Problem 32

Question

Evaluating the integral $$ \int \sin ^{2} x d x $$ requires two steps. First, write $$ \sin ^{2} x=(\sin x)(\sin x) $$ and integrate by parts to show that $$ \int \sin ^{2} x d x=-\sin x \cos x+\int \cos ^{2} x d x $$ Then, use $\sin ^{2} x+\cos ^{2} x=1$ to replace $\cos ^{2} x$ in the integral on the right-hand side, and complete the integration of $\int \sin ^{2} x d x$.

Step-by-Step Solution

Verified

Answer

$\int \sin^2 x \, dx = \frac{x}{2} - \frac{\sin x \cos x}{2} + C$.

1Step 1: Understand the Integration by Parts

The integral of the squared sine function, $\int \sin^2 x \, dx$, can be tackled using integration by parts. Recall that integration by parts is given by $\int u \, dv = uv - \int v \, du$. Here, set $u = \sin x$ and $dv = \sin x \, dx$. Calculate $du = \cos x \, dx$ and $v = -\cos x$ by integrating $dv$. Plug these into the formula: $\int \sin^2 x \, dx = uv - \int v \, du = (-\sin x \cos x) + \int \cos^2 x \, dx$.

2Step 2: Substitute Using Pythagorean Identity

To solve $\int \cos^2 x \, dx$, use the identity $\sin^2 x + \cos^2 x = 1$. This implies $\cos^2 x = 1 - \sin^2 x$. Substitute this back into the equation: \[\int \sin^2 x \, dx = -\sin x \cos x + \int (1 - \sin^2 x) \, dx.\] This simplifies to $-\sin x \cos x + \int 1 \, dx - \int \sin^2 x \, dx$.

3Step 3: Solve for the Integral

Move the integral of $\sin^2 x$ to one side to isolate it: $\int \sin^2 x \, dx + \int \sin^2 x \, dx = -\sin x \cos x + \int 1 \, dx$. This becomes $2 \int \sin^2 x \, dx = -\sin x \cos x + x$. Divide through by 2 to obtain $\int \sin^2 x \, dx = \frac{-\sin x \cos x + x}{2} + C$.

4Step 4: Final Simplification

The final integral expression is $\int \sin^2 x \, dx = \frac{x}{2} - \frac{\sin x \cos x}{2} + C$, where $C$ is the constant of integration. This is the required result for the given integration problem.

Key Concepts

Integration by PartsTrigonometric IdentitiesPythagorean Identity

Integration by Parts

Integration by parts is a technique used in calculus to integrate products of functions. The formula for integration by parts is \[ \int u \, dv = uv - \int v \, du \] This means that we select two parts, "$u$" and "$dv$", from the function we want to integrate. We derive "$du$" from "$u$" and find "$v$" by integrating "$dv$". Then, we substitute everything back into the formula.

In the given problem, we need to integrate $\sin^2 x$. We start by rewriting it as $(\sin x)(\sin x)$ and choose $u = \sin x$ and $dv = \sin x \, dx$. This leads to $du = \cos x \, dx$ and integrating $dv$ gives $v = -\cos x$.

Substituting into the integration by parts formula yields:

$-\sin x \cos x + \int \cos^2 x \, dx$

Understanding each step carefully is crucial when dealing with 'Integration by Parts' to ensure accurate results.

Trigonometric Identities

Trigonometric identities are equations that involve trigonometric functions that are true for every value of the variable within its domain. These identities are incredibly useful in integrating and simplifying expressions that contain trigonometric functions.

In the current exercise, we encounter the identity $\cos^2 x = 1 - \sin^2 x$, which is derived from the Pythagorean Identity (another core trigonometric identity). This substitution helps to simplify the integral by replacing $\cos^2 x$.

It transforms the equation to:

\[\int \sin^2 x \, dx = -\sin x \cos x + \int (1 - \sin^2 x) \, dx\]

Substituting identities effectively can significantly simplify solving calculus problems.

Pythagorean Identity

The Pythagorean identities are foundational in trigonometry. The most common one is $\sin^2 x + \cos^2 x = 1$.

For calculus problems dealing with trigonometric integrals, this identity comes in handy, providing alternate expressions for $\cos^2 x$ and $\sin^2 x$. In our solved problem, it allows: