Evaluate $\int_0^{\pi/2} [9\sin(3x)]\,dx$

Key takeaway: This integral is solved by finding an antiderivative of $\sin(3x)$ and then evaluating it at the endpoints.

Factor out the constant:

$\int_0^{\pi/2} 9\sin(3x)\,dx = 9\int_0^{\pi/2} \sin(3x)\,dx.$

An antiderivative of $\sin(3x)$ is

$\int \sin(3x)\,dx = -\frac{\cos(3x)}{3}.$

So,

$9\int_0^{\pi/2} \sin(3x)\,dx = 9\left[-\frac{\cos(3x)}{3}\right]_0^{\pi/2} = \left[-3\cos(3x)\right]_0^{\pi/2}.$

Now evaluate:

$-3\cos\left(3\cdot\frac{\pi}{2}\right)-\left(-3\cos(0)\right).$

Since $\cos(3\pi/2)=0$ and $\cos(0)=1$,

$0-(-3)=3.$

Answer: $3$

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Pitfalls the pros know 👇 A common mistake is forgetting the chain rule factor when integrating $\sin(3x)$. Another error is evaluating $\cos(3\pi/2)$ incorrectly; it equals $0$, not $-1$.

What if the problem changes? If the integrand were $9\sin(kx)$, the antiderivative would be $-\frac{9}{k}\cos(kx)$, as long as $k\neq 0$. If the upper limit changed, you would evaluate the same antiderivative at the new bound and subtract the value at the lower bound.

Tags: trigonometric integral, chain rule, antiderivative