Q.Integrate the following function: $\int (ax^2 + bx + c) dx$

Step 1 — The key identity (and why it works)

We want to show, with $\theta = \dfrac{x}{2}$:

Multiply the right side by $\sin\theta$ and use $2\cos A \sin B = \sin(A+B) - \sin(A-B)$:

• $\sin\theta \cdot 1 = \sin\theta$
• $2\cos 2\theta \sin\theta = \sin 3\theta - \sin\theta$
• $2\cos 4\theta \sin\theta = \sin 5\theta - \sin 3\theta$

Add them — the $\sin\theta$ and $\sin3\theta$ terms telescope away:

Step 2 — Rewrite the integrand

With $\theta = \dfrac{x}{2}$, so $2\theta = x$ and $4\theta = 2x$:

The quotient — which looked like it needed a substitution — is just a polynomial in cosines. That is the whole trick.

Step 3 — Integrate term by term …

The problem asks for the indefinite integral of $\sqrt{\csc x - \sin x}$. At first glance, the expression inside the square root looks messy — cosecant minus sine. But the moment you see $\csc x$, you should think: write everything in terms of $\sin x$ and $\cos x$. That’s almost always the first step when dealing with trigonometric integrals in Indian exams. Once you do that, the expression often simplifies to something you can integrate directly.

Let’s go through it step by step.

1. Rewrite the integrand in terms of sine and cosine. Recall that $\csc x = \frac{1}{\sin x}$. So

Using the identity $\sin^2 x + \cos^2 x = 1$, we have $1 - \sin^2 x = \cos^2 x$. Therefore

2. Take the square root.

In indefinite integration, we typically work over intervals where $\cos x \geq 0$ (e.g., $0 < x < \frac{\pi}{2}$) so that $|\cos x| = \cos x$. The constant $C$ will absorb any sign adjustments for other intervals. So we take

3. Set up the integral.

4. Use substitution. Let $t = \sin x$. Then $dt = \cos x \, dx$. The integral becomes $$$$

Step 1 — Clean up the integrand.

That lone $\cos x$ is the signal: it is exactly the derivative of $\sin x$, so a substitution is available.

Step 2 — Substitute.

Let $t = \sin x \;\Rightarrow\; dt = \cos x\,dx$. Then

Step 3 — Use the standard exponential form.

The standard result is

Here the bracket is $(t+1)$. Take $f(t)=t$; then $f'(t)=1$ and indeed $f(t)+f'(t) = t+1$. ✓ Hence

$I = e^{t}\cdot t + c$ …

The greatest-integer function $[x^2]$ is piecewise constant: it changes value only where $x^2$ hits an integer. So the integral is a sum of rectangles, each of height equal to that integer and width equal to the length of the interval where $x^2$ stays between consecutive integers.

1. Find the breakpoints.

   $x^2 = n$ gives $x = \sqrt n$. For $n = 1,2,3,4$ we get $x = 1, \sqrt2, \sqrt3, 2$, so the interval splits as $0 < 1 < \sqrt2 < \sqrt3 < 2$.

2. Value of $[x^2]$ on each piece.

   • $[0,1)$: $0 \le x^2 < 1 \Rightarrow [x^2] = 0$
   • $[1,\sqrt2)$: $1 \le x^2 < 2 \Rightarrow [x^2] = 1$
   • $[\sqrt2,\sqrt3)$: $2 \le x^2 < 3 \Rightarrow [x^2] = 2$
   • $[\sqrt3,2)$: $3 \le x^2 < 4 \Rightarrow [x^2] = 3$
   Watch out

   At the single point $x=2$, $x^2 = 4$ so $[x^2]=4$, but one isolated point has zero width and does not affect the integral. Use $[x^2]=3$ across $[\sqrt3, 2)$.

3. Write the integral as a sum.

4. Simplify. …

We start with the given inequality involving a definite integral. The key is to evaluate the integral, then solve the resulting quadratic inequality. This is a classic "evaluate, then solve" problem — no tricks, just careful algebra.

1. Evaluate the integral. The integral $\int_0^a x\,dx$ is the area under the line $y=x$ from $0$ to $a$. Its value is

This holds for any real $a$ (if $a<0$, the integral gives a negative area, which is fine).

2. Set up the inequality. The problem states

Multiply both sides by $2$ (positive, so inequality direction stays the same):

3. Rearrange into standard quadratic form. Bring all terms to one side:

4. Factor the quadratic. We look for two numbers that multiply to $-8$ and add to $-2$: those are $-4$ and $+2$. So

Thus the inequality becomes

5. Solve the product inequality. The product of two factors is $\le 0$ when one factor is non-positive and the other non-negative. The critical points are $a = -2$ and $a = 4$. Testing intervals:
   • For $a < -2$: both $(a-4)$ and $(a+2)$ are negative, product positive → not $\le 0$. …