Q.Evaluate the definite integral: $\int_{0}^{\pi/4} (2 \sec^2 x + x^3 + 2) \, dx$

The key insight here is that the integrand looks messy, but it contains a hidden structure: the polynomial $x^3 + 3x^2 + 3x + 3$ is almost a perfect cube expansion, and the term $(x+1)\cos(x+1)$ suggests a shift that might produce symmetry. When you see an integral over an interval that isn't symmetric about zero, but the integrand has terms like $(x+1)$, it's often smart to shift the variable so that the interval becomes symmetric. Symmetry then lets you kill odd functions instantly.

Let’s set $t = x + 1$. Then $x = t - 1$, and when $x = -2$, $t = -1$; when $x = 0$, $t = 1$. The differential $dx = dt$, so the integral becomes

Now expand the polynomial part. Compute $(t-1)^3 = t^3 - 3t^2 + 3t - 1$. Then $3(t-1)^2 = 3(t^2 - 2t + 1) = 3t^2 - 6t + 3$. Next $3(t-1) = 3t - 3$. Add the constant $+3$. Summing all these:

• $t^3$ term: $t^3$
• $t^2$ terms: $-3t^2 + 3t^2 = 0$
• $t$ terms: $3t - 6t + 3t = 0$
• constants: $-1 + 3 - 3 + 3 = 2$

So the polynomial simplifies beautifully to $t^3 + 2$. The whole integrand is therefore $t^3 + 2 + t\cos t$.

Now integrate from $-1$ to $1$:

Here’s where symmetry does the heavy lifting. The function $t^3$ is odd — its integral over $[-1,1]$ is zero. The function $t\cos t$ is also odd (product of odd $t$ and even $\cos t$), so its integral over $[-1,1]$ is zero too. Only the constant $2$ survives:

Step 1 — Complete the square.

The radicand has no linear-free form, so we force one:

Put $t = x - 1$ (so $dt = dx$) and $a = 2$. The integral becomes

Step 2 — The standard formula (and why it works).

It is derived by integration by parts (taking $\sqrt{t^2+a^2}$ as the first function and $1$ as the second); the leftover integral reproduces the original one plus $\int \frac{a^2}{\sqrt{t^2+a^2}}dt$, whose value is the logarithmic term.

Step 3 — Substitute $a = 2$.

so

Step 4 — Back-substitute $t = x-1$ and $t^2 + 4 = x^2 - 2x + 5$.

$\int \sqrt{5-2x+x^{2}}\,dx = \frac{x-1}{2}\sqrt{5-2x+x^{2}} + 2\log\left|(x-1) + \sqrt{5-2x+x^{2}}\right| + C$ …

The problem gives you an integral with $\sin^2 x$ and $\cos^2 x$ in the denominator, and the answer choices all involve $\tan^{-1}$ of something with $\tan x$. That’s a strong hint: integrals of the form $\int \frac{dx}{a + b\sin^2 x}$ (or with $\cos^2 x$) are tamed by the substitution $t = \tan x$, because $\sin^2 x$ and $\cos^2 x$ become rational functions of $t$. But here the denominator already has both squares, so the first step is to simplify it into a single term.

Notice $3\sin^2 x + 8\cos^2 x$ is not a constant — but we can rewrite it using $\sin^2 x = 1 - \cos^2 x$ or $\cos^2 x = 1 - \sin^2 x$. Either works; let’s use $\sin^2 x = 1 - \cos^2 x$ to combine terms:

$1 + 3\sin^2 x + 8\cos^2 x = 1 + 3(1 - \cos^2 x) + 8\cos^2 x = 1 + 3 - 3\cos^2 x + 8\cos^2 x = 4 + 5\cos^2 x$.

That’s cleaner: the denominator becomes $4 + 5\cos^2 x$. Now the integral is $\int \frac{dx}{4 + 5\cos^2 x}$.

1. Divide numerator and denominator by $\cos^2 x$. This is the classic move when you see $\cos^2 x$ in the denominator — it turns $\frac{1}{\cos^2 x}$ into $\sec^2 x$, which is the derivative of $\tan x$.

But $\sec^2 x = 1 + \tan^2 x$, so:

2. Substitute $t = \tan x$. Then $dt = \sec^2 x \, dx$, and the integral becomes:

3. Factor to match the standard arctan form. Recall $\int \frac{du}{u^2 + a^2} = \frac{1}{a}\tan^{-1}\left(\frac{u}{a}\right) + C$. Here $4t^2 + 9 = 4\left(t^2 + \frac{9}{4}\right) = 4\left(t^2 + \left(\frac{3}{2}\right)^2\right)$. So: $\int \frac{dt}{4t^2 + 9} = \frac{1}{4} \int \frac{dt}{t^2 + \left(\frac{3}{2}\right)^2}.$ …

Step 1 — Definition (integral as the limit of a sum).

Step 2 — Apply to $f(x)=x^2$. With $f(a+rh)=(a+rh)^2=a^2+2arh+r^2h^2$,

Using $nh=b-a$ and letting $n\to\infty$:

$\int_a^b x^2dx=a^2(b-a)+a(b-a)^2+\frac{(b-a)^3}{3}=\frac{b^3-a^3}{3}.$ …

Step 1 — Spot the substitution

We need

Notice $x^6=(x^3)^2$ and the numerator $x^2\,dx$ is (up to a constant) the differential of $x^3$. So set

Step 2 — Rewrite the integral

Also write $a^6=(a^3)^2$. Then

Step 3 — Apply the standard result

The standard integral is

Why this form and not $\log|u-\sqrt{\cdot}|$: differentiating $\log|u+\sqrt{u^2+k^2}|$ gives …

Step 1 — Factor the denominator.

Treating $x^{6}=(x^{2})^{3}$ and using $a^{3}+b^{3}=(a+b)(a^{2}-ab+b^{2})$ with $a=x^{2},b=1$:

The presence of the factor $(1+x^{2})$ is the hint that a $\tan^{-1}x$ term is coming.

Step 2 — Split the numerator to match that factor.

We want one piece to cancel $(1-x^{2}+x^{4})$. Note the algebraic identity

Hence

Step 3 — Integrate the first piece.

Step 4 — Integrate the second piece.

Put $t=x^{3}\Rightarrow dt=3x^{2}dx\Rightarrow x^{2}dx=\tfrac{dt}{3}$, and $x^{6}=t^{2}$:

Step 5 — Combine. …