Integral Of Sin 2x Cos X Trick Students Need Now
Integral of sin 2x cos x: solved in seconds with a practical approach
The integral $$\displaystyle \int \sin(2x)\cos(x)\,dx$$ can be evaluated quickly by using a trigonometric identity, then integrating a simple expression. A concise, reliable method is to rewrite $$\sin(2x)$$ as $$2\sin x\cos x$$, leading to an integrand that is a product of sine and cosine powers which folds cleanly into a single-variable polynomial in $$\cos x$$. This yields an exact antiderivative with a straightforward substitution.
First, express the integrand with a standard identity: $$\sin(2x) = 2\sin x\cos x$$. Thus
$$\displaystyle \int \sin(2x)\cos x\,dx = \int 2\sin x\cos x \cos x\,dx = 2\int \sin x \cos^2 x\,dx.$$
Next, use the substitution $$u = \cos x$$. Then $$du = -\sin x\,dx$$, so $$-du = \sin x\,dx$$. The integral becomes
$$\displaystyle 2\int \sin x \cos^2 x\,dx = -2\int u^2\,du = -\frac{2}{3}u^3 + C = -\frac{2}{3}\cos^3 x + C.$$
Therefore, the evaluated integral is
$$\displaystyle \int \sin(2x)\cos x\,dx = -\frac{2}{3}\cos^3 x + C.$$
For verification, differentiate the result:
$$\displaystyle \frac{d}{dx}\left(-\frac{2}{3}\cos^3 x\right) = -2\cos^2 x(-\sin x) = 2\sin x\cos^2 x,$$
which matches the transformed integrand $$2\sin x\cos^2 x$$ obtained earlier, confirming the correctness.
Alternative route via a product-to-sum identity
Alternatively, apply the product-to-sum identity: $$\sin A \cos B = \frac{1}{2}[\sin(A+B) + \sin(A-B)]$$. With $$A=2x$$ and $$B=x$$,
$$\displaystyle \sin(2x)\cos(x) = \frac{1}{2}[\sin(3x) + \sin(x)].$$
Integrating term-by-term gives
$$\displaystyle \int \sin(2x)\cos(x)\,dx = \frac{1}{2}\int \sin(3x)\,dx + \frac{1}{2}\int \sin(x)\,dx = -\frac{1}{6}\cos(3x) - \frac{1}{2}\cos(x) + C.$$
Both expressions are equivalent up to a constant, since $$-\frac{2}{3}\cos^3 x$$ and $$-\frac{1}{6}\cos(3x) - \frac{1}{2}\cos(x)$$ differ by a constant when simplified using triple-angle identities. The choice of form can depend on the context or the desired simplicity for a given application.
Practical notes for educators and administrators
- Consistency matters: choose a single antiderivative form to minimize confusion in lesson materials.
- Real-world checks: verify differentiation of the chosen antiderivative to ensure alignment with the original integrand.
- Visual aids: graph the two equivalent antiderivative forms to illustrate how constants can absorb differences between expressions.
- Application focus: recognizing quick transformations supports time-efficient problem solving in exams and classroom drills.
- Step 1: Rewrite using $$\sin(2x) = 2\sin x\cos x$$.
- Step 2: Substitute $$u=\cos x$$ to linearize the integral in $$u$$.
- Step 3: Integrate and back-substitute to obtain the antiderivative.
- Step 4: Optionally cross-check with the product-to-sum route for additional verification.
Frequently asked questions
| Method | ||
|---|---|---|
| Substitution | Use $$\sin(2x)=2\sin x\cos x$$; let $$u=\cos x$$; integrate $$u^2$$ | $$-\frac{2}{3}\cos^3 x + C$$ |
| Product-to-sum | $$\sin(2x)\cos x = \frac{1}{2}[\sin(3x)+\sin x]$$; integrate termwise | $$-\frac{1}{6}\cos 3x - \frac{1}{2}\cos x + C$$ |
Helpful tips and tricks for Integral Of Sin 2x Cos X Trick Students Need Now
Why can we substitute $$u=\cos x$$ in this integral?
Because the differential $$du = -\sin x\,dx$$ matches the remaining $$\sin x\,dx$$ factor after factoring $$\sin(2x)$$ as $$2\sin x\cos x$$. This substitution reduces the integral to a straightforward power integral in $$u$$.
Are both antiderivative forms always equivalent?
Yes. They differ by a constant, as can be shown via standard trigonometric identities like the triple-angle formula and algebraic rearrangements.
Which form is preferable for teaching?
Many educators prefer the $$ -\frac{2}{3}\cos^3 x + C $$ form for its compactness, while others favor the product-to-sum form, $$ -\frac{1}{6}\cos(3x) - \frac{1}{2}\cos x + C $$, to illustrate multiple solution pathways and reinforce identity fluency.
