Q.By using the properties of definite integrals, evaluate the integral $\int_{0}^{\pi/2}\cos^2 x\,dx$

Step 1 — Spot the structure.

Everything except $\cos x\,dx$ is a function of $\sin x$ — the classic cue for $t = \sin x$.

Step 2 — Substitute, changing the limits too.

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

• When $x = 0$: $t = \sin 0 = 0$.
• When $x = \dfrac{\pi}{2}$: $t = \sin\dfrac{\pi}{2} = 1$.

Step 3 — Handle the improper rational function.

The degree of the numerator equals that of the denominator, so split it first:

Step 4 — Integrate and apply the limits.

Step 5 — Sanity check.

$\log 2 \approx 0.693$, so $I \approx 0.307$ — positive, as it must be since the integrand is $\ge 0$ on $[0, \pi/2]$. This also rules out (B) $-\log 2$ and (D) $\log 2 - 1$, both negative. ✓

ANSWER: C

The key insight here is symmetry. When you integrate an odd function over an interval symmetric about zero, the positive and negative contributions cancel exactly, giving zero. This is one of the most powerful shortcuts in definite integration — it saves you from doing any messy polynomial expansion or trigonometric substitution.

Let’s check if the integrand is odd.

1. Recall the definition of an odd function A function $f(x)$ is odd if $f(-x) = -f(x)$ for all $x$ in its domain. For such a function,

2. Examine the integrand Let

Replace $x$ with $-x$:

Since $\sin(-x) = -\sin x$ and $\cos(-x) = \cos x$, we get

3. Conclusion about parity

   Because $f(-x) = -f(x)$, the function is odd.

4. Apply the symmetry property

   The integration limits are $-\frac{\pi}{2}$ to $\frac{\pi}{2}$, which is symmetric about zero. Therefore,

Concept: if f is an odd function, the integral of f over a symmetric interval [-a, a] is zero.

sin(-x) = -sin(x), so sin x is odd, and the interval [-pi/2, pi/2] is symmetric about 0.

Direct check: integral of sin x dx = -cos x, evaluated from -pi/2 to pi/2:

(-cos(pi/2)) - (-cos(-pi/2)) = -0 + 0 = 0.

ANSWER: C

The relevant concept here is the odd/even function trick for definite integrals, but with a twist. $\cot^{-1}x$ is neither odd nor even. However, it has a useful symmetry: for any $x$, $\cot^{-1}(-x) = \pi - \cot^{-1}x$. This lets us rewrite the integral over $[-3, 3]$ in a way that cancels the "odd part" and leaves only a constant times the length of the interval.

Let’s work through it step by step.

1. Set up the integral and split the interval. We have $I = \int_{-3}^{3} \cot^{-1}x \, dx$. A standard trick for symmetric limits is to split at $0$:

2. Use the substitution $x \to -t$ on the first integral. Let $x = -t$, so $dx = -dt$. When $x = -3$, $t = 3$; when $x = 0$, $t = 0$. Then

3. Apply the symmetry property. For any $t$, $\cot^{-1}(-t) = \pi - \cot^{-1}t$. So the first integral becomes

4. Combine with the second integral. The original $I$ is now

The two $\int_{0}^{3} \cot^{-1}t \, dt$ terms cancel exactly.

So we are left with

Step 1 — Simplify the integrand.

Method 1 — King's property (the elegant route).

The property states $\displaystyle\int_0^a f(x)\,dx=\int_0^a f(a-x)\,dx$. With $a=1$:

But $\log\dfrac{x}{1-x}=-\log\dfrac{1-x}{x}$, so the second expression is $-I$. Hence

Method 2 — Evaluate the two pieces directly (the check).

For the second, integrate by parts:

using $\lim_{x\to0^+}x\log x=0$.

For the first, substitute $u=1-x$ (so $du=-dx$, and the limits swap):

Therefore

Both the integrals are improper (the integrand blows up at $x=0$ and $x=1$) but each converges, so the cancellation is legitimate.

ANSWER: B

Step 1 — The concept

$|f(x)|$ equals $f(x)$ where $f \ge 0$ and $-f(x)$ where $f < 0$. So before integrating we must find where each expression inside the modulus changes sign within the limits $[1, 5]$, and use additivity of the integral to split there.

• $|x - 3|$ changes sign at $x = 3$, which is inside $[1,5]$ → must split.
• $|1 - x|$ changes sign at $x = 1$, which is the left endpoint → no interior split needed; on all of $(1,5]$ we have $x > 1$, so $1 - x < 0$ and $|1-x| = x - 1$.

Step 2 — First integral: $\displaystyle\int_1^5 |x-3|\,dx$

(Geometric check: two right triangles, each of base $2$ and height $2$, area $\tfrac12(2)(2) = 2$ each. ✓)

Step 3 — Second integral: $\displaystyle\int_1^5 |1-x|\,dx$

On $[1,5]$, $x \ge 1$, so $|1 - x| = x - 1$:

(Geometric check: a right triangle of base $4$ and height $4$, area $\tfrac12(4)(4) = 8$. ✓)

Step 4 — Add

ANSWER: A

Step 1 — Recall the odd-function property

If $g(-x) = -g(x)$ for all $x$ in $[-2,2]$, then $\displaystyle\int_{-2}^{2} g(x)\,dx = 0$.

Step 2 — Test option (A)

So (A) is odd, and its integral over $[-2,2]$ is $0$.

Step 3 — Rule out the remaining options

(B) $\sin(2+x)$ is not odd about $x=0$. (C) $2x^3+2x+1$ has the even constant term $1$, contributing $\int_{-2}^{2}1\,dx = 4 \ne 0$. (D) $2x\tan x$ is a product of two odd functions, hence even, and does not integrate to zero in general.