Animated Solution for Mathematics - Indefinite Integration: Evaluate $\int (\sqrt{\tan x}+\sqrt{\cot x})dx$

Imagine you are standing before a complex, intimidating integral: $I=\int (\sqrt{\tan x}+\sqrt{\cot x})dx$. It looks like a tangled knot of trigonometric functions, but in the world of JEE Advanced, every knot has a loose thread.

The first step is to simplify our perspective. We know that $\tan x$ and $\cot x$ are reciprocals. By rewriting them as $\frac{\sin x}{\cos x}$ and $\frac{\cos x}{\sin x}$, we transform the expression into:

This is the first step of our journey: moving from the unfamiliar to the fundamental.

Now, let's combine these terms. By taking the least common multiple, we get:

Look at that numerator! It is $\sin x+\cos x$. This is a massive hint. In calculus, whenever you see a sum of sine and cosine, you should immediately think about the derivative of their difference.

Before we jump to that, we need to clean up the denominator. We have $\sqrt{\sin x\cos x}$. If we multiply and divide by $\sqrt{2}$, we get:

Since $2\sin x\cos x=\sin 2x$, our integral becomes:

Substitution

Now, let's perform the magic. We need a substitution that will turn our numerator into $dt$. As we suspected, let $t=\sin x-\cos x$.

Differentiating both sides, we get $\frac{dt}{dx}=\cos x+\sin x$, which means $dt=(\sin x+\cos x)dx$. Our numerator is perfectly accounted for.

Now, we express $\sin 2x$ in terms of $t$. If we square our substitution:

Using the identity ${\sin }^{2}x+{\cos }^{2}x=1$ and $2\sin x\cos x=\sin 2x$, we find $t^2=1-\sin 2x$, or $\sin 2x=1-t^2$. Substituting these into our integral, we get:

This is a standard integral. The integral of $\frac{1}{\sqrt{1-t^2}}$ is ${\sin }^{-1}t$. So, our result is:

Substituting back $t=\sin x-\cos x$, we get:

While this is a perfectly valid answer, we can express it in terms of ${\tan }^{-1}$. Using the identity ${\sin }^{-1}y={\tan }^{-1}\left(\frac{y}{\sqrt{1-y^2}}\right)$, we convert our result. Since $1-y^2=\sin 2x$, the argument becomes $\frac{\sin x-\cos x}{\sqrt{\sin 2x}}$.

Dividing the numerator and denominator by $\sqrt{\sin x\cos x}$, we arrive at the elegant final form:

With patience and the right tools, even the most terrifying integrals bow to the beauty of mathematics.