1. **State the problem:** We want to evaluate or simplify the integral $$\int \frac{\cos^n x - \delta_i^n x}{\sqrt{1 + \cos^n x}} \, dn$$ with respect to $n$. 2. **Analyze the integrand:** The numerator is $\cos^n x - \delta_i^n x$. Here, $\cos^n x$ means $(\cos x)^n$, i.e., cosine of $x$ raised to the power $n$. $\delta_i^n x$ looks like a term involving Kronecker delta or some function indexed by $i$ and $n$ applied to $x$. Without further context or definition of $\delta_i^n x$, it's unclear how to simplify it. 3. **Nature of integration variable:** The integral is with respect to $n$. Usually, $n$ is an integer index (like in sums or discrete powers), but here it acts as a continuous variable. 4. **Complications:** - The integrand depends on $n$ in powers: $\cos^n x$ and $\delta_i^n x$. - The denominator depends on $n$ as $\sqrt{1 + \cos^n x}$. 5. **Approach:** - Without explicit definition of $\delta_i^n x$, simplify assuming $\delta_i^n x = 0$ (or no term), to illustrate process. - Then the integral reduces to $$\int \frac{\cos^n x}{\sqrt{1 + \cos^n x}} \, dn$$ 6. **Substitution:** Let $u = \cos^n x$, so $du/dn = \cos^n x \ln(\cos x)$. But since integral is w.r.t $n$, make substitution: $$u = \cos^n x = e^{n \ln(\cos x)}$$ Then: $$\frac{du}{dn} = \ln(\cos x) e^{n \ln(\cos x)} = u \ln(\cos x)$$ thus: $$dn = \frac{du}{u \ln(\cos x)}$$ 7. **Rewrite integral in $u$:** $$\int \frac{u}{\sqrt{1+u}} \cdot dn = \int \frac{u}{\sqrt{1+u}} \cdot \frac{du}{u \ln(\cos x)} = \frac{1}{\ln(\cos x)} \int \frac{1}{\sqrt{1+u}} du$$ 8. **Integrate:** $$\int \frac{1}{\sqrt{1+u}} du = 2 \sqrt{1+u} + C$$ 9. **Final answer:** $$\int \frac{\cos^n x}{\sqrt{1 + \cos^n x}} \, dn = \frac{2}{\ln(\cos x)} \sqrt{1 + \cos^n x} + C$$ 10. **Note:** The original integral has an additional $- \delta_i^n x$ term whose effect can't be analyzed without explicit definition. Hence the solution depends on that term's nature. If known, integrate similarly term-by-term.