Q.Find

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

Key idea: Use the power-reduction identity $\cos^2 x = \frac{1+\cos 2x}{2}$.

$\int \cos^2 x\, dx = \int \frac{1+\cos 2x}{2}\, dx = \frac12 \int dx + \frac12 \int \cos 2x\, dx$ …

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

The square of a cosine is not directly integrable in that form. But the double-angle identity lets us replace $\cos^2 x$ with something linear in $\cos 2x$, which is easy to integrate.

1. Apply the identity.

2. Integrate term by term. The integral of $1$ is $x$, and the integral of $\cos 2x$ is $\frac{\sin 2x}{2}$ (by the reverse chain rule: derivative of $\sin 2x$ is $2\cos 2x$, so we divide by 2).

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

A product of sines and cosines of different angles is not directly integrable. The product-to-sum identity converts it into a sum of two sines (or cosines), each of which integrates cleanly.

1. Apply the identity with $A = 2x$, $B = 3x$.

Since $\sin(-\theta) = -\sin\theta$, this simplifies to:

2. Integrate.

(Recall $\int \sin kx\, dx = -\frac{\cos kx}{k}$.)

3. Simplify.

--- …

Method: Power-Reduction and Product-to-Sum Identities

Use this for integrals of even powers of sine/cosine, or products of two different-frequency trig functions, which have no direct antiderivative in their given form.

Steps

Step 1: Choose the right identity.

For an even power, use power reduction: $\cos^2x=\dfrac{1+\cos2x}{2}$, $\sin^2x=\dfrac{1-\cos2x}{2}$. For a product of different frequencies, use product-to-sum: $\sin A\cos B=\tfrac12[\sin(A+B)+\sin(A-B)]$. For an odd power like $\sin^3x$, write $\sin^3x=\sin x(1-\cos^2x)$.

Step 2: Rewrite the integrand as a sum of simple terms. …

Common Mistakes

Mistake 1: Writing $\int\cos^2x\,dx=\dfrac{\cos^3x}{3}$.

Why it's wrong: that would be $\int\cos^2x\,d(\cos x)$, which is not what the integral is. Correct approach: reduce with $\cos^2x=\frac{1+\cos2x}{2}$ first.

Mistake 2: Forgetting the $\dfrac1k$ factor when integrating $\cos 2x$ or $\cos 5x$.

Why it's wrong: $\int\cos kx\,dx=\frac1k\sin kx$, so $\int\cos5x\,dx=\frac15\sin5x$, not $\sin5x$. Correct approach: divide by the frequency $k$. …