Further Maths/Lessons/Methods in Calculus (Year 2)/Inverse Trig Functions
Core Pure · Methods in Calculus · Year 2

Differentiating & Integrating Inverse Trigonometric Functions

Implicit differentiation unlocks derivatives for arcsin, arccos and arctan — and running the results backwards gives two integrals you'll use constantly.

14 min read 4 worked examples

Differentiating the inverse trig functions

Theorem 3.3 — derivatives of arcsin, arccos, arctan

$$\frac{d}{dx}(\arcsin x)=\frac1{\sqrt{1-x^2}}$$

$$\frac{d}{dx}(\arccos x)=-\frac1{\sqrt{1-x^2}}$$

$$\frac{d}{dx}(\arctan x)=\frac1{1+x^2}$$

Notice that $\arcsin x$ and $\arccos x$ have derivatives that are negatives of each other — consistent with $\arcsin x+\arccos x=\dfrac\pi2$ being constant, so their derivatives must sum to zero.

Image pending
inverse-trig-graphs.png
The graphs of $y=\arcsin x$, $y=\arccos x$ and $y=\arctan x$, each restricted to the domain/range that makes it a genuine function.
ProofDerivative of arcsin x, by implicit differentiation
1
Let $y=\arcsin x$, so $\sin y=x$, with $-\dfrac\pi2\leq y\leq\dfrac\pi2$.
Rewrite the inverse function as an equation you can differentiate implicitly.
2
Differentiating both sides with respect to $x$: $\cos y\cdot\dfrac{dy}{dx}=1$, so $\dfrac{dy}{dx}=\dfrac1{\cos y}$.
Implicit differentiation, treating $y$ as a function of $x$.
3
Since $\cos^2y+\sin^2y=1$, $\cos y=\pm\sqrt{1-\sin^2y}=\pm\sqrt{1-x^2}$.
Use the Pythagorean identity and substitute $\sin y=x$.
4
For $-\dfrac\pi2\leq y\leq\dfrac\pi2$, $\cos y\geq0$, so we must take the positive root: $\cos y=\sqrt{1-x^2}$.
The restricted range of $\arcsin$ rules out the negative root.
5
Therefore $\dfrac{dy}{dx}=\dfrac1{\sqrt{1-x^2}}$.
Substitute back into the result from step 2.
∎
Example 1 · Chain rule with an inverse trig function
Given $y=\arctan(3x)$, find $\dfrac{dy}{dx}$.

Let $u=3x$, so $y=\arctan u$. Then $\dfrac{dy}{du}=\dfrac1{1+u^2}$ and $\dfrac{du}{dx}=3$.

By the chain rule: $\dfrac{dy}{dx}=\dfrac{dy}{du}\times\dfrac{du}{dx}=\dfrac3{1+9x^2}$.

Example 2 · A product involving arcsin
Given $y=x^2\arcsin x$, find $\dfrac{dy}{dx}$.

By the product rule: $\dfrac{dy}{dx}=2x\arcsin x+x^2\times\dfrac1{\sqrt{1-x^2}}=2x\arcsin x+\dfrac{x^2}{\sqrt{1-x^2}}$.

Integrating to give inverse trig functions

Theorem 3.4 — standard integral results

Reversing the derivatives above (with a scaling constant $a>0$) gives two integrals worth memorising:

$$\int\frac1{a^2+x^2}\,dx=\frac1a\arctan\left(\frac xa\right)+c$$

$$\int\frac1{\sqrt{a^2-x^2}}\,dx=\arcsin\left(\frac xa\right)+c, \; |x|\lt a$$

Example 3 · Matching a denominator to the standard form
Find $\displaystyle\int\frac9{16+x^2}\,dx$.

Compare $16+x^2$ with $a^2+x^2$: here $a=4$.

$\displaystyle\int\frac9{16+x^2}\,dx=9\int\frac1{16+x^2}\,dx=9\times\frac14\arctan\left(\frac x4\right)+c=\frac94\arctan\left(\frac x4\right)+c$.

Watch out — factoring out a coefficient of $x^2$ first

If the coefficient of $x^2$ isn't $1$, factor it out before matching to the standard form — e.g. $9+4x^2=4\left(\dfrac94+x^2\right)$, so $a=\dfrac32$ and there's an overall factor of $\dfrac14$ to carry through the integral.

Example 4 · Rewriting before matching, and a split numerator
Find $\displaystyle\int\frac{x+3}{\sqrt{4-x^2}}\,dx$.

Split the numerator: $\displaystyle\int\frac x{\sqrt{4-x^2}}\,dx+3\int\frac1{\sqrt{4-x^2}}\,dx$.

For the first integral, use the substitution $u=4-x^2$, $du=-2x\,dx$: $\displaystyle\int\frac x{\sqrt{4-x^2}}\,dx=-\frac12\int u^{-1/2}\,du=-\sqrt u+c=-\sqrt{4-x^2}+c$.

For the second, $a=2$: $3\displaystyle\int\frac1{\sqrt{4-x^2}}\,dx=3\arcsin\left(\frac x2\right)+c$.

Combining: $\displaystyle\int\frac{x+3}{\sqrt{4-x^2}}\,dx=-\sqrt{4-x^2}+3\arcsin\left(\frac x2\right)+c$.

Exam-style questions

Question 1 5 marks

Given $y=\arccos(2x)$, find $\dfrac{dy}{dx}$, and state the values of $x$ for which the derivative is defined.

Let $u=2x$: $\dfrac{dy}{du}=-\dfrac1{\sqrt{1-u^2}}$, $\dfrac{du}{dx}=2$. (M1, A1)

$\dfrac{dy}{dx}=-\dfrac2{\sqrt{1-4x^2}}$. (M1, A1)

Defined for $|2x|\lt1$, i.e. $-\dfrac12\lt x\lt\dfrac12$. (A1)

Question 2 6 marks

Find $\displaystyle\int\frac5{9+25x^2}\,dx$.

Factor: $9+25x^2=25\left(\dfrac9{25}+x^2\right)$, so with $a=\dfrac35$: $\displaystyle\int\frac1{9+25x^2}\,dx=\frac1{25}\int\frac1{\frac9{25}+x^2}\,dx$. (M1, A1)

$=\dfrac1{25}\times\dfrac1{3/5}\arctan\left(\dfrac x{3/5}\right)+c=\dfrac1{15}\arctan\left(\dfrac{5x}3\right)+c$. (M1, A1)

So $\displaystyle\int\frac5{9+25x^2}\,dx=5\times\dfrac1{15}\arctan\left(\dfrac{5x}3\right)+c=\dfrac13\arctan\left(\dfrac{5x}3\right)+c$. (A1, A1)

Question 3 6 marks

Evaluate $\displaystyle\int_0^1\frac1{\sqrt{4-x^2}}\,dx$, leaving your answer in terms of $\pi$.

With $a=2$: $\displaystyle\int\frac1{\sqrt{4-x^2}}\,dx=\arcsin\left(\frac x2\right)+c$. (M1, A1)

$\Big[\arcsin\left(\dfrac x2\right)\Big]_0^1=\arcsin\left(\dfrac12\right)-\arcsin(0)=\dfrac\pi6-0=\dfrac\pi6$. (M1, A1, A1, A1)

← Previous
Improper Integrals & Mean Value