Q.If ∫ 𝑥3 𝑠𝑖𝑛4(𝑥4) cos(𝑥4) 𝑑𝑥 = 𝑎 𝑠𝑖𝑛5(𝑥4) + C, then 𝑎 is equal to
(A) − 1 10
(B) 1 20
(C) 1 4
(D) 1 5

Concept: U Substitution — the integrand contains a composite function whose derivative (up to a constant) is also present.

Let $u = \sin(x^4)$. Then $du = \cos(x^4) \cdot 4x^3 \, dx$, so $x^3 \cos(x^4) \, dx = \frac{du}{4}$.

The integral becomes: …

We have the integral $\int x^3 \sin^4(x^4) \cos(x^4) \, dx$ and are told it equals $a \sin^5(x^4) + C$. The task is to find $a$.

The key insight is that the integrand contains a composition of functions: $\sin^4(x^4)$ and $\cos(x^4)$, multiplied by $x^3$. The derivative of $x^4$ is $4x^3$, and we see $x^3$ sitting there — a perfect setup for substitution. When you see a function and its derivative (or a constant multiple) nearby, substitution is the natural path.

Let’s work through it step by step.

1. Choose the substitution.

   The inner function $x^4$ appears inside both sine and cosine. Let $u = x^4$. Then $du = 4x^3 \, dx$, so $x^3 \, dx = \frac{du}{4}$.

2. Rewrite the integral in terms of $u$.

   The integral becomes:

3. Now handle the $u$-integral. We have $\sin^4(u) \cos(u)$. Notice that the derivative of $\sin(u)$ is $\cos(u)$. So let $v = \sin(u)$. Then $dv = \cos(u) \, du$. The integral becomes:

4. Back-substitute. First $v = \sin(u)$, then $u = x^4$:

5. Compare with the given form. …

Method: Finding an unknown coefficient by substitution

Use this when an integral of a composite function is given in the form (constant) $\times$ (some function) $+C$, and you must identify the constant. You do not need to "guess" — integrate honestly and compare.

Steps

Step 1: Spot the inner function whose derivative is present.

Look for a chunk $g(x)$ sitting inside another function, with $g'(x)$ (up to a numerical factor) also appearing in the integrand. Here powers of $x$ next to a $\sin/\cos$ of $x^4$ signal $g(x)=x^4$ (or directly $g(x)=\sin(x^4)$).

Step 2: Substitute $u=g(x)$ and convert $dx$.

Solve for the exact group that appears, e.g. $x^3\cos(x^4)\,dx=\tfrac14\,du$. The numerical factor from $du$ is exactly what produces the unknown coefficient. …

Common Mistakes

Mistake 1: Dropping the constant from $du$.

Why it's wrong: with $u=\sin(x^4)$, $du=4x^3\cos(x^4)\,dx$, so $x^3\cos(x^4)\,dx=\tfrac14\,du$ — writing it as $du$ loses the $\tfrac14$ and gives $a=\tfrac15$ instead of $\tfrac{1}{20}$. Correct approach: always compute $du$ fully and solve for the exact differential group.

Mistake 2: Choosing the wrong inner function.

Why it's wrong: substituting $u=x^4$ leaves a $\sin^4 u\cos u$ that still needs a second step; not realising this makes students stop early. Correct approach: either substitute $u=\sin(x^4)$ in one move, or carry the second substitution $v=\sin u$ through to the end. …