Assertion (A): $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{(\sin x)^{\sqrt{2}} d x}{(\sin…
Assertion (A): $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{(\sin x)^{\sqrt{2}} d x}{(\sin x)^{\sqrt{2}}+(\cos x)^{\sqrt{2}}}=\frac{\pi}{12}$
Reason (R): $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{f(x) d x}{f(x)+f\left(\frac{\pi}{2}-x\right)}=\frac{\pi}{12}$
- A is true, $\mathrm{R}$ is true and $\mathrm{R}$ is the correct explanation of A
- A is true, $\mathrm{R}$ is true but $\mathrm{R}$ is not the correct explanation of $\mathrm{A}$
- $\mathrm{A}$ is true, $\mathrm{R}$ is false
- A is false, $\mathrm{R}$ is true
Solution
Let $\int_{\pi / 6}^{\pi / 3} \frac{(\sin x)^{\sqrt{2}} d x}{(\sin x)^{\sqrt{2}}+(\cos x)^{\sqrt{2}}}...(1)$
Since $\int_a^b f(x) d x=\int_a^b f(a+b-x) d x$
Hence $I=\int_{\pi / 6}^{\pi / 3} \frac{(\cos x)^{\sqrt{2}} \cdot d x}{(\sin x)^{\sqrt{2}}+(\cos x)^{\sqrt{2}}}...(2)$
$
\begin{aligned}
& \text { Equation (1)+(2), } \\
& \Rightarrow I \int_{1 / 3}^{/ 3} \frac{(\sin x)^{\sqrt{2}}(\cos x)^{\sqrt{2}}}{(\sin x)^{\sqrt{2}}(\cos x)^{\sqrt{2}}} d x \\
& \Rightarrow I=\frac{\pi}{12}
\end{aligned}
$
Hence Assertion ' $A$ ' is correct.
$
\begin{aligned}
& \text { Let } P=\int_{\pi / 6}^{\pi / 3} \frac{f(x) d x}{f(x)+f\left(\frac{\pi}{2}-x\right)}...(3) \\
& \Rightarrow P=\int_{\pi / 6}^{\pi / 3} \frac{f\left(\frac{\pi}{2}-x\right) d x}{f(x)+f\left(\frac{\pi}{2}-x\right)}...(4)
\end{aligned}
$
Equation (3) + (4),
$
2 P=\int_{\pi / 6}^{\pi / 3} d x \Rightarrow P=\frac{\pi}{12}
$
Hence Reason $R$ is also true putting $f(x)=(\sin x)^{\sqrt{2}}$ in Reason ' $R$ ' we get$\int_{\pi / 6}^{\pi / 3} \frac{(\sin x)^{\sqrt{2}} d x}{(\sin x)^{\sqrt{2}}+(\cos x)^{\sqrt{2}}}=\frac{\pi}{12}$
Hence $R$ is correct explanation of ' $A$ '
Asked in: AP EAMCET 2023 (15 May Shift 1)
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