2.5, 3.5 - Using identities to integrate

2.5, 3.5 - Using identities to integrate

Some trigonometric integrands do not look like any standard derivative until you rewrite them. In 2.5 and 3.5, double-angle and reciprocal identities turn squared functions into forms you can integrate. You will then choose suitable identities for exact definite integrals and distinguish a signed integral from an area.

Squared sine and cosine

How can you integrate sin⁡2x\sin^2x? Treating it like x2x^2 would suggest sin⁡3x/3\sin^3x/3, but differentiation gives sin⁡2xcos⁡x\sin^2x\cos x. That extra factor means the suggestion is wrong.

Instead, change the integrand before integrating. Rearrange cos⁡(2x)=1−2sin⁡2x\cos(2x)=1-2\sin^2x to get sin⁡2x=(1−cos⁡(2x))/2\sin^2x=(1-\cos(2x))/2. A square has become a constant and an ordinary cosine.

Reducing squared sine and cosine

For real uu,

sin⁡2u=1−cos⁡(2u)2,cos⁡2u=1+cos⁡(2u)2.\sin^2u=\frac{1-\cos(2u)}2, \qquad \cos^2u=\frac{1+\cos(2u)}2.

These are rearrangements of the double-angle identities in the formula booklet; the rearranged forms are not listed separately.

The angle doubles as well as the function losing its square. For cos⁡2(2x)\cos^2(2x), the new cosine is cos⁡(4x)\cos(4x). Integration must then undo its inner derivative 44. Throughout this lesson, angles are in radians.

Worked example 1: squared sine

Question. Find ∫sin⁡2x dx\int\sin^2x\,\mathrm dx.

Plan. Replace the square using the double-angle identity, then integrate term by term.

∫sin⁡2x dx=12∫[1−cos⁡(2x)] dx=x2−14sin⁡(2x)+C\begin{aligned} \int\sin^2x\,\mathrm dx &=\frac12\int[1-\cos(2x)]\,\mathrm dx\\ &=\frac{x}{2}-\frac14\sin(2x)+C \end{aligned}

The coefficient 1/41/4 combines the outside 1/21/2 with division by the inner derivative 22.

Answer. x2−14sin⁡(2x)+C\frac{x}{2}-\frac14\sin(2x)+C.

Check. Differentiation gives 12−12cos⁡(2x)=sin⁡2x\frac12-\frac12\cos(2x)=\sin^2x.

Change only sine to cosine. The identity's sign changes, while the coefficients stay the same.

Worked example 2: squared cosine

Question. Find ∫cos⁡2x dx\int\cos^2x\,\mathrm dx.

Plan. Use the plus-sign version of the squared-function identity.

∫cos⁡2x dx=12∫[1+cos⁡(2x)] dx=x2+14sin⁡(2x)+C\begin{aligned} \int\cos^2x\,\mathrm dx &=\frac12\int[1+\cos(2x)]\,\mathrm dx\\ &=\frac{x}{2}+\frac14\sin(2x)+C \end{aligned}

Answer. x2+14sin⁡(2x)+C\frac{x}{2}+\frac14\sin(2x)+C.

Check. Its derivative is 12+12cos⁡(2x)=cos⁡2x\frac12+\frac12\cos(2x)=\cos^2x. Adding this derivative to the previous one gives 11, as sin⁡2x+cos⁡2x=1\sin^2x+\cos^2x=1 requires.

An identity changes the appearance of a function without changing its values. It gives us access to a standard integral; it does not create a new integration rule for arbitrary squared functions.

Watch out

cos⁡2(2x)=(1+cos⁡(4x))/2\cos^2(2x)=(1+\cos(4x))/2, not (1+cos⁡(2x))/2(1+\cos(2x))/2. Double the whole original angle before integrating.

Now change the angle in the second example.

Rearranging reciprocal identities

The square of a tangent needs a different identity. Since 1+tan⁡2u=sec⁡2u1+\tan^2u=\sec^2u, rewrite tan⁡2u\tan^2u as sec⁡2u−1\sec^2u-1. Both terms now have known antiderivatives.

For cotangent, 1+cot⁡2u=cosec⁡2u1+\cot^2u=\operatorname{cosec}^2u gives the matching rearrangement. The formula booklet supplies
ddx(cot⁡x)=−cosec⁡2x\frac{\mathrm d}{\mathrm dx}(\cot x)=-\operatorname{cosec}^2x. Reversing it gives ∫cosec⁡2x dx=−cot⁡x+C\int\operatorname{cosec}^2x\,\mathrm dx=-\cot x+C. For a linear angle ax+bax+b, divide by aa as before.

Integrating squared tangent and cotangent

For a≠0a\ne0, on an interval where the integrand is defined,

∫tan⁡2(ax+b) dx=1atan⁡(ax+b)−x+C∫cot⁡2(ax+b) dx=−1acot⁡(ax+b)−x+C\begin{aligned} \int\tan^2(ax+b)\,\mathrm dx&=\frac1a\tan(ax+b)-x+C\\ \int\cot^2(ax+b)\,\mathrm dx&=-\frac1a\cot(ax+b)-x+C \end{aligned}

The reciprocal identities and the derivatives of tan⁡x\tan x and cot⁡x\cot x are in the formula booklet. These squared-function integrals are not listed separately: reconstruct them from those results.

There are two signs to track in the cotangent result. Its derivative supplies the negative coefficient of cot⁡\cot; the −x-x comes from subtracting 11 in the identity. Differentiating the answer distinguishes their jobs.

Worked example 3: squared tangent

Question. Find ∫tan⁡2(2x) dx\int\tan^2(2x)\,\mathrm dx.

Plan. Write the integrand as sec⁡2(2x)−1\sec^2(2x)-1, then integrate.

∫tan⁡2(2x) dx=∫[sec⁡2(2x)−1] dx=12tan⁡(2x)−x+C\begin{aligned} \int\tan^2(2x)\,\mathrm dx &=\int[\sec^2(2x)-1]\,\mathrm dx\\ &=\frac12\tan(2x)-x+C \end{aligned}

Answer. 12tan⁡(2x)−x+C\frac12\tan(2x)-x+C, on an interval where cos⁡(2x)≠0\cos(2x)\ne0.

Check. Its derivative is sec⁡2(2x)−1=tan⁡2(2x)\sec^2(2x)-1=\tan^2(2x).

Change tangent to cotangent. Keep the angle 2x2x, but use the negative derivative of cotangent and check its different undefined inputs.

Worked example 4: squared cotangent

Question. Find ∫cot⁡2(2x) dx\int\cot^2(2x)\,\mathrm dx.

Plan. Write cot⁡2(2x)=cosec⁡2(2x)−1\cot^2(2x)=\operatorname{cosec}^2(2x)-1 and reverse the cotangent derivative.

∫cot⁡2(2x) dx=∫[cosec⁡2(2x)−1] dx=−12cot⁡(2x)−x+C\begin{aligned} \int\cot^2(2x)\,\mathrm dx &=\int[\operatorname{cosec}^2(2x)-1]\,\mathrm dx\\ &=-\frac12\cot(2x)-x+C \end{aligned}

Answer. −12cot⁡(2x)−x+C-\frac12\cot(2x)-x+C, on an interval where sin⁡(2x)≠0\sin(2x)\ne0.

Check. Differentiating the first term gives cosec⁡2(2x)\operatorname{cosec}^2(2x); subtracting 11 gives the required cot⁡2(2x)\cot^2(2x).

Continue this lesson.

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