If int_0^pi {x,f({{cos }^2}x + {{tan }^4}x),d | vedclass

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(B) Let $I = \int_0^\pi x f(\cos^2 x + 	an^4 x) dx$ ... $(i)$Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we get:$I = \int_0^\pi (\pi -

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$\pi $

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(B) Let $I = \int_0^\pi x f(\cos^2 x + 	an^4 x) dx$ ... $(i)$Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we get:$I = \int_0^\pi (\pi - x) f(\cos^2(\pi - x) + 	an^4(\pi - x)) dx$Since $\cos(\pi - x) = -\cos x$ and $	an(\pi - x) = -	an x$,we have $\cos^2(\pi - x) = \cos^2 x$ and $	an^4(\pi - x) = 	an^4 x$.Thus,$I = \int_0^\pi (\pi - x) f(\cos^2 x + 	an^4 x) dx$ ... $(ii)$Adding $(i)$ and $(ii)$:$2I = \int_0^\pi (x + \pi - x) f(\cos^2 x + 	an^4 x) dx$$2I = \pi \int_0^\pi f(\cos^2 x + 	an^4 x) dx$Using the property $\int_0^{2a} f(x) dx = 2 \int_0^a f(x) dx$ if $f(2a-x) = f(x)$:Since $f(\cos^2(\pi-x) + 	an^4(\pi-x)) = f(\cos^2 x + 	an^4 x)$,the integral satisfies this condition.$2I = \pi \cdot 2 \int_0^{\pi/2} f(\cos^2 x + 	an^4 x) dx$$I = \pi \int_0^{\pi/2} f(\cos^2 x + 	an^4 x) dx$Comparing this with the given equation $I = k \int_0^{\pi/2} f(\cos^2 x + 	an^4 x) dx$,we find $k = \pi$.

If $\int_0^\pi {x\,f({{\cos }^2}x + {{\tan }^4}x)\,dx} = k\int_0^{\pi /2} {f({{\cos }^2}x + {{\tan }^4}x)\,dx,}$ then the value of $k$ is

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