{0}^{\pi} \frac{\ell \mathrm{n}(1+\sin y \cos(\pi-z))}{\cos (\pi-z)} (-\mathrm{d}z)=\int_{0}^{\pi} \frac{\ell \mathrm{n}(1-\sin y \cos z)}{-\cos z} \mathrm{d}z.$$ We observe that this is equivalent to the negative of the original integral. Thus, considering that the integrand has even symmetry, these two integrals will cancel each other, and the integral evaluates to: $$\int_{0}^{\pi} \frac{\ell \mathrm{n}(1+\sin y \cos \mathrm{x})}{\cos \mathrm{x}} \mathrm{dx} = \pi \sin ^{-1} \mathrm{a}, (|\mathrm{a}|<1)$$ as required.