Q.Find the following integrals:

(i) $\int \sin^3 x \cos^2 x\, dx$

Key idea: Use $\sin^2 x = 1 - \cos^2 x$ and substitute $u = \cos x$.

Write $\sin^3 x = \sin x \cdot \sin^2 x = \sin x (1 - \cos^2 x)$.

Then the integral becomes

$\int \sin x (1 - \cos^2 x) \cos^2 x\, dx$.

Let $u = \cos x$, so $du = -\sin x\, dx$. The integral transforms to

$-\int (1 - u^2) u^2\, du = -\int (u^2 - u^4)\, du = -\left( \frac{u^3}{3} - \frac{u^5}{5} \right) + C$. …

(i) $\int \sin^3 x \cos^2 x\, dx$

Concept and intuition:

When you see odd powers of sine (or cosine) multiplied by the other function, the classic trick is to peel off one factor of the odd-powered trig function and use the Pythagorean identity to express the rest in terms of the other trig function. Here, $\sin^3 x = \sin^2 x \cdot \sin x = (1 - \cos^2 x)\sin x$. That lone $\sin x$ is perfect for the substitution $u = \cos x$, because $du = -\sin x\, dx$.

Step-by-step:

1. Rewrite the integrand

2. Substitute $u = \cos x$, so $du = -\sin x\, dx$ and $\sin x\, dx = -du$. The integral becomes:

3. Integrate term by term

4. Back-substitute $u = \cos x$: $\frac{\cos^5 x}{5} - \frac{\cos^3 x}{3} + C$ …

Method: Trig Identities Before Substitution (Products & Quotients of Sine/Cosine)

Use this for integrals of trigonometric products or quotients where no direct antiderivative exists: convert the integrand with an identity so a clean substitution or standard form appears.

Steps

Step 1: Classify the integrand.

For a product like $\sin^m x\cos^n x$ with an odd power, peel off one factor of the odd-powered function and convert the rest with $\sin^2x+\cos^2x=1$. For a quotient like $\dfrac{\sin x}{\sin(x+a)}$, rewrite the numerator angle in terms of the denominator angle, e.g. $\sin x=\sin\big((x+a)-a\big)$ and expand.

Step 2: Substitute or split. …

Common Mistakes

Mistake 1: Substituting $u=\cos x$ but forgetting the sign of $du$.

Why it's wrong: $du=-\sin x\,dx$, so the peeled-off $\sin x\,dx$ becomes $-du$; dropping the minus flips the answer. Correct approach: track $du=-\sin x\,dx$ carefully.

Mistake 2: Trying to integrate $\dfrac{\sin x}{\sin(x+a)}$ term-by-term as a quotient.

Why it's wrong: there is no quotient rule for integrals. Correct approach: rewrite $\sin x=\sin((x+a)-a)$ and expand into $\cos a-\cot(x+a)\sin a$ style terms. …