Sample 2025-26 Class 12 Maths Q12 Solution: a in ∫ x³ sin⁴(x⁴) cos(x⁴) dx = a sin⁵(x⁴) + C

The question

If $\int x^3 \sin^4(x^4)\cos(x^4)\,dx = a\sin^5(x^4) + C$ ⁠, then $a$ is equal to

(A) $-\frac{1}{10}$

(B) $\frac{1}{20}$

(C) $\frac{1}{4}$

(D) $\frac{1}{5}$

Exam answer

Let $I = \int x^3 \sin^4(x^4)\cos(x^4)\,dx$ ⁠.

Put $\sin(x^4) = t$ ⁠. Then $4x^3\cos(x^4)\,dx = dt$ ⁠.

Why?

Choose $t = \sin(x^4)$ because its derivative is, apart from the 4, the rest of the integrand. By the chain rule,

\frac{d}{dx}\sin(x^4) = \cos(x^4) \cdot 4x^3

Divide both sides of $4x^3\cos(x^4)\,dx = dt$ by 4 to get the next line.

So $x^3\cos(x^4)\,dx = \frac{1}{4}\,dt$ ⁠.

\begin{aligned} I &= \frac{1}{4}\int t^4\,dt = \frac{t^5}{20} + C \\ &= \frac{1}{20}\sin^5(x^4) + C \end{aligned}

Why?

$\sin^4(x^4)$ becomes $t^4$ ⁠. Group the integrand into the two parts:

\begin{aligned} I &= \int \underbrace{\sin^4(x^4)}_{t^4}\,\underbrace{x^3\cos(x^4)\,dx}_{\frac14\,dt} \\ &= \int t^4 \cdot \frac14\,dt = \frac14\int t^4\,dt \end{aligned}

The power rule:

\int t^n\,dt = \frac{t^{n + 1}}{n + 1} + C

So $\int t^4\,dt = \frac{t^5}{5}$ ⁠, and $\frac{1}{4} \cdot \frac{t^5}{5} = \frac{t^5}{20}$ ⁠. Then put $t = \sin(x^4)$ back.

$\therefore\ a = \frac{1}{20}$

Why?

Compare with $a\sin^5(x^4) + C$ ⁠: $a$ is the number in front of $\sin^5(x^4)$ ⁠.

Watch out: $\frac{1}{5}$ (D) forgets the $\frac{1}{4}$ from $dt$ ⁠, and $\frac{1}{4}$ (C) forgets the $\frac{1}{5}$ from the power rule.

Answer: (B) $\dfrac{1}{20}$

Where this comes from

Chapter 7, Integrals: Integration by substitution.

Question type: Substitution

How to spot it: click to see

Type 2. Substitution: when the derivative is hiding in the question

Spot it: a function and its derivative both appear, like $\log x$ and $\frac{1}{x}$ ⁠, or $x^2 + 1$ and $x$ ⁠.

Here: $\sin^4(x^4)$ sits beside $x^3\cos(x^4)$ ⁠, which is the derivative of $\sin(x^4)$ apart from the 4. So put $\sin(x^4) = t$ ⁠, and the integral is $\frac14\int t^4\,dt$ ⁠.

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Q12: aaa in ∫x3sin⁡4(x4)cos⁡(x4) dx=asin⁡5(x4)+C\int x^3\sin^4(x^4)\cos(x^4)\,dx = a\sin^5(x^4) + C∫x3sin4(x4)cos(x4)dx=asin5(x4)+C

Q12: $a$ in $\int x^3\sin^4(x^4)\cos(x^4)\,dx = a\sin^5(x^4) + C$