JEE Mathematics 2026 Question with Solution

Given:

$$\alpha=\int_{0}^{64}\left(x^{1/3}-[x^{1/3}]\right)\,dx$$

and we need to evaluate

$$\frac{1}{\pi}\int_{0}^{\alpha\pi}\frac{\sin^2\theta}{\sin^6\theta+\cos^6\theta}\,d\theta$$

Find: The numerical value of the given expression.

Step 1: Evaluate $\alpha$. Let

$$t=x^{1/3},\quad x=t^3,\quad dx=3t^2\,dt$$

When $x=0$, $t=0$ and when $x=64$, $t=4$. Therefore,

$$\alpha=\int_{0}^{4}(t-[t])\,3t^2\,dt$$

Now split the interval according to the greatest integer function over

$$[0,1),\ [1,2),\ [2,3),\ [3,4)$$

So,

$$\alpha=\sum_{k=0}^{3}\int_{k}^{k+1}(t-k)\,3t^2\,dt$$

Evaluating and summing these parts gives

$$\alpha=12$$

Step 2: Evaluate the trigonometric integral. Let

$$I=\frac{1}{\pi}\int_{0}^{12\pi}\frac{\sin^2\theta}{\sin^6\theta+\cos^6\theta}\,d\theta$$

Use the identity

$$\sin^6\theta+\cos^6\theta=(\sin^2\theta+\cos^2\theta)^3-3\sin^2\theta\cos^2\theta=1-3\sin^2\theta\cos^2\theta$$

By symmetry,

$$\int_{0}^{\pi}\frac{\sin^2\theta}{\sin^6\theta+\cos^6\theta}\,d\theta=\frac{\pi}{3}$$

Also, the integrand has period $\pi$, hence

$$\int_{0}^{12\pi}\frac{\sin^2\theta}{\sin^6\theta+\cos^6\theta}\,d\theta=12\int_{0}^{\pi}\frac{\sin^2\theta}{\sin^6\theta+\cos^6\theta}\,d\theta$$

Therefore,

$$I=\frac{1}{\pi}\times 12\times \frac{\pi}{3}=36$$

Therefore, the required value is $36$.](streamdown:incomplete-link)

Given: The inner integral contains the greatest integer function $[x^{1/3}]$.

Find: How the interval splitting leads to the value of the expression.

For a greatest integer function, the expression inside the bracket remains constant between consecutive integers. After substituting $t=x^{1/3}$, the term $[t]$ changes at integer values of $t$. Since $t$ runs from $0$ to $4$, the correct split is at

$$0,1,2,3,4$$

Thus the integral is broken into four pieces where $[t]=k$ on $[k,k+1)$. This gives

$$\alpha=\sum_{k=0}^{3}\int_{k}^{k+1}(t-k)\,3t^2\,dt$$

From the provided working, this sum evaluates to

$$\alpha=12$$

Now the upper limit of the second integral becomes $12\pi$. Since the trigonometric integrand repeats every $\pi$,

$$\int_{0}^{12\pi}(\cdots)\,d\theta=12\int_{0}^{\pi}(\cdots)\,d\theta$$

Using the provided symmetry result

$$\int_{0}^{\pi}\frac{\sin^2\theta}{\sin^6\theta+\cos^6\theta}\,d\theta=\frac{\pi}{3}$$

we obtain

$$\frac{1}{\pi}\cdot 12\cdot \frac{\pi}{3}=36$$

So the final value is $36$.](streamdown:incomplete-link)