Mathematics

If $$f(x)=\displaystyle\int^x_0(\cos (\sin t)+\cos(\cos t)dt$$, then $$f(x+\pi)$$ is?


ANSWER

$$=f(\pi)+2f(\dfrac{\pi}{2})$$


SOLUTION
$$f(x)+f(\pi)$$
$$f(x+\pi)=\displaystyle\int^{x+\pi}_0(\cos(\sin t)+\cos(\cos t))dt$$
$$=\displaystyle\int^{\pi}_0(\cos(\sin t)+\cos (\cos t))dt$$
$$+\displaystyle\int^{x+\pi}_{\pi}(\cos (\sin t)+\cos (\cos t))dt$$
$$=f(\pi)+\displaystyle\int^x_0(\cos (\sin t)+\cos(\cos t))dt$$
$$[\therefore$$ for $$g(x)=\cos(\sin x)+\cos(\cos x), f(x+\pi)=f(x)]$$
$$=f(\pi)+f(x)$$
$$=f(\pi)+2f(\dfrac{\pi}{2})$$
$$[\therefore g(x)$$ has period $$\pi/2]$$.
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Single Correct Medium Published on 17th 09, 2020
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