Trigonometric Integrals

In this section, we'll revisit u-substitution, but specifically for trigonometric functions.

Specifically, we'll solve problems of the form ∫sin⁡mx cos⁡nx dx\int \sin^m x\ \cos^n x\ dx and ∫sec⁡mx tan⁡nx dx.\int \sec^m x\ \tan^n x\ dx.

To do all the examples in this section, we'll need to review some trig identities, as well as trig derivatives, but before we do, we'll start with a few simple examples.

Evaluate ∫sin⁡2x cos⁡x dx\displaystyle\int \sin^2 x\ \cos x \ dx.

Solution

If we let u=sin⁡xu=\sin x, then du=cos⁡x dxdu = \cos x\ dx, so ∫sin⁡2x cos⁡x dx=∫u2 du=13u3+C=13sin⁡3x+C\begin{aligned} \int \sin^2 x\ \cos x \ dx &= \int u^2 \ du\ &= \dfrac{1}{3}u^3 + C\ &= \ans{\dfrac{1}{3}\sin^3 x + C} \end{aligned}

Evaluate ∫sin⁡xcos⁡4x dx\displaystyle\int \dfrac{\sin x}{\cos^4 x}\ dx.

Solution

This time, let u=cos⁡xu=\cos x so that du=−sin⁡x dxdu = -\sin x \ dx, or −du=sin⁡x dx-du = \sin x \ dx.

Then ∫sin⁡xcos⁡4x dx=∫−1u4 du=∫−u−4 du=13u−3+C=13cos⁡3x+C\begin{aligned} \int \dfrac{\sin x}{\cos^4 x} \ dx &= \int -\dfrac{1}{u^4} \ du\ &= \int -u^{-4}\ du\ &= \dfrac{1}{3}u^{-3} + C\ &= \ans{\dfrac{1}{3 \cos^3 x} + C} \end{aligned}

Evaluate ∫tan⁡x dx\displaystyle\int \tan x\ dx.

Solution

Once we rewrite tan⁡x\tan x as sin⁡xcos⁡x\dfrac{\sin x}{\cos x}, this problem looks a lot like the previous one.

We can let u=cos⁡xu=\cos x again, so that −du=sin⁡x dx-du = \sin x \ dx: ∫tan⁡x dx=∫sin⁡xcos⁡x dx=∫−1u du=−ln⁡u+C=−ln⁡∣cos⁡x∣+C\begin{aligned} \int \tan x\ dx = \int \dfrac{\sin x}{\cos x} \ dx &= \int -\dfrac{1}{u} \ du\ &= -\ln u + C\ &= \ans{-\ln |\cos x| + C} \end{aligned}

Before we do more examples, let's review a few trigonometric identities.

Reviewing Identities

The most important trigonometric identity of all is sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1

Dividing that identity by cos⁡2x\cos^2 x gives another variation: tan⁡2x+1=sec⁡2x\tan^2 x + 1 = \sec^2 x

Besides that, we need the half-angle identities: sin⁡2x=12−12cos⁡(2x)cos⁡2x=12+12cos⁡(2x)\begin{aligned} \sin^2 x &= \dfrac{1}{2}-\dfrac{1}{2}\cos(2x)\ \cos^2 x &= \dfrac{1}{2}+\dfrac{1}{2}\cos(2x) \end{aligned}

Finally, recall the following derivatives: ddx[sec⁡x]=sec⁡xtan⁡xddx[tan⁡x]=sec⁡2x\begin{aligned} \dfrac{d}{dx}[\sec x] &= \sec x \tan x\ \dfrac{d}{dx}[\tan x] &= \sec^2 x \end{aligned}

∫sin⁡mx cos⁡nx dx\displaystyle\int \sin^m x\ \cos^n x\ dx

In this form, uu will either be the sine or cosine function; we just need to figure out which one.

Take a look at the first examples we did: we want dudu to be the one that appears once. For instance, to evaluate ∫sin⁡2x cos⁡x dx\int \sin^2 x\ \cos x\ dx we would want cos⁡x\cos x to be dudu, so we'd let u=sin⁡xu=\sin x. On the other hand, to evaluate ∫sin⁡x cos⁡2x dx\int \sin x\ \cos^2 x\ dx we would let u=cos⁡xu=\cos x so that dudu would involve sin⁡x\sin x.

What about ∫sin⁡3x cos⁡2x dx?\int \sin^3 x\ \cos^2 x\ dx?

If u=sin⁡xu=\sin x and du=cos⁡x dxdu=\cos x\ dx, there would be a leftover cos⁡x\cos x after substituting, which is a problem. On the other hand, if u=cos⁡xu=\cos x and −du=sin⁡x dx-du = \sin x \ dx, after substituting we'd have ∫−sin⁡2x u2 du\int -\sin^2 x\ u^2\ du

We can deal with this leftover sin⁡2x\sin^2 x by using the identity sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1 ∫−(1−cos⁡2x)u2 du=∫−(1−u2)u2 du\int -(1-\cos^2 x) u^2\ du = \int -(1-u^2)u^2\ du

Finally, we can integrate this: ∫−u2+u4 du=−13u3+15u5+C=−13cos⁡3x+15cos⁡5x+C\begin{aligned} \int -u^2+u^4\ du &= -\dfrac{1}{3}u^3+\dfrac{1}{5}u^5 + C\ &= -\dfrac{1}{3}\cos^3 x + \dfrac{1}{5}\cos^5 x + C \end{aligned}

Evaluate ∫sin⁡2x cos⁡3x dx\displaystyle\int \sin^2 x\ \cos^3 x\ dx.

Solution

Following what we just saw, we want to split off a single cos⁡x\cos x - that'll serve as dudu eventually, and the leftover cos⁡2x\cos^2 x can be replaced with 1−sin⁡2x1-\sin^2 x. ∫sin⁡2x cos⁡3x dx=∫sin⁡2x cos⁡2x cos⁡x dx=∫sin⁡2x (1−sin⁡2x) cos⁡x dx\begin{aligned} \int \sin^2 x\ \cos^3 x\ dx &= \int \sin^2 x\ \cos^2 x\ \cos x\ dx\ &= \int \sin^2 x\ (1-\sin^2 x)\ \cos x\ dx \end{aligned}

Now let u=sin⁡xu=\sin x so that du=cos⁡x dxdu = \cos x \ dx. ∫sin⁡2x (1−sin⁡2x) cos⁡x dx=∫u2(1−u2) du=∫u2−u4 du=13u3−15u5+C=13sin⁡3x−15sin⁡5x+C\begin{aligned} \int \sin^2 x\ (1-\sin^2 x)\ \cos x\ dx &= \int u^2(1-u^2)\ du\ &= \int u^2-u^4\ du\ &= \dfrac{1}{3}u^3 - \dfrac{1}{5}u^5 + C\ &= \ans{\dfrac{1}{3}\sin^3 x - \dfrac{1}{5}\sin^5 x + C} \end{aligned}

Evaluate ∫sin⁡3x dx\displaystyle\int \sin^3 x\ dx.

Solution

This time, split off a sin⁡x\sin x; the remaining sin⁡2x\sin^2 x can be rewritten in terms of cos⁡x\cos x and we'll make u=cos⁡xu=\cos x.

∫sin⁡3x dx=∫sin⁡2x sin⁡x dx=∫(1−cos⁡2x) sin⁡x dx=∫−(1−u2) du=∫−1+u2 du=−u+13u3+C=−cos⁡x+13cos⁡3x+C\begin{aligned} \int \sin^3 x\ dx &= \int \sin^2 x\ \sin x\ dx\ &= \int (1-\cos^2 x) \ \sin x\ dx\ &= \int -(1-u^2)\ du\ &= \int -1+u^2\ du\ &= -u + \dfrac{1}{3}u^3 + C\ &= \ans{-\cos x + \dfrac{1}{3}\cos^3 x + C} \end{aligned}

Conclusion: how to integrate ∫sin⁡mx cos⁡nx dx\displaystyle\int \sin^m x\ \cos^n x\ dx

Even Powers

Evaluate ∫cos⁡2x dx\displaystyle\int \cos^2 x\ dx.

Solution

This one only has an even power, so we need to use a half-angle identity. Remember that cos⁡2x=12+12cos⁡(2x),\cos^2 x = \dfrac{1}{2} + \dfrac{1}{2}\cos(2x), so we can rewrite the integral as ∫12+12cos⁡(2x) dx.\int \dfrac{1}{2} + \dfrac{1}{2}\cos(2x)\ dx.

The first part we can deal with directly, and for the second part, we can use u-substitution, with u=2xu=2x. Since ∫cos⁡(2x) dx=12sin⁡(2x),\int \cos (2x)\ dx = \dfrac{1}{2}\sin (2x), we can get the final answer:

∫cos⁡2x dx=∫12+12cos⁡(2x) dx=12x+14sin⁡(2x)+C\begin{aligned} \int \cos^2 x\ dx &= \int \dfrac{1}{2} + \dfrac{1}{2}\cos (2x) \ dx\ &= \ans{\dfrac{1}{2}x + \dfrac{1}{4}\sin(2x) + C} \end{aligned}

  1. ∫sin⁡3x cos⁡2x dx\displaystyle\int \sin^3 x\ \cos^2 x\ dx

    2. Let u=cos⁡x⟶−du=sin⁡x dxu=\cos x \longrightarrow -du = \sin x\ dx

    ∫sin⁡3x cos⁡2x dx=∫sin⁡2x cos⁡2x sin⁡x dx=∫(1−cos⁡2x) cos⁡2x sin⁡x dx=∫−(1−u2)u2 du=∫−u2+u4 du=−13u3+15u5+C=−13cos⁡3x+15cos⁡5x+C\begin{aligned} \int \sin^3 x \ \cos^2 x \ dx &= \int \sin^2 x \ \cos^2 x \ \sin x \ dx\ &= \int (1-\cos^2 x) \ \cos^2 x \ \sin x \ dx\ &= \int -(1-u^2)u^2\ du\ &= \int -u^2+u^4\ du\ &= -\dfrac{1}{3}u^3 + \dfrac{1}{5}u^5 + C\ &= \ans{-\dfrac{1}{3}\cos^3 x + \dfrac{1}{5}\cos^5 x + C} \end{aligned}

  2. ∫cos⁡5(πx) dx\displaystyle\int \cos^5 (\pi x)\ dx

    3. Let u=sin⁡(πx)⟶1πdu=cos⁡(πx) dxu=\sin (\pi x) \longrightarrow \dfrac{1}{\pi}du = \cos (\pi x)\ dx

    ∫cos⁡5(πx) dx=∫cos⁡2(πx) cos⁡2(πx) cos⁡(πx) dx=∫(1−sin⁡2(πx))(1−sin⁡2(πx)) cos⁡(πx) dx=∫1π(1−u2)(1−u2) du=∫1pi−2πu2+1πu4 du=1πu−23πu3+15πu5+C=1πsin⁡(πx)−23πsin⁡3(πx)+15πsin⁡5(πx)+C\begin{aligned} \int \cos^5 (\pi x) \ dx &= \int \cos^2 (\pi x) \ \cos^2 (\pi x) \ \cos (\pi x) \ dx\ &= \int (1-\sin^2 (\pi x)) (1-\sin^2 (\pi x)) \ \cos (\pi x) \ dx\ &= \int \dfrac{1}{\pi}(1-u^2)(1-u^2)\ du\ &= \int \dfrac{1}{pi}-\dfrac{2}{\pi}u^2+\dfrac{1}{\pi}u^4\ du\ &= \dfrac{1}{\pi}u - \dfrac{2}{3\pi}u^3 + \dfrac{1}{5\pi}u^5 + C\ &= \ans{\dfrac{1}{\pi}\sin(\pi x) - \dfrac{2}{3\pi} \sin^3 (\pi x) + \dfrac{1}{5\pi}\sin^5 (\pi x) + C} \end{aligned}

  3. ∫sin⁡2x cos⁡2x dx\displaystyle\int \sin^2 x\ \cos^2 x\ dx

    4. ∫sin⁡2x cos⁡2x dx=∫(12−12cos⁡(2x))(12+12cos⁡(2x)) dx=∫14−14cos⁡2(2x) dx=∫14−14(12+12cos⁡(4x)) dx=∫18−18cos⁡(4x) dx=18x−132sin⁡(4x)+C\begin{aligned} \int \sin^2 x\ \cos^2 x\ dx &= \int \left(\dfrac{1}{2} - \dfrac{1}{2}\cos(2x)\right)\left(\dfrac{1}{2} + \dfrac{1}{2}\cos(2x)\right)\ dx\ &= \int \dfrac{1}{4} - \dfrac{1}{4}\cos^2(2x)\ dx\ &= \int \dfrac{1}{4} - \dfrac{1}{4}\left(\dfrac{1}{2} + \dfrac{1}{2}\cos(4x)\right)\ dx\ &= \int \dfrac{1}{8} - \dfrac{1}{8}\cos(4x)\ dx\ &= \ans{\dfrac{1}{8}x - \dfrac{1}{32} \sin (4x) + C} \end{aligned}

  4. ∫sin⁡2(2θ) dθ\displaystyle\int \sin^2 (2\theta)\ d\theta

    5. ∫sin⁡2(2θ) dθ=∫12−12cos⁡(4θ) dθ=12θ−18sin⁡(4θ)+C\begin{aligned} \int \sin^2 (2\theta)\ d\theta &= \int \dfrac{1}{2} - \dfrac{1}{2} \cos (4\theta)\ d\theta\ &= \ans{\dfrac{1}{2}\theta - \dfrac{1}{8} \sin (4\theta) + C} \end{aligned}

∫tan⁡mx sec⁡nx dx\displaystyle\int \tan^m x\ \sec^n x\ dx

These examples will work similarly. Just like sine and cosine are a matched pair of derivatives, secant and tangent match each other's derivatives: ddx[sec⁡x]=sec⁡x tan⁡xddx[tan⁡x]=sec⁡2x\begin{aligned} \dfrac{d}{dx}[\sec x] &= \sec x \ \tan x\ \dfrac{d}{dx}[\tan x] &= \sec^2 x \end{aligned}

We're either going to make u=sec⁡xu=\sec x, in which case du=sec⁡x tan⁡x dxdu = \sec x \ \tan x \ dx, or we'll make u=tan⁡xu=\tan x, so that du=sec⁡2xdu = \sec^2 x.

Thus, we'll either split off sec⁡x tan⁡x\sec x \ \tan x or sec⁡2x\sec^2 x to serve as dudu, then rewrite the rest of the integral in terms of the appropriate function.

This boils down to the following:

Conclusion: how to integrate ∫tan⁡mx sec⁡nx dx\displaystyle\int \tan^m x \ \sec^n x \ dx

Evaluate ∫tan⁡6x sec⁡4x dx\displaystyle\int \tan^6 x \ \sec^4 x\ dx.

Solution

The power of secant is even, so let u=tan⁡x⟶du=sec⁡2x dxu=\tan x \longrightarrow du = \sec^2 x \ dx.

∫tan⁡6x sec⁡4x dx=∫tan⁡6x sec⁡2x sec⁡2x dx=∫tan⁡6x (1+tan⁡2x) sec⁡2x dx=∫u6(1+u2) du=∫u6+u8 du=17u7+19u9+C=17tan⁡7x+19tan⁡9x+C\begin{aligned} \int \tan^6 x \ \sec^4 x \ dx &= \int \tan^6 x \ \sec^2 x \ \sec^2 x \ dx\ &= \int \tan^6 x \ (1+\tan^2 x) \ \sec^2 x \ dx\ &= \int u^6(1+u^2) \ du\ &= \int u^6 + u^8 \ du\ &= \dfrac{1}{7}u^7 + \dfrac{1}{9}u^9 + C\ &= \ans{\dfrac{1}{7}\tan^7 x + \dfrac{1}{9}\tan^9 x + C} \end{aligned}

Evaluate ∫sec⁡2x tan⁡x dx\displaystyle\int \sec^2 x \ \tan x \ dx.

This one has both an even power of secant and an odd power of tangent, so we can actually solve it either way.

Solution 1

Let u=tan⁡xu=\tan x, so du=sec⁡2x dxdu=\sec^2 x\ dx. ∫sec⁡2x tan⁡x dx=∫u du=12u2+C=12tan⁡2x+C\begin{aligned} \int \sec^2 x \ \tan x \ dx &= \int u \ du\ &= \dfrac{1}{2}u^2 + C\ &= \ans{\dfrac{1}{2}\tan^2 x + C} \end{aligned}

Solution 2

Let u=sec⁡xu = \sec x, so du=sec⁡x tan⁡x dxdu = \sec x \ \tan x \ dx. ∫sec⁡2x tan⁡x dx=∫u du=12u2+C=12sec⁡2x+C\begin{aligned} \int \sec^2 x \ \tan x \ dx &= \int u \ du\ &= \dfrac{1}{2}u^2 + C\ &= \ans{\dfrac{1}{2}\sec^2 x + C} \end{aligned}

That's odd: we got two different answers, but if we did everything correctly, both solutions should end up with the same answer. So what's the deal?

Remember, based on the identity 1+tan⁡2x=sec⁡2x1 + \tan^2 x = \sec^2 x the only difference between tan⁡2x\tan^2 x and sec⁡2x\sec^2 x is a constant. The arbitrary constants in the answers absorb that difference.

  1. ∫tan⁡5x sec⁡7x dx\displaystyle\int \tan^5 x \ \sec^7 x \ dx

    2. Let u=sec⁡x⟶du=sec⁡x tan⁡x dxu=\sec x \longrightarrow du = \sec x \ \tan x \ dx ∫tan⁡5x sec⁡7x dx=∫tan⁡4x sec⁡6x sec⁡x tan⁡x dx=∫(sec⁡2x−1)2 sec⁡6x sec⁡x tan⁡x dx=∫(u2−1)2u6 du=∫u10−2u8+u6 du=111u11−29u9+17u7+C=111sec⁡11x−29sec⁡9x+17sec⁡7x+C\begin{aligned} \int \tan^5 x \ \sec^7 x \ dx &= \int \tan^4 x \ \sec^6 x \ \sec x \ \tan x \ dx\ &= \int (\sec^2 x - 1)^2 \ \sec^6 x \ \sec x \ \tan x \ dx\ &= \int (u^2-1)^2 u^6 \ du\ &= \int u^{10}-2u^8+u^6 \ du\ &= \dfrac{1}{11}u^{11} - \dfrac{2}{9}u^9 + \dfrac{1}{7}u^7 + C\ &= \ans{\dfrac{1}{11}\sec^{11} x - \dfrac{2}{9}\sec^9 x + \dfrac{1}{7}\sec^7 x + C} \end{aligned}

  2. ∫tan⁡5x sec⁡4x dx\displaystyle\int \tan^5 x \ \sec^4 x \ dx

    3. Let u=tan⁡x⟶du=sec⁡2x dxu=\tan x \longrightarrow du = \sec^2 x \ dx ∫tan⁡5x sec⁡4x dx=∫tan⁡5x sec⁡2x sec⁡2x dx=∫tan⁡5x (tan⁡2x+1) sec⁡2x dx=∫u5(u2+1) du=∫u7+u5 du=18u8+16u6+C=18tan⁡8x+16tan⁡6x+C\begin{aligned} \int \tan^5 x \ \sec^4 x \ dx &= \int \tan^5 x \ \sec^2 x \ \sec^2 x \ dx\ &= \int \tan^5 x \ (\tan^2 x + 1) \ \sec^2 x \ dx\ &= \int u^5(u^2+1) \ du\ &= \int u^7+u^5 \ du\ &= \dfrac{1}{8}u^{8} + \dfrac{1}{6}u^6 + C\ &= \ans{\dfrac{1}{8}\tan^{8} x + \dfrac{1}{6}\tan^6 x + C} \end{aligned}

  3. ∫sec⁡6t dt\displaystyle\int \sec^6 t \ dt

    4. Let u=tan⁡t⟶du=sec⁡2t dtu=\tan t \longrightarrow du = \sec^2 t \ dt ∫sec⁡6t dt=∫sec⁡4t sec⁡2t dt=∫(tan⁡2t+1)2 sec⁡2t dt=∫(u2+1)2 du=∫u4+2u2+1 du=15u5+23u3+u+C=15tan⁡5t+23tan⁡3t+tan⁡t+C\begin{aligned} \int \sec^6 t \ dt &= \int \sec^4 t \ \sec^2 t \ dt\ &= \int (\tan^2 t + 1)^2 \ \sec^2 t \ dt\ &= \int (u^2+1)^2 \ du\ &= \int u^4+2u^2+1 \ du\ &= \dfrac{1}{5}u^{5} + \dfrac{2}{3}u^3 + u + C\ &= \ans{\dfrac{1}{5}\tan^{5} t + \dfrac{2}{3}\tan^3 t + \tan t + C} \end{aligned}