Pc6 - Integrals giving inverse trigonometric functions | OCR A-Level Further Mathematics B (MEI)

Recognise integrals that reverse the derivatives of arcsine and arctangent, including scaled inputs. Pc6 also uses sine and tangent substitutions for related integrals, with careful choices of angle range, transformed limits and square-root signs.

Two shapes to recognise

How can an area under $\sqrt{4-x^2}$ produce a multiple of $\pi$ ? The graph is a circle's upper half, but the same answer can be found algebraically by turning $4-x^2$ into a squared cosine. First, recognise the simpler inverse-trigonometric integrals that make that method work.

You already know

$d(\arcsin u)/dx=u'/\sqrt{1-u^2}$ and $d(\arctan u)/dx=u'/(1+u^2)$ . The chain rule means that a scaled input needs a compensating constant when these derivatives are reversed.

For example, differentiating $\arcsin(x/3)$ gives $1/\sqrt{9-x^2}$ . Differentiating $\arctan(x/3)$ gives $3/(9+x^2)$ , so its coefficient must be divided by $3$ to integrate $1/(9+x^2)$ .

Standard inverse-trigonometric integrals

For $a>0$ ,

\begin{aligned} \int\frac{dx}{\sqrt{a^2-x^2}}&=\arcsin\frac xa+C && |x|<a \\ \int\frac{dx}{a^2+x^2}&=\frac1a\arctan\frac xa+C && x\in\mathbb R \end{aligned}

In the formula booklet.

The square root distinguishes the two shapes. A difference under a square root suggests arcsine; a positive sum of squares in the denominator suggests arctangent. A denominator $a^2-x^2$ without a root is a different integral, generally handled by partial fractions.

Worked example 1: recognising both shapes

Question. Find

\int\left(\frac2{\sqrt{9-x^2}}+\frac3{9+x^2}\right)\,dx

Plan. Integrate the two terms separately, using $a=3$ in both standard formulae.

\begin{aligned} \int\frac2{\sqrt{9-x^2}}\,dx&=2\arcsin\frac x3 && \text{arcsine shape} \\ \int\frac3{9+x^2}\,dx&=3\left(\frac13\arctan\frac x3\right) && \text{arctangent shape} \\ &=\arctan\frac x3 \end{aligned}

Why this step? The arctangent formula has an outside factor $1/a$ ; the arcsine formula does not. They look similar but scale differently.

Answer. $2\arcsin(x/3)+\arctan(x/3)+C$ , for $|x|<3$ .

Check. Differentiating gives $2/\sqrt{9-x^2}+3/(9+x^2)$ , the original integrand.

Change only the input from $x$ to $2x$ throughout the two kernels. The substitution $u=2x$ contributes an extra factor $1/2$ .

Worked example 2: a scaled variable

Question. Find

\int\left(\frac2{\sqrt{9-4x^2}}+\frac3{9+4x^2}\right)\,dx

Plan. Put $u=2x$ , so $dx=du/2$ , then use the previous recognition rules.

\begin{aligned} I&=\frac12\int\left(\frac2{\sqrt{9-u^2}}+\frac3{9+u^2}\right)\,du && \text{change variable} \\ &=\frac12\left(2\arcsin\frac u3+\arctan\frac u3\right)+C && \text{standard forms} \\ &=\arcsin\frac{2x}3+\frac12\arctan\frac{2x}3+C && \text{return to }x \end{aligned}

Why this step? Replacing $x$ by $2x$ in an antiderivative would double its derivative. The factor $1/2$ compensates for that chain factor.

Answer. $\arcsin(2x/3)+\frac12\arctan(2x/3)+C$ , for $|x|<3/2$ .

Check. At $x=0$ , the integrand is $2/3+1/3=1$ ; the derivative of the answer is also $2/3+1/3=1$ .

Completing the square can reveal the same forms after a translation. For example, $x^2-2x+5=(x-1)^2+4$ , so its reciprocal integrates to $\frac12\arctan((x-1)/2)+C$ . The standard formula is supplied; spotting and making the translation is your job.

The same idea applies to a quadratic under a square root. In this example the completed square reveals both the translation and the appropriate radius.

Worked example 3: a hidden difference of squares

Question. Evaluate $\int_1^2\dfrac{dx}{\sqrt{3+2x-x^2}}$ exactly.

Plan. Complete the square, put $u=x-1$ , and use the standard arcsine integral with transformed limits.

\begin{aligned} 3+2x-x^2&=4-(x-1)^2 && \text{complete the square} \\ u&=x-1,\quad du=dx && \text{translate the input} \\ x=1,2&\ \Longrightarrow\ u=0,1 && \text{change both limits} \\ I&=\int_0^1\frac{du}{\sqrt{4-u^2}} && \text{arcsine shape} \\ &=\left[\arcsin\frac u2\right]_0^1=\frac\pi6 \end{aligned}

Why this step? The negative coefficient of $x^2$ signals a difference of squares after completing the square; it does not prevent a real-valued square root on the given interval.

Answer. $\pi/6$ .

Check. The integrand increases from $1/2$ to $1/\sqrt3$ over an interval of width $1$ , so its integral lies between those numbers; $\pi/6$ does.

After recognising the completed square, the integration is the same standard form as before. The extra work is algebra and careful translation of the endpoints.

Watch out

Wrong: $\int dx/(9+x^2)=\arctan(x/3)+C$ .

Right: $\frac13\arctan(x/3)+C$ . Differentiate a proposed answer to catch a missing scale factor immediately.

The next pair changes the variable scale again, while keeping the two standard shapes.

Sine substitution removes a difference of squares

When $\sqrt{a^2-x^2}$ appears in a more complicated integrand, put $x=a\sin\theta$ , with $-\pi/2\le\theta\le\pi/2$ . Then

dx=a\cos\theta\,d\theta,\qquad \sqrt{a^2-x^2}=a\sqrt{1-\sin^2\theta}=a\cos\theta

The last equality uses $a>0$ and $\cos\theta\ge0$ on the chosen range. Without that restriction, $\sqrt{\cos^2\theta}=|\cos\theta|$ , not automatically $\cos\theta$ .

[DIAGRAM: asset_name: A triangle for the sine substitution; asset_slug: h645_b040_sine_triangle; caption: Measuring theta from the vertical gives x equals two sin theta and the square root equals two cos theta; description: The first-quadrant part of the circle x squared plus y squared equals four contains the point one comma square root three. A right triangle joins the origin, the point and its projection on the x-axis. Its hypotenuse is two, horizontal side is x equals one, and vertical side is square root of four minus x squared equals square root three. Theta is measured clockwise from the positive vertical axis to the radius and is pi over six. A side panel connects the triangle with x equals two sin theta and the square root equals two cos theta, with non-negative cosine on the principal sine interval.]
Measuring theta from the vertical gives x equals two sin theta and the square root equals two cos theta

The angle in this triangle is measured from the vertical, so the horizontal coordinate is the opposite side. That makes it a picture of the sine substitution rather than the usual cosine-coordinate parametrisation.

Sine substitution

For $a>0$ and $-a\le x\le a$ , use

x=a\sin\theta,\qquad -\frac\pi2\le\theta\le\frac\pi2

so that $\sqrt{a^2-x^2}=a\cos\theta$ . Convert $dx$ , the entire integrand and any limits consistently.

Not in the formula booklet: learn it.

The substitution is useful because $1-\sin^2\theta=\cos^2\theta$ removes the square root. The remaining squared cosine can be integrated using the booklet's double-angle identities.

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Pc6 - Integrals giving inverse trigonometric functions

Two shapes to recognise

Sine substitution removes a difference of squares

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