Standard integrals of trigonometric functions
In chapter 6 we established the following results for differentiating trigonometric functions:
| $\frac{d}{dx}\sin\left(f\left(x\right)\right)$ddxsin(f(x)) | $=$= | $f'\left(x\right)\cos\left(f\left(x\right)\right)$f′(x)cos(f(x)) |
| $\frac{d}{dx}\cos\left(f\left(x\right)\right)$ddxcos(f(x)) | $=$= | $-f'\left(x\right)\sin\left(f\left(x\right)\right)$−f′(x)sin(f(x)) |
| $\frac{d}{dx}\tan\left(f\left(x\right)\right)$ddxtan(f(x)) | $=$= | $f'\left(x\right)\sec^2\left(f\left(x\right)\right)$f′(x)sec2(f(x)) |
Reversing these results, we get the following set of rules for integrating trigonometric functions:
$\int\cos x\ dx=\sin x+C$∫cosx dx=sinx+C, for some constant $C$C
$\int\sin x\ dx=-\cos x+C$∫sinx dx=−cosx+C, for some constant $C$C
$\int\sec^2x\ dx=\tan x+C$∫sec2x dx=tanx+C, for some constant $C$C
Worked examples
Example 1
If $f(x)=\sec^2x$f(x)=sec2x, find an antiderivative $F(x)=\int\sec^2x\ dx$F(x)=∫sec2x dx.
Think: We know from having previously differentiated the tangent function, that $\frac{d}{dx}\tan x=\sec^2x$ddxtanx=sec2x.
Do: Reversing this process we get:
| $F(x)$F(x) | $=$= | $\int\sec^2x\ dx$∫sec2x dx |
| $=$= | $\tan x+C$tanx+C for any constant $C$C |
Example 2
Evaluate $\int_0^{\frac{\pi}{2}}\cos t\ dt$∫π20cost dt.
Think: To find a definite integral, we first find an antiderivative. Then, we subtract its value at the lower terminal of integration from its value at the upper terminal.
Do: The antiderivative of $\cos t$cost is $\sin t$sint. So, we determine the integral as follows:
| $\int_0^{\frac{\pi}{2}}\cos t\ dt$∫π20cost dt | $=$= | $[\sin t]_0^{\frac{\pi}{2}}$[sint]π20 |
| $=$= | $\sin\frac{\pi}{2}-\sin0$sinπ2−sin0 | |
| $=$= | $1$1 |
Practice questions
Question 1
Find a primitive function for $-7\cos x$−7cosx.
You may use $C$C as the constant of integration.
Question 2
Find the integral of $\sec^2\left(x\right)$sec2(x).
You may use $C$C as the constant of integration.
Integrals of $\sin\left(ax+b\right)$sin(ax+b), $\cos\left(ax+b\right)$cos(ax+b) and $\sec^2\left(ax+b\right)$sec2(ax+b)
Extending the standard integrals for trigonometric functions by replacing the argument $x$x with a linear function $ax+b$ax+b, we get:
$\int\sin\left(ax+b\right)dx=-\frac{1}{a}\cos x+C$∫sin(ax+b)dx=−1acosx+C, for some constant $C$C
$\int\cos\left(ax+b\right)dx=\frac{1}{a}\sin x+C$∫cos(ax+b)dx=1asinx+C, for some constant $C$C
$\int\sec^2\left(ax+b\right)dx=\frac{1}{a}\tan x+C$∫sec2(ax+b)dx=1atanx+C, for some constant $C$C
Practice questions
Question 3
Integrate $\sin\left(5x-\frac{\pi}{4}\right)$sin(5x−π4).
You may use $C$C as the constant of integration.
Question 4
Evaluate $\int_{-\frac{\pi}{2}}^{\pi}\cos\left(4x-\frac{\pi}{2}\right)dx$∫π−π2cos(4x−π2)dx.
Trigonometric integrals involving the reverse chain rule
The integrals considered in the following examples make use of the chain rule, also known as the function-of-a-function rule. The key with these types of problems is to determine if a multiple of the derivative of the argument of the trigonometric function is present in the expression being integrated.
$\int\ f'\left(x\right)\ f\left(x\right)^n\ dx=\frac{1}{n+1}f\left(x\right)^{n+1}+c$∫ f′(x) f(x)n dx=1n+1f(x)n+1+c for $n\ne-1$n≠−1
Worked examples
Example 3
Find the primitive function of $y=2x\cos\left(x^2\right)$y=2xcos(x2).
Think: The primitive of this function can be found because $2x$2x is the derivative of $x^2$x2.
Do: The $\cos$cos function is the derivative of the $\sin$sin function and we observe that differentiating $\sin\left(x^2\right)$sin(x2) gives $2x\cos\left(x^2\right)$2xcos(x2). So:
$\int\ 2x\cos\left(x^2\right)\ \mathrm{d}x=\sin\left(x^2\right)+C$∫ 2xcos(x2) dx=sin(x2)+C
Alternatively we can find $\int\ 2x.\cos\left(x^2\right)\ \mathrm{d}x$∫ 2x.cos(x2) dx by first putting $u=x^2$u=x2 and then:
$\frac{\mathrm{d}u}{\mathrm{d}x}=2x$dudx=2x
On making the substitutions, the integral becomes:
$\int\ \cos u.\frac{\mathrm{d}u}{\mathrm{d}x}\ \mathrm{d}x$∫ cosu.dudx dx.
It can be shown rigorously that this is the same as $\int\ \cos u\ \mathrm{d}u$∫ cosu du. We recognise this as $\sin u+C$sinu+C which is just $\sin\left(x^2\right)+C$sin(x2)+C.
Example 4
Find the definite integral $I=\int_0^{\pi}\ \left(3x^2+1\right)\sin\left(x^3+x\right)\ \mathrm{d}x$I=∫π0 (3x2+1)sin(x3+x) dx.
Think: If we consider the argument of the $\sin$sin function, $x^3+x$x3+x, we can see that the derivative of this is $3x^2+1$3x2+1.
Do: Using the reverse chain rule we can write:
$I=\left[-\cos\left(x^3+x\right)\right]_0^{\pi}$I=[−cos(x3+x)]π0
This gives:
$I=-\cos\left(\pi^3+\pi\right)+\cos0\approx1.917$I=−cos(π3+π)+cos0≈1.917.
Practice questions
Question 5
Consider the function $y=\cos\left(6x^2-\frac{\pi}{3}\right)$y=cos(6x2−π3).
Find the derivative $\frac{dy}{dx}$dydx.
Hence find the exact value of $\int_{\sqrt{\frac{\pi}{18}}}^{\sqrt{\frac{\pi}{6}}}\left(-12x\sin\left(6x^2-\frac{\pi}{3}\right)\right)dx$∫√π6√π18(−12xsin(6x2−π3))dx.
Question 6
Find the exact value of $I$I, where $I=\int_0^{\frac{\pi^2}{9}}\frac{\sin\sqrt{x}}{\sqrt{x}}dx$I=∫π290sin√x√xdx.
Trigonometric integrals requiring identities
Integrals can be difficult to find and may not exist in a form that can be written in terms of simple functions. Manipulating the expression to be integrated and using trigonometric identities or substitutions can assist in obtaining a solution.
Practice questions
Question 7
Find the exact value of $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{3-\cos^3\left(x\right)}{\cos^2\left(x\right)}dx$∫π3π63−cos3(x)cos2(x)dx.
Question 8
Given that $\sin3t=3\sin t-4\sin^3\left(t\right)$sin3t=3sint−4sin3(t), find the indefinite integral $\int3\sin^3\left(t\right)dt$∫3sin3(t)dt.
You may use $C$C as the constant of integration.
Primitives of $\tan x$tanx and $\cot x$cotx
The primitives of $\tan x$tanx and $\cot x$cotx can be found by using the identities $\tan x=\frac{\sin x}{\cos x}$tanx=sinxcosx, $\cot x=\cos x\sin x$cotx=cosxsinx and the integral for a reciprocal function $\int\frac{f'\left(x\right)}{f\left(x\right)}dx=\ln\left|f\left(x\right)\right|+C$∫f′(x)f(x)dx=ln|f(x)|+C.
Worked example
Example 5
Evaluate $\int_0^{\frac{\pi}{4}}\tan t\ dt$∫π40tant dt.
Think: We can rewrite $\tan t$tant as $\frac{\sin t}{\cos t}$sintcost.
Do: Using the identity above, we can write:
$\int_0^{\frac{\pi}{4}}\frac{\sin t}{\cos t}$∫π40sintcost
Observe that $\frac{\sin t}{\cos t}$sintcost is almost in the form $\frac{f'\left(x\right)}{f\left(x\right)}$f′(x)f(x), and we can put it in exactly this form by rewriting it as $-\frac{-\sin t}{\cos t}$−−sintcost, since $-\sin t$−sint is the derivative of $\cos t$cost.
Therefore, the antiderivative is $-\ln\left|\cos t\right|$−ln|cost| (which should be checked by differentiating) and we calculate:
| $-\left[\ln\left|\cos t\right|\right]_0^{\frac{\pi}{4}}$−[ln|cost|]π40 | $=$= | $-\left[\ln\left(\frac{1}{\sqrt{2}}\right)-\ln1\right]$−[ln(1√2)−ln1] |
| $=$= | $-\ln\left(\frac{1}{\sqrt{2}}\right)$−ln(1√2) | |
| $\approx$≈ | $0.3466$0.3466 |
The primitive of $\cot x$cotx is developed in a very similar way and results in $\ln\left|\sin x\right|+C$ln|sinx|+C.