}\right]_0^{64}=192 \& \\ & \int_0^{64}\left[X^{1 / 3}\right] d x=\int_0^1\left[X^{1 / 3}\right] d x+\int_1^8\left[X^{1 / 3}\right] d x+\int_8^{27}\left[X^{1 / 3}\right] d x \int_{27}^{64}\left[X^{1 / 3}\right] d x=156 \end{aligned} $$

So $\alpha=192-156=36$

Now $E=\frac{1}{\pi} \int_0^{36 \pi} \frac{\sin ^2{ }_\theta}{\sin ^6 \theta+\cos ^6 \theta} d \theta=\frac{36}{\pi} \cdot 2 \int_0^\pi \frac{\sin ^2 \theta}{\sin ^6 \theta+\cos ^6 \theta} d \theta$

$$ \Rightarrow E=\frac{36}{\pi} \int_0^{\pi / 2} \frac{\sin ^2 \theta}{\sin ^6 \theta+\cos ^6 \theta} d \theta $$

Let $J=\int_0^{\pi / 2} \frac{\sin ^2 \theta}{\sin ^6 \theta+\cos ^6 \theta} d \theta \ldots \ldots \ldots$

Applying King $J=\int_0^{\pi / 2} \frac{\cos ^2 \theta}{\sin ^6 \theta+\cos ^6 \theta} d \theta \ldots \ldots \ldots .$.

Now $2 J=\int_0^{\pi / 2} \frac{1}{\sin ^6 \theta+\cos ^6 \theta} d \theta(\operatorname{add}(1) \&(2))=\int_0^{\pi / 2} \frac{\sec ^6 \theta}{\tan ^6 \theta+1} d \theta$

$$ \begin{aligned} &\text { Put } \tan \theta=\lambda\\ &\begin{aligned} & \int_0^{\infty} \frac{\left(1+\lambda^2\right)^2}{\lambda^4-\lambda^2+1} d \lambda \\ & \int_0^{\infty} \frac{1+\frac{1}{\lambda^2}}{\lambda^2-1+\frac{1}{\lambda^2}} d \lambda=\pi \\ & \Rightarrow J=\frac{\pi}{2} \\ & E=\frac{36}{\pi} \cdot 2 J=36 \end{aligned} \end{aligned} $$

19:T60a,

$$ \begin{aligned} & \int_0^{64} X^{\frac{1}{3}} d x=\frac{3}{4} \cdot\left[X^{\frac{4}{3}}\right]_0^{64}=192 \& \\ & \int_0^{64}\left[X^{1 / 3}\right] d x=\int_0^1\left[X^{1 / 3}\right] d x+\int_1^8\left[X^{1 / 3}\right] d x+\int_8^{27}\left[X^{1 / 3}\right] d x \int_{27}^{64}\left[X^{1 / 3}\right] d x=156 \end{aligned} $$

अतः $\alpha=192-156=36$

अब $E=\frac{1}{\pi} \int_0^{36 \pi} \frac{\sin ^2{ }_\theta}{\sin ^6 \theta+\cos ^6 \theta} d \theta=\frac{36}{\pi} \cdot 2 \int_0^\pi \frac{\sin ^2 \theta}{\sin ^6 \theta+\cos ^6 \theta} d \theta$

$$ \Rightarrow E=\frac{36}{\pi} \int_0^{\pi / 2} \frac{\sin ^2 \theta}{\sin ^6 \

Definite Integration

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