Integration by Parts: Trigonometric and Inverse Trig Functions

This lesson works through integration by parts for products that involve trigonometric and inverse trigonometric functions. Throughout we use the ILATE rule (Inverse, Logarithmic, Algebraic, Trigonometric, Exponential) to decide which factor is the first function (the one we differentiate) and which is the second (the one we integrate), and the by-parts formula

$\int u\,v\,dx = u\int v\,dx - \int\left(\frac{du}{dx}\int v\,dx\right)dx.$

Integral of $x\sin 3x$

Here $x$ is algebraic and $\sin 3x$ is trigonometric, so by ILATE $x$ is the first function.

$\int x\sin 3x\,dx = x\left(-\tfrac{\cos 3x}{3}\right) - \int 1\cdot\left(-\tfrac{\cos 3x}{3}\right)dx.$

$= -\tfrac{x}{3}\cos 3x + \tfrac{1}{3}\int\cos 3x\,dx = -\tfrac{x}{3}\cos 3x + \tfrac{1}{9}\sin 3x + c.$

Integral of $\dfrac{x\sin^{-1}x}{\sqrt{1-x^2}}$

Now $\sin^{-1}x$ is inverse trigonometric, so it is the first function and $\dfrac{x}{\sqrt{1-x^2}}$ is the second.

Inner integral. With $1-x^2 = t$, so $-2x\,dx = dt$ and $x\,dx = -\tfrac{1}{2}dt$:

$\int\frac{x}{\sqrt{1-x^2}}\,dx = -\tfrac{1}{2}\int t^{-1/2}\,dt = -\sqrt{t} = -\sqrt{1-x^2}.$

By parts.

$\int\frac{x\sin^{-1}x}{\sqrt{1-x^2}}\,dx = \sin^{-1}x\left(-\sqrt{1-x^2}\right) - \int\frac{1}{\sqrt{1-x^2}}\left(-\sqrt{1-x^2}\right)dx.$

$= -\sqrt{1-x^2}\,\sin^{-1}x + \int 1\,dx = -\sqrt{1-x^2}\,\sin^{-1}x + x + c.$

Integral of $x\sec^2 x$

Here $x$ is the first function and $\sec^2 x$ the second, since $\int\sec^2 x\,dx = \tan x$.

$\int x\sec^2 x\,dx = x\tan x - \int 1\cdot\tan x\,dx = x\tan x - \log\lvert\sec x\rvert + c.$

Integral of $x\sin^{-1}x$

This time substitute first. Put $\sin^{-1}x = t$, so $x = \sin t$ and $dx = \cos t\,dt$:

$\int x\sin^{-1}x\,dx = \int t\,\sin t\cos t\,dt = \tfrac{1}{2}\int t\,\sin 2t\,dt.$

Now apply by parts with $t$ first and $\sin 2t$ second:

$\tfrac{1}{2}\int t\sin 2t\,dt = \tfrac{1}{2}\left[t\left(-\tfrac{\cos 2t}{2}\right) - \int\left(-\tfrac{\cos 2t}{2}\right)dt\right].$

$= \tfrac{1}{4}\left(-t\cos 2t + \tfrac{\sin 2t}{2}\right).$

Write $\cos 2t = 1 - 2\sin^2 t$ and $\sin 2t = 2\sin t\cos t$, then put $t = \sin^{-1}x$, $\sin t = x$, $\cos t = \sqrt{1-x^2}$:

$= \tfrac{2x^2-1}{4}\sin^{-1}x + \tfrac{x\sqrt{1-x^2}}{4} + c.$

Integral of $\tan^{-1}x$

With a single function, multiply by $1$ and take $\tan^{-1}x$ as the first function, $1$ as the second.

$\int\tan^{-1}x\,dx = x\tan^{-1}x - \int\frac{x}{1+x^2}\,dx.$

For the remaining integral put $1+x^2 = t$, so $2x\,dx = dt$:

$\int\frac{x}{1+x^2}\,dx = \tfrac{1}{2}\int\frac{dt}{t} = \tfrac{1}{2}\log\lvert 1+x^2\rvert.$

$\int\tan^{-1}x\,dx = x\tan^{-1}x - \tfrac{1}{2}\log\lvert 1+x^2\rvert + c.$

Integral of $\sin^{-1}x$

Again multiply by $1$, with $\sin^{-1}x$ first.

$\int\sin^{-1}x\,dx = x\sin^{-1}x - \int\frac{x}{\sqrt{1-x^2}}\,dx.$

The inner integral is $-\sqrt{1-x^2}$ (from the substitution used earlier), so

$\int\sin^{-1}x\,dx = x\sin^{-1}x + \sqrt{1-x^2} + c.$

Integral of $(\sin^{-1}x)^2$

Put $\sin^{-1}x = \theta$, so $x = \sin\theta$ and $dx = \cos\theta\,d\theta$:

$\int(\sin^{-1}x)^2\,dx = \int\theta^2\cos\theta\,d\theta.$

By parts, with $\theta^2$ first and $\cos\theta$ second:

$= \theta^2\sin\theta - \int 2\theta\sin\theta\,d\theta = \theta^2\sin\theta - 2\int\theta\sin\theta\,d\theta.$

Apply by parts once more to $\int\theta\sin\theta\,d\theta = -\theta\cos\theta + \sin\theta$:

$= \theta^2\sin\theta + 2\theta\cos\theta - 2\sin\theta.$

Finally put $\theta = \sin^{-1}x$, $\sin\theta = x$, $\cos\theta = \sqrt{1-x^2}$:

$\int(\sin^{-1}x)^2\,dx = x(\sin^{-1}x)^2 + 2\sqrt{1-x^2}\,\sin^{-1}x - 2x + c.$

Key takeaways

• Use ILATE to choose the first function: an inverse trig factor is differentiated, a trig factor is usually integrated.
• When the inner integral is awkward, a substitution (often $1-x^2=t$ or $1+x^2=t$) clears it up before you finish the by-parts step.
• For inverse trig functions on their own, multiply by $1$ so by parts can be applied; harder powers like $(\sin^{-1}x)^2$ need a substitution followed by two rounds of parts.