Q.Integrate the following function: $\frac{1}{\sin x \cos^3 x}$

The key idea is to rewrite the integrand in terms of $\tan x$ so that a U Substitution works cleanly.

First, split the fraction:

Now write $\sec^4 x = \sec^2 x \cdot \sec^2 x = (1 + \tan^2 x) \sec^2 x$.

Let $u = \tan x$, so $du = \sec^2 x \, dx$. The integral becomes: …

Why This Approach Works

When you see a product of $\sin x$ and powers of $\cos x$ in the denominator, your first instinct might be to try splitting it into partial fractions or using a Weierstrass substitution. But there's a cleaner path.

Notice that $\frac{1}{\sin x \cos^3 x}$ can be rewritten by dividing numerator and denominator by $\cos^4 x$. Why $\cos^4 x$? Because $\frac{1}{\cos^4 x} = \sec^4 x$, and $\frac{\sin x}{\cos x} = \tan x$. This transforms the integrand into something involving $\tan x$ and $\sec^2 x$ — and $\sec^2 x$ is the derivative of $\tan x$. That's the signal for a $u$-substitution.

Step-by-Step Solution

1. Rewrite the integrand

Start with:

Divide numerator and denominator by $\cos^4 x$:

2. Express $\sec^4 x$ in terms of $\sec^2 x$

Recall that $\sec^2 x = 1 + \tan^2 x$. So:

Thus:

3. Perform the substitution

Let $u = \tan x$. Then $du = \sec^2 x \, dx$. The integral becomes:

4. Simplify and integrate

Split the fraction: …

Method: Insert $\sin^2+\cos^2=1$ to split $\dfrac{1}{\sin x\cos^3 x}$

For a reciprocal of a $\sin$–$\cos$ product, replace the constant $1$ in the numerator with $\sin^2 x+\cos^2 x$ so the fraction breaks into pieces that each integrate.

Steps

Step 1: Rewrite the numerator.

Step 2: Split and simplify each fraction.

The first piece is $\tan x\sec^2 x$; the second is $\dfrac{1}{\sin x\cos x}$.

Step 3: Integrate each standard form. …

Common Mistakes

Mistake 1: Not thinking to insert $\sin^2 x+\cos^2 x$ in the numerator.

Why it's wrong: as $\dfrac{1}{\sin x\cos^3 x}$ the integral has no visible route; the identity trick is what breaks it apart. Correct approach: write $1=\sin^2 x+\cos^2 x$ so it splits into $\tan x\sec^2 x+\dfrac{1}{\sin x\cos x}$.

Mistake 2: Mishandling $\displaystyle\int\frac{dx}{\sin x\cos x}$. …