Let $I=\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{4} x}{\sin ^{4} x+\cos ^{4} x} d x$ ...(i) Then, by using, $\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x$, we get $I=\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{4}\left(\frac{\pi}{2}-x\right)}{\sin ^{4}\left(\frac{\pi}{2}-x\right)+\cos ^{4}\left(\frac{\pi}{2}-x\right)} d x$ = $\int_{0}^{\frac{\pi}{2}} \frac{\cos ^{4} x}{\cos ^{4} x+\sin ^{4} x} d x$ ...(ii) Adding (i) and (ii), we get $2 \mathrm{I}=\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{4} x+\cos ^{4} x}{\sin ^{4} x+\cos ^{4} x} d x=\int_{0}^{\frac{\pi}{2}} 1 \cdot d x=[x]_{0}^{\frac{\pi}{2}}=\frac{\pi}{2}$ Hence, $I=\frac{\pi}{4}$
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