Q.Integrate the following function: $\sin^2 (2x + 5)$

The key idea is to use the power-reduction identity (a form of the double-angle formula) to rewrite $\sin^2 \theta$ before integrating.

Step 1: Recall the identity $\sin^2 \theta = \frac{1 - \cos 2\theta}{2}$.

Here $\theta = 2x + 5$, so

Step 2: Integrate term by term:

Step 3: The first integral is $\frac{x}{2}$. For the second, use a quick substitution $u = 4x+10$ (or simply note $\int \cos(ax+b)\,dx = \frac{1}{a}\sin(ax+b)$):

Step 4: Combine the results:

Why this approach works

When you see a squared trigonometric function like $\sin^2(\text{something})$, your first instinct might be to try a substitution. But substitution alone won't help here — the square is the real obstacle. The cleanest path is to use the power-reduction identity (also called the half-angle formula):

This identity comes straight from the double-angle formula for cosine: $\cos 2\theta = 1 - 2\sin^2 \theta$. Rearranging gives the form above. It transforms a square (hard to integrate directly) into a simple linear combination of a constant and a cosine (easy to integrate).

Once we apply this, the integral breaks into two elementary pieces. The constant term integrates to a linear function, and the cosine term integrates to a sine — with a chain-rule factor from the inner function $4x+10$.

Let's work through it.

1. Apply the power-reduction identity

   Set $\theta = 2x + 5$. Then:

So the integral becomes:

2. Split into two simpler integrals

   Factor out the constant $\frac12$:

The first integral is trivial: $\int 1 \, dx = x$.

3. Handle the cosine integral with a substitution

   For $\int \cos(4x + 10) \, dx$, let $u = 4x + 10$. Then $du = 4 \, dx$, so $dx = \frac{du}{4}$. This gives:

4. Combine the pieces

   Putting it all together:

5. Check by differentiating (optional but good practice)

   Differentiate your answer:

Factor $\frac12$:

It matches. Always a good feeling.

Method: Power reduction for an even power of $\sin$ or $\cos$

An even power such as $\sin^2(\text{linear})$ has no direct antiderivative — lower the power first using a double-angle identity, then integrate the resulting cosine.

Steps

Step 1: Replace the squared term with its power-reduction identity.

Here $\theta$ is the whole linear argument (for $\sin^2(2x+5)$, take $\theta=2x+5$, so $2\theta=4x+10$).

Step 2: Integrate the constant and the cosine separately.

The constant $\tfrac12$ integrates to $\tfrac{x}{2}$. For the cosine, use

Step 3: Keep the $\dfrac1k$ chain-rule factor.

The single most common slip is writing $\int\cos(4x+10)\,dx=\sin(4x+10)$ without the $\tfrac14$. The linear coefficient must divide the result.

For any even power, apply the identity once (or repeatedly for 4th/6th powers) until only first-degree cosines remain.

Common Mistakes

Mistake 1: Writing $\displaystyle\int\sin^2(2x+5)\,dx=\frac{\sin^3(2x+5)}{3}$.

Why it's wrong: the power rule needs the derivative of the inside present; here $\cos(2x+5)\cdot 2$ is missing, so that step is invalid. Correct approach: use power reduction, $\sin^2\theta=\dfrac{1-\cos 2\theta}{2}$.

Mistake 2: Forgetting the $\tfrac14$ when integrating $\cos(4x+10)$.

Why it's wrong: $\displaystyle\int\cos(4x+10)\,dx=\dfrac{\sin(4x+10)}{4}$, because the argument has slope $4$. Dropping this factor makes the $\sin$ term four times too large. Correct approach: divide by the coefficient of $x$, giving $-\dfrac{\sin(4x+10)}{8}$ overall.