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Q.
Let $f(x)=x^4+|x|$ and $I_1=\int\limits_0^\pi f(\cos x) d x$ and $I_2=\int\limits_0^{\frac{\pi}{2}} f(\sin x) d x$, then the value of $\frac{I_1}{I_2}$ is
Integrals
Solution:
$I_1=\int\limits_0^\pi\left(\cos ^4 x+|\cos x|\right) d x$
$I _1=2 \int\limits_0^{\frac{\pi}{2}} \cos ^4 x +|\cos x | dx $
$\downarrow \text { king } $
$I _1=2 \int\limits_0^{\frac{\pi}{2}}\left(\sin ^4 x +|\sin x |\right) dx =2 I _2 \Rightarrow \frac{ I _1}{ I _2}=2 $