MCQ 12 Marks
If $I =\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{\sin ^4 x}{\sin ^4 x+\cos ^4 x} d x$, then the value of $I$ is :
- A$\frac{\pi}{3}$
- B$\pi$
- C$\frac{\pi}{2}$
- D$\frac{\pi}{4}$
Answer
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$\frac{1}{2}$
$I=\int_0^k \frac{1}{2\left(1+(2 x)^2\right)} d x=\frac{\pi}{16}$
$\therefore \frac{1}{2} \times \frac{1}{2}\left[\tan ^{-1}(2 x)\right]_0^k=\frac{\pi}{16}$
$\tan ^{-1} 2 k-\tan ^{-1} 0=\frac{\pi}{4}$
$2 k=1$
$k=\frac{1}{2}$
$\begin{aligned} & \int_0^\alpha 3 x^2 d x=8 \\ & \Rightarrow\left[\frac{3 x^3}{3}\right]_0^\alpha=8 \\ & \Rightarrow\left[x^3\right]_0^\alpha=8 \\ & \Rightarrow \alpha ^3=8 \\ & \therefore \alpha=2\end{aligned}$
$\begin{aligned} & \int_0^\alpha\left(3 x^2+2 x+1\right) d x=14 \\ & {\left[x^3+x^2+x\right]_0^\alpha=14} \\ & \alpha^3+\alpha^2+\alpha-14=0 \\ & (\alpha-2)\left(\alpha^2+3 \alpha+7\right)=0\end{aligned}$
But $\alpha^2+3 \alpha+7=0$ does not have real roots
$\alpha=2$
$\begin{aligned} & \text { Let } I=\int_4^9 \frac{1}{\sqrt{x}} d x \\ & =\int_4^9 x^{-\left(\frac{1}{2}\right)} d x \\ & =\left[\frac{x^{\frac{1}{2}}}{\frac{1}{2}}\right]_4^9=2[\sqrt{x}]_4^9 \\ & =2(\sqrt{9}-\sqrt{4}) \\ & =2(3-2) \\ & \therefore I=2\end{aligned}$