Two Sure Questions from Integrals

This lesson works through two integrals that come up often in Class 12 exams. Each one uses a substitution to turn an awkward expression into a standard form we already know how to integrate.

First integral: roots of cot and tan

We want to evaluate

$I = \int \left( \sqrt{\cot x} + \sqrt{\tan x} \right)\, dx.$

Write $\sqrt{\cot x} = \tfrac{1}{\sqrt{\tan x}}$ and factor $\sqrt{\tan x}$ out of both terms:

$I = \int \sqrt{\tan x}\left( \frac{1}{\tan x} + 1 \right) dx = \int \sqrt{\tan x}\,(\cot x + 1)\, dx.$

The substitution

Put $\tan x = t^2$. Differentiating, $\sec^2 x\, dx = 2t\, dt$, so

$dx = \frac{2t}{\sec^2 x}\, dt = \frac{2t}{1 + \tan^2 x}\, dt = \frac{2t}{1 + t^4}\, dt.$

With $\tan x = t^2$ we have $\sqrt{\tan x} = t$ and $\cot x + 1 = \tfrac{1}{t^2} + 1 = \tfrac{t^2 + 1}{t^2}$. Substituting,

$I = \int t \cdot \frac{t^2 + 1}{t^2} \cdot \frac{2t}{1 + t^4}\, dt = 2 \int \frac{t^2 + 1}{t^4 + 1}\, dt.$

Dividing through by $t^2$

Divide the top and bottom by $t^2$:

$I = 2 \int \frac{1 + \tfrac{1}{t^2}}{t^2 + \tfrac{1}{t^2}}\, dt.$

Use the identity $t^2 + \tfrac{1}{t^2} = \left( t - \tfrac{1}{t} \right)^2 + 2$. Now put $u = t - \tfrac{1}{t}$, so that $du = \left( 1 + \tfrac{1}{t^2} \right) dt$, which is exactly the numerator. Then

$I = 2 \int \frac{du}{u^2 + 2} = 2 \int \frac{du}{u^2 + (\sqrt{2})^2}.$

Standard form

Using $\displaystyle \int \frac{dx}{a^2 + x^2} = \frac{1}{a}\tan^{-1}\frac{x}{a}$ with $a = \sqrt{2}$,

$I = 2 \cdot \frac{1}{\sqrt{2}} \tan^{-1}\frac{u}{\sqrt{2}} = \sqrt{2}\,\tan^{-1}\frac{u}{\sqrt{2}}.$

Resubstitute $u = t - \tfrac{1}{t}$ and then $t = \sqrt{\tan x}$:

$I = \sqrt{2}\,\tan^{-1}\!\left( \frac{\tan x - 1}{\sqrt{2\,\tan x}} \right) + C.$

Second integral: a root in the denominator

Now evaluate

$I = \int \frac{\sin 2x \cos 2x}{\sqrt{9 - \cos^4 2x}}\, dx.$

Write $9 = 3^2$ and $\cos^4 2x = (\cos^2 2x)^2$, so

$I = \int \frac{\sin 2x \cos 2x}{\sqrt{3^2 - (\cos^2 2x)^2}}\, dx.$

The substitution

Put $t = \cos^2 2x$. Then $\sin 2x \cos 2x\, dx = -\tfrac{1}{4}\, dt$; keeping the constant aside, the integral reduces to

$I = \int \frac{\tfrac{1}{2}\, dt}{\sqrt{3^2 - t^2}}.$

Standard form

Using $\displaystyle \int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\frac{x}{a}$ with $a = 3$,

$I = \frac{1}{2}\sin^{-1}\frac{t}{3}.$

Resubstitute $t = \cos^2 2x$:

$I = \frac{1}{2}\sin^{-1}\!\left( \frac{\cos^2 2x}{3} \right) + C.$

Key takeaways

• Factoring a sum of root trig functions into one root times a bracket exposes a clean substitution.
• The substitution $\tan x = t^2$ clears the roots and, after dividing by $t^2$ and completing the square with $t - \tfrac{1}{t}$, gives a standard inverse tangent integral.
• Spotting $a^2 - x^2$ under a square root signals the inverse sine form $\sin^{-1}\tfrac{x}{a}$.