Q.Integrate the following function: $\sin^3 (2x + 1)$

Concept: U Substitution — reduce the odd power using $\sin^2\theta = 1 - \cos^2\theta$, then substitute for the cosine.

Step 1: Let $u = 2x + 1$, so $du = 2\,dx$ and $dx = \frac{du}{2}$.

The integral becomes

Step 2: Write $\sin^3 u = \sin u \cdot \sin^2 u = \sin u (1 - \cos^2 u)$.

Then

Step 3: Substitute $t = \cos u$, $dt = -\sin u \, du$. The integral becomes …

When you see an odd power of sine (or cosine), your first instinct should be to peel off one factor. Why? Because the remaining even power can be rewritten using the Pythagorean identity, turning the integral into a form ready for substitution.

Here, the function is $\sin^3(2x+1)$. The argument $(2x+1)$ is linear, so the chain rule will eventually give us a factor of $\frac{1}{2}$ after substitution. Let’s walk through it.

1. Separate one sine factor Write $\sin^3(2x+1) = \sin(2x+1) \cdot \sin^2(2x+1)$. This lets us use the identity $\sin^2\theta = 1 - \cos^2\theta$ on the squared part:

2. Choose the substitution

   The expression now contains $\sin(2x+1)$ and $\cos(2x+1)$. If we set $u = \cos(2x+1)$, then the derivative is $du = -2\sin(2x+1)\,dx$.

   This is perfect: the $\sin(2x+1)\,dx$ in our integral will be replaced by $-\frac{1}{2}\,du$.

   Watch out

   A common mistake is forgetting the factor of $2$ from the chain rule when differentiating $\cos(2x+1)$. Always check: derivative of $\cos(2x+1)$ is $-\sin(2x+1) \cdot 2$, so $du = -2\sin(2x+1)\,dx$.

3. Rewrite the integral

   The original integral is

Substitute $u = \cos(2x+1)$ and $dx = -\frac{du}{2\sin(2x+1)}$:

The $\sin(2x+1)$ cancels neatly, leaving:

$\int (1 - u^2) \cdot \left(-\frac{1}{2}\right) du = -\frac{1}{2} \int (1 - u^2)\, du.$ …

Method: Odd power of sine — save one factor and substitute

When an odd power of $\sin$ (of any linear argument) is integrated, peel off one sine factor to pair with $dx$ and convert the rest to cosine, so a $u=\cos(\cdot)$ substitution works.

Steps

Step 1: Split the odd power, keeping one sine aside.

using $\sin^2\theta=1-\cos^2\theta$. The lone $\sin\theta$ is what pairs with the differential.

Step 2: Substitute $u=\cos\theta$.

With $\theta$ the linear argument (e.g. $\theta=2x+1$),

so the spare $\sin\theta\,dx$ becomes a constant times $du$. …

Common Mistakes

Mistake 1: Writing $\displaystyle\int\sin^3(2x+1)\,dx=\frac{\sin^4(2x+1)}{4}$.

Why it's wrong: the power rule does not apply to $\sin^3$ without a matching $\cos(2x+1)$ differential. Correct approach: split $\sin^3\theta=(1-\cos^2\theta)\sin\theta$ and substitute $u=\cos(2x+1)$.

Mistake 2: Forgetting the $\tfrac12$ from $du=-2\sin(2x+1)\,dx$. …