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Three cases decided by two parities: peel an odd sine, peel an odd cosine, or halve the powers with $\sin^2\theta = \tfrac12(1 - \cos 2\theta)$.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to read the parities of the two exponents and name the method they force, peel a single factor off an odd power and use the Pythagorean identity on what is left, substitute for the function you did not peel and carry the minus sign of $du$ correctly, apply the half-angle identities repeatedly when both powers are even, change the limits of a definite integral with the variable, and say why the even case produces an answer containing $\pi$ and the odd case does not.
You know $\dfrac{d}{dx}\sin x = \cos x$ and $\dfrac{d}{dx}\cos x = -\sin x$, and you know $\sin^2 x + \cos^2 x = 1$.
Those two facts together are the whole of this lesson. The derivative rules say that a spare factor of $\sin x$ is exactly what $u = \cos x$ needs for its $du$; the identity says that any even power of sine can be rewritten in cosines. Put them together and an integrand with an odd power collapses into a polynomial.
To peel a factor is to write $\sin^{2k+1}x = \sin^{2k}x \cdot \sin x$ and keep the last factor aside for the $du$. It is only possible when the power is odd, because an even power has no spare factor once it is written as a power of $\sin^2$.
The half-angle identities are $\sin^2\theta = \dfrac{1 - \cos 2\theta}{2}$ and $\cos^2\theta = \dfrac{1 + \cos 2\theta}{2}$. They trade a squared function for a function of double the angle, which is the trade that makes an even power integrable at all.
For $\displaystyle\int \sin^m x\,\cos^n x\,dx$ there are exactly three cases, and which one you are in is decided by the parities of $m$ and $n$.
$m$ odd. Peel one $\sin x$; the rest is an even power of sine, which $\sin^2 = 1 - \cos^2$ turns into cosines. Substitute $u = \cos x$, $du = -\sin x\,dx$. The integral becomes $-\int (1 - u^2)^{(m-1)/2}u^n\,du$: a polynomial.
$n$ odd. The mirror image: peel one $\cos x$, use $\cos^2 = 1 - \sin^2$, substitute $u = \sin x$, $du = \cos x\,dx$.
Both even. There is no factor to peel, so no substitution exists. Use $$\sin^2\theta = \frac{1 - \cos 2\theta}{2}, \qquad \cos^2\theta = \frac{1 + \cos 2\theta}{2}$$ to halve every power; repeat until only first powers of cosines of multiple angles are left, and integrate those directly.
Both odd. Either peel works. Take whichever leaves the smaller polynomial.
The same three cases govern $\displaystyle\int \tan^m x \sec^n x\,dx$, with $\sec^2 = 1 + \tan^2$ in place of the Pythagorean identity: an even power of $\sec$ lets you peel $\sec^2$ for $u = \tan x$, and an odd power of $\tan$ lets you peel $\sec\tan$ for $u = \sec x$.
Another way: picture
$\sin^2 x$ and $\cos^2 x$ drawn on the same axes: two identical waves, one the other shifted by a quarter period, both oscillating between $0$ and $1$ about the height $\tfrac12$. They sum to the constant $1$, which is why each averages exactly $\tfrac12$, and the half-angle identity is nothing more than that picture written as a formula.
Another way: steps
It is tempting to think the even case is just a harder version of the odd case. It is not; it is a different kind of problem.
With an odd power there is a spare factor, and that factor is the derivative of the thing you substitute for. The substitution exists because the integrand contains its own $du$. With both powers even there is no spare factor anywhere, and no substitution in $\sin$ or $\cos$ can work — not because it is difficult, but because the necessary $du$ is not present in the integrand.
So the even case does something else entirely: it lowers the power using an identity, at the cost of doubling the angle. $\sin^4 x$ becomes a combination of $1$, $\cos 2x$ and $\cos^2 2x$; applying the identity again turns the last of these into $1$ and $\cos 4x$. Each pass halves the powers and doubles the angles, and after enough passes only first powers remain, which integrate on sight.
That is also why the even case produces answers containing $\pi$ over a quarter or half turn while the odd case produces rational numbers: the constant term of the expansion is what survives the integration over a full period, and only even powers have one.
Rewriting before peeling. Applying $\sin^2 = 1 - \cos^2$ to the whole odd power leaves a stray $\sin x$ inside a bracket where it does you no good. Peel first, then rewrite what is left.
Losing the minus sign of $du$. $u = \cos x$ gives $du = -\sin x\,dx$. In a definite integral the minus and the reversed limits cancel, which is tidy — but only if both are written down.
Substituting for the same function you peeled. If you peel a sine, you substitute for the cosine. Peeling and substituting for the same function leaves the integral untouched.
Trying to peel an even power. $\sin^2 x = \sin x \cdot \sin x$ is true and useless: the rest is $\sin x$, an odd power, and you are going in circles.
Half-angle applied to the angle rather than the power. The identity halves the exponent and doubles the angle. Writing $\sin^2 x = \dfrac{1 - \cos x}{2}$ is the same formula with the $2$ in the wrong place, and it is wrong at every value of $x$ but one.
These integrals are filed under trigonometry, and learners approach them looking for identities to apply — as many as possible, as early as possible. That is exactly backwards, and it is why a correct method often produces an unmanageable mess.
In the odd case almost no trigonometry is used at all. One factor is peeled, one identity is applied to what is left, and from that moment the problem is $\int (1 - u^2)^k u^n\,du$: a polynomial, integrated by the power rule you learned before any of this. A learner who expands with product-to-sum formulas first will get a correct answer eventually, by a much longer road.
The useful instinct is the opposite of the expected one: do as little trigonometry as the parities force you to do. Peel, substitute, and stop. Reach for the half-angle identities only in the case where no substitution exists, because that is the only case in which they are the shortest route rather than a detour.
$\displaystyle\int \sin^5 x\,dx$: peel one sine, leaving $\sin^4 x = (1 - \cos^2 x)^2$.
The odd power is what makes a spare factor available.
With $u = \cos x$: $-\displaystyle\int (1 - u^2)^2 du = -\int (1 - 2u^2 + u^4)du = -u + \dfrac{2u^3}{3} - \dfrac{u^5}{5} + C$.
A polynomial, expanded and integrated term by term.
Back in $x$: $-\cos x + \dfrac{2\cos^3 x}{3} - \dfrac{\cos^5 x}{5} + C$. Differentiating it returns $\sin^5 x$, which is the check worth doing once.
$\displaystyle\int_0^{\pi} \sin^4 x\,dx$: no peel is available, so $\sin^4 x = \left(\dfrac{1 - \cos 2x}{2}\right)^2 = \dfrac{1 - 2\cos 2x + \cos^2 2x}{4}$.
First pass: powers halved, angle doubled.
$\cos^2 2x = \dfrac{1 + \cos 4x}{2}$, so the integrand is $\dfrac{3}{8} - \dfrac{\cos 2x}{2} + \dfrac{\cos 4x}{8}$.
Second pass: only first powers left.
Over $[0, \pi]$ both cosine terms integrate to zero, leaving $\dfrac{3\pi}{8}$. Only the constant term survived — which is the general rule for an even power over a whole number of half-turns, and it agrees with the reduction formula of the last lesson.
The cosine power is odd, so peel one $\cos x$ and write $\cos^2 x = 1 - \sin^2 x$.
Which parity is odd decides which function you substitute for.
$u = \sin x$, $du = \cos x\,dx$; the limits become $0$ and $1$, with no sign change this time.
The cosine substitution keeps the limits in order.
$\displaystyle\int_0^1 u^2(1 - u^2)\,du = \dfrac{1}{3} - \dfrac{1}{5} = \dfrac{2}{15}$.
How would you start $\int \sin^4 x\,dx$?
Put the steps of evaluating $\displaystyle\int \sin^{3}x\,\cos^{6}x\,dx$ into the order you do them.
Number the steps in order (write the number in the box):
Evaluate $\displaystyle\int_0^{\pi/2} \sin^{3}x\,\cos^{7}x\,dx$.
The integral equals answer.
Match each integral to what you would substitute.
| $u = \cos x$ | $u = \sin x$ | No substitution — halve the angle instead | |
|---|---|---|---|
| $\displaystyle\int \sin^{7}x\,dx$ | |||
| $\displaystyle\int \cos^{7}x\,dx$ | |||
| $\displaystyle\int \sin^{2}x\cos^{2}x\,dx$ | |||
| $\displaystyle\int \sin^{6}x\,dx$ |
For each integrand, say which power is odd.
| Which power is odd? | |
|---|---|
| $\sin^{7}x\,\cos^{2}x$ | |
| $\sin^{2}x\,\cos^{7}x$ | |
| $\sin^{2}x\,\cos^{4}x$ |
An alternating current is $i(t) = 5\sin(\omega t)$ amperes. Over one full period, what is the average value of $i^{2}$?
The average value is answer square amperes.
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
How would you start $\int \sin^2 x \cos^2 x\,dx$?
You can decide a trigonometric integral's method from its exponents alone, before doing any algebra. Say in your own words why two even powers admit no substitution at all. Next: the substitutions that go the other way, putting trigonometry into an integral that had none.
10. Your turn: $\displaystyle\int_0^{\pi/2}\sin^{2}x\,\cos^{3}x\,dx$, step 3
A polynomial again.