Integrating Trigonometric Functions

Cosine integrates to sine

Integration undoes differentiation, so every derivative of a trigonometric function, read backwards, is an integral. Sine differentiates to cosine, so cosine integrates to sine: $\int \cos x \,dx = \sin x + C$.

Check by differentiating the answer. sin x + C differentiates to cos x, and the C, a constant, differentiates to 0. That returns the integrand exactly, so the integral is right.

The gold curve is y = sin x and the plain curve is y = cos x, with x in radians. At x = 0, sine climbs with gradient 1, and cos 0 = 1. At $x = \frac{\pi}{2}$, sine is at its top and has gradient 0, and $\cos\left(\frac{\pi}{2}\right) = 0$. At $x = \pi$, sine falls with gradient −1, and $\cos \pi = -1$. The height of the cosine is the gradient of the sine everywhere.

Walking the ring backwards

Differentiating four times brings sine back to itself: sin x goes to cos x, cos x goes to −sin x, −sin x goes to −cos x, and −cos x goes back to sin x. Written in a ring, differentiation is one step forwards.

Integration is one step backwards. The step before sin x on the ring is −cos x, so $\int \sin x \,dx = -\cos x + C$. That is where the minus sign comes from: cos x differentiates to −sin x, so it is −cos x that differentiates to sin x. Check: −cos x differentiates to −(−sin x) = sin x.

The other two follow the same way. The step before −sin x is cos x, so $\int \left(-\sin x\right) \,dx = \cos x + C$, and the step before −cos x is −sin x, so $\int \left(-\cos x\right) \,dx = -\sin x + C$. Each one checks by differentiating, and the minus signs land where the ring puts them.

Radians only

These results hold when x is in radians. Only there does sin x differentiate to exactly cos x. Measured in degrees, an angle of x° is $\frac{\pi x}{180}$ radians, so by the chain rule the derivative of sin(x°) is $\left(\frac{\pi}{180}\right) \cos\left(x^\circ\right)$, and every integral would carry the factor $\frac{180}{\pi}$ the other way.

So in every integral of sine or cosine, the limits and the angles are in radians: $\pi$ for half a turn, $\frac{\pi}{2}$ for a quarter.

A multiple of x inside

When the angle is 3x rather than x, the chain rule makes sin 3x differentiate to 3 cos 3x, three times too much. So divide by the 3: $\int \cos 3x \,dx = \frac{{\sin 3x}}{3} + C$. Check: $\frac{{\sin 3x}}{3}$ differentiates to $\frac{{3 \cos 3x}}{3} = \cos 3x$.

Any linear bracket works the same way, dividing by the multiplier of x: $\int \sin\left(2x + 1\right) \,dx = - \frac{{\cos\left(2x + 1\right)}}{2} + C$. Check: $- \frac{{\cos\left(2x + 1\right)}}{2}$ differentiates to $- \frac{{-2 \sin\left(2x + 1\right)}}{2} = \sin\left(2x + 1\right)$.

One arch of sine

From x = 0 to $x = \pi$, the curve y = sin x makes one arch above the x-axis. Its area is the definite integral of sin x from 0 to $\pi$, and the antiderivative −cos x gives it at once: $\left[-\cos x\right]_{0}^{\pi} = \left(-\cos \pi\right) - \left(-\cos 0\right)$.

Now $\cos \pi = -1$, so $-\cos \pi = 1$, and cos 0 = 1, so −cos 0 = −1. The area is 1 − (−1) = 2, exactly.

The picture agrees. The arch sits inside a rectangle $\pi$ wide and 1 tall, of area $\pi \approx 3.14$, and it contains the triangle with the same base and a peak at $\left(\frac{\pi}{2} , 1\right)$, of area $\frac{\pi}{2} \approx 1.57$. The area 2 lies between the two.

One arch of y = sin x, shaded from x = 0 to $x = \pi$, with its top at $\left(\frac{\pi}{2} , 1\right)$. The shaded area is $\left[-\cos x\right]_{0}^{\pi} = 1 - \left(-1\right) = 2$.

Other limits

The area under y = cos x from 0 to $\frac{\pi}{2}$ is one quarter-wave: $\left[\sin x\right]_{0}^{\frac{\pi}{2}} = \sin\left(\frac{\pi}{2}\right) - \sin 0 = 1 - 0 = 1$.

With the multiple inside, the integral of cos 3x from 0 to $\frac{\pi}{6}$ is $\left[\frac{{\sin 3x}}{3}\right]_{0}^{\frac{\pi}{6}}$. At $\frac{\pi}{6}$ the angle 3x is $\frac{\pi}{2}$, so the value is $\frac{1}{3} - 0 = \frac{1}{3}$. This quarter-wave of cos 3x is three times narrower than the quarter-wave of cos x from 0 to $\frac{\pi}{2}$, and its area is a third as much.

Over a whole wave, from 0 to $2\pi$, the integral of sin x is $\left[-\cos x\right]_{0}^{2\pi} = \left(-1\right) - \left(-1\right) = 0$. The second arch lies below the x-axis, so it counts as −2, and it cancels the first.

A whole wave of y = sin x, shaded from x = 0 to $x = 2\pi$. The arch above the axis counts as 2 and the arch below it as −2, so the integral over the wave is 0.

The usual mistakes

The minus on the wrong function. $\int \cos x \,dx = \sin x + C$ has no minus; it is $\int \sin x \,dx = -\cos x + C$ that carries one. Differentiating the answer settles it every time.

Taking a function as its own integral. cos x differentiates to −sin x, not to cos x, so $\int \cos x \,dx$ is not cos x + C. Only $e^{x}$ is its own derivative.

Forgetting to divide by the multiplier. $\int \cos 3x \,dx$ is $\frac{{\sin 3x}}{3} + C$, not sin 3x + C, which differentiates to 3 cos 3x.

Working in degrees. The integral of cos x from 0 to 90 is not 1; 90 radians is not a right angle. Use $\frac{\pi}{2}$.