\(1\mid={ }^{1} \int_{-1}\left(x^{7}+\cos ^{-1} x\right) d x\) then \(\cos \mid\) is equal to....... (a) 1 (b) 0 (c) \(-1\) (d) \((1 / 2)\)

First, we need to integrate the expression inside the integral: \(x^7 + \cos^{-1}(x)\). Let's break it down into two integrals, and integrate each term separately. 1. Integrate \(x^7\): This is a simple power rule integration, which is given by \(\int x^n dx = \frac{1}{n+1}x^{n+1}\). Substituting n = 7, we get: \[\int x^7 dx = \frac{1}{8}x^8\]. 2. Integrate \(\cos^{-1}(x)\): Integrating \(\cos^{-1}(x)\) is more complex. Here we will use integration by parts: Take: \[u = \cos^{-1}(x), \hspace{1cm} dv = dx\] Then: \[du = \frac{-1}{\sqrt{1 - x^2}} dx, \hspace{1cm} v = x\] Applying integration by parts: \[\int \cos^{-1}(x) dx = x \cdot \cos^{-1}(x) - \int x \cdot \frac{-1}{\sqrt{1 - x^2}} dx\] Simplifying and integrating the remaining part: \[\Rightarrow \int \cos^{-1}(x) dx = x \cdot \cos^{-1}(x) + \int \frac{x}{\sqrt{1 - x^2}} dx\] Now, using the substitution method: Let: \[t = 1 - x^2, \hspace{1cm} dt = -2x dx\] Integrating, we get: \[\int \frac{x}{\sqrt{1 - x^2}} dx = -\frac{1}{2} \int \frac{1}{\sqrt{t}} dt = -\frac{1}{2} \cdot 2 \sqrt{t} + C\] Substituting back the value of \(t\), we get: \[\int \cos^{-1}(x) dx = x \cdot \cos^{-1}(x) - \sqrt{1 - x^2} + C\]. Now, combining both integrations: \[\int_{} (x^7 + \cos^{-1} x) dx = \frac{1}{8}x^8 + x \cdot \cos^{-1}(x) - \sqrt{1 - x^2} + C\]

Now, we need to evaluate the integral within the given limits, which are from -1 to 1. Using the fundamental theorem of calculus: \[F(1) - F(-1) = \left(\frac{1}{8}(1)^8 + 1 \cdot \cos^{-1}(1) - \sqrt{1 - (1)^2}\right) - \left(\frac{1}{8}(-1)^8 + (-1) \cdot \cos^{-1}{(-1)} - \sqrt{1 - (-1)^2}\right)\]

Now let's simplify and solve the expression: \[F(1) - F(-1) = \left(\frac{1}{8} + 0 - 0\right) - \left(\frac{1}{8} - \pi + 0\right) = \pi\] So we have: \[1\mid = \pi\] Now, comparing the result with the given expression: \[1\mid = {}^{1} \int_{-1}\left(x^{7}+\cos ^{-1} x\right) d x \longrightarrow \cos \mid = \pi\] However, none of the given options matches the value of \(\pi\). This could be an issue with the exercise itself or the limitations of the possible answers provided.