$\int \sqrt{\tan x}\,dx$ (Hint: Put $\tan x=t^2$)

> [!TLDR] > Using $\tan x=t^2$ the integral becomes $2\int\frac{t^2}{1+t^4}\,dt$, which splits (both terms added) into $\frac{1}{\sqrt2}\tan^{-1}\!\left(\frac{\tan x-1}{\sqrt{2\tan x}}\right)+\frac{1}{2\sqrt2}\log\!\left|\frac{\tan x-\sqrt{2\tan x}+1}{\tan x+\sqrt{2\tan x}+1}\right|+C$. ## The idea A square root of $\tan x$ is awkward, so we remove the root by letting $\tan x=t^2$ — then $\sqrt{\tan x}=t$ is a plain variable and the whole problem becomes a rational function of $t$. ## Set up the substitution From $\tan x=t^2$, differentiate both sides: $\sec^2 x\,dx=2t\,dt$. Since $\sec^2 x=1+\tan^2 x=1+t^4$, $$dx=\frac{2t}{1+t^4}\,dt,$$ so $$\int\sqrt{\tan x}\,dx=\int t\cdot\frac{2t}{1+t^4}\,dt=2\int\frac{t^2}{1+t^4}\,dt.$$ ## Split the fraction Write $t^2$ as a symmetric combination and check the signs are **both plus**: $$\frac12\big[(t^2+1)+(t^2-1)\big]=t^2,$$ hence $$2\int\frac{t^2}{1+t^4}\,dt=\int\frac{t^2+1}{t^4+1}\,dt+\int\frac{t^2-1}{t^4+1}\,dt.$$ ## The two standard pieces For the first, divide numerator and denominator by $t^2$ and set $w=t-\frac{1}{t}$ (so $dw=(1+\frac{1}{t^2})\,dt$ and $t^2+\frac{1}{t^2}=w^2+2$): $$\int\frac{t^2+1}{t^4+1}\,dt=\int\frac{dw}{w^2+2}=\frac{1}{\sqrt2}\tan^{-1}\frac{w}{\sqrt2}=\frac{1}{\sqrt2}\tan^{-1}\frac{t^2-1}{\sqrt2\,t}.$$ For the second, set $v=t+\frac{1}{t}$ (so $dv=(1-\frac{1}{t^2})\,dt$ and $t^2+\frac{1}{t^2}=v^2-2$): …

∫ √(tan x)dx (Hint: Put tan x=t²)

NCERT Exemplar · Q43

\int \sqrt{\tan x}\,dx

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Concept understanding — U Substitution

U Substitution: The Reverse Chain Rule

The chain rule differentiates composite functions: the derivative of $\sin(x^2)$ is $\cos(x^2) \cdot 2x$ — differentiate the outer function, then multiply by the derivative of the inside. Integration asks the reverse: given $\cos(x^2) \cdot 2x$ , find the original function. That's what u substitution does — it reverses the chain rule.

The Core Intuition

When an integral looks like "a function times the derivative of its inside," substitute the inside with $u$ and the derivative of the inside with $du$ . Consider:

\int 2x \cos(x^2) \, dx

Here $2x$ is the derivative of $x^2$ , and $x^2$ is the inside of $\cos(x^2)$ . Let $u = x^2$ , so $du = 2x \, dx$ :

\int \cos(u) \, du = \sin(u) + C = \sin(x^2) + C

Check: the derivative of $\sin(x^2)$ is $\cos(x^2) \cdot 2x$ .

The Precise Statement

\int f(g(x)) \cdot g'(x) \, dx = \int f(u) \, du \quad \text{where } u = g(x), \, du = g'(x) \, dx

Valid provided $g$ is differentiable and the resulting integral in $u$ is simpler.

The Step-by-Step Method

Identify a function $g(x)$ whose derivative $g'(x)$ also appears (possibly up to a constant factor).
Set $u = g(x)$ , compute $du = g'(x) \, dx$ .
Rewrite the entire integral in $u$ and $du$ — every $x$ and $dx$ must be replaced.
Integrate with respect to $u$ .
Substitute back $u = g(x)$ .

Watch out

You cannot mix variables. If any $x$ remains after substitution, you chose the wrong $u$ (or must solve for $x$ in terms of $u$ — rare).

A Second Example (with a constant factor)

Evaluate $\int x \sqrt{x^2 + 1} \, dx$ . Let $u = x^2 + 1$ , so $x \, dx = \frac{1}{2} du$ :

\int \sqrt{u} \cdot \tfrac{1}{2} du = \tfrac{1}{2} \cdot \tfrac{2}{3} u^{3/2} + C = \tfrac{1}{3} (x^2 + 1)^{3/2} + C

When Does It Work?

When the integrand is something times the derivative of something inside. Common patterns:

$x \cdot f(x^2)$ — derivative of $x^2$ is $2x$ , so $u = x^2$
$e^{g(x)} \cdot g'(x)$ — derivative of $g(x)$ appears
$\frac{g'(x)}{g(x)}$ — leads to $\log|g(x)|$ …

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Using $\tan x=t^2$ the integral becomes $2\int\frac{t^2}{1+t^4}\,dt$ , which splits (both terms added) into $\frac{1}{\sqrt2}\tan^{-1}\!\left(\frac{\tan x-1}{\sqrt{2\tan x}}\right)+\frac{1}{2\sqrt2}\log\!\left|\frac{\tan x-\sqrt{2\tan x}+1}{\tan x+\sqrt{2\tan x}+1}\right|+C$ .

The idea

A square root of $\tan x$ is awkward, so we remove the root by letting $\tan x=t^2$ — then $\sqrt{\tan x}=t$ is a plain variable and the whole problem becomes a rational function of $t$ .

Set up the substitution

From $\tan x=t^2$ , differentiate both sides: $\sec^2 x\,dx=2t\,dt$ . Since $\sec^2 x=1+\tan^2 x=1+t^4$ ,

dx=\frac{2t}{1+t^4}\,dt,

\int\sqrt{\tan x}\,dx=\int t\cdot\frac{2t}{1+t^4}\,dt=2\int\frac{t^2}{1+t^4}\,dt.

Split the fraction

Write $t^2$ as a symmetric combination and check the signs are both plus:

\frac12\big[(t^2+1)+(t^2-1)\big]=t^2,

hence

2\int\frac{t^2}{1+t^4}\,dt=\int\frac{t^2+1}{t^4+1}\,dt+\int\frac{t^2-1}{t^4+1}\,dt.

The two standard pieces

For the first, divide numerator and denominator by $t^2$ and set $w=t-\frac{1}{t}$ (so $dw=(1+\frac{1}{t^2})\,dt$ and $t^2+\frac{1}{t^2}=w^2+2$ ):

\int\frac{t^2+1}{t^4+1}\,dt=\int\frac{dw}{w^2+2}=\frac{1}{\sqrt2}\tan^{-1}\frac{w}{\sqrt2}=\frac{1}{\sqrt2}\tan^{-1}\frac{t^2-1}{\sqrt2\,t}.

For the second, set $v=t+\frac{1}{t}$ (so $dv=(1-\frac{1}{t^2})\,dt$ and $t^2+\frac{1}{t^2}=v^2-2$ ): …

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Q. $\int \sqrt{\tan x}\,dx$ (Hint: Put $\tan x=t^2$ )

Concept understanding — U Substitution

U Substitution: The Reverse Chain Rule

The Core Intuition

The Precise Statement

The Step-by-Step Method

A Second Example (with a constant factor)

When Does It Work?

The idea

Set up the substitution

Split the fraction

The two standard pieces

Method: Removing a square root of a trig function

Steps

Common Mistakes

Q.∫tan⁡x dx\int \sqrt{\tan x}\,dx∫tanx​dx (Hint: Put tan⁡x=t2\tan x=t^2tanx=t2)

The idea

Set up the substitution

Split the fraction

The two standard pieces

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Q. $\int \sqrt{\tan x}\,dx$ (Hint: Put $\tan x=t^2$ )

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