Evaluate the integral: I = integral dx/(sin²x cos x)

Evaluate the integral: $$ I = \int \frac{dx}{\sin^2x\,\cos x}. $$

Step 1 · Set up the substitution

Multiplying numerator and denominator by $\cos x$:

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

Let $u=\sin x$, so $du=\cos x\,dx$:

$$ I = \int \frac{du}{u^2(1-u^2)}. $$

Step 2 · Partial fraction decomposition

Write $1-u^2=(1-u)(1+u)$ and decompose:

$$ \frac{1}{u^2(1-u)(1+u)} = \frac{A}{u} + \frac{B}{u^2} + \frac{C}{1-u} + \frac{D}{1+u}. $$

Clearing denominators and matching coefficients (or evaluating at $u=0,\,1,\,-1$) gives $A=0$, $B=1$, $C=\tfrac12$, $D=\tfrac12$. Thus:

$$ \frac{1}{u^2(1-u^2)} = \frac{1}{u^2} + \frac{1/2}{1-u} + \frac{1/2}{1+u}. $$

Step 3 · Integrate each term

$$ I = \int \left(\frac{1}{u^2} + \frac{1/2}{1-u} + \frac{1/2}{1+u}\right) du = -\frac1u - \frac12\ln|1-u| + \frac12\ln|1+u| + C. $$

Step 4 · Simplify and substitute back

Combining the logarithms:

$$ I = -\frac1u + \frac12\ln\left|\frac{1+u}{1-u}\right| + C. $$

Substituting back $u=\sin x$:

$$ I = -\frac{1}{\sin x} + \frac12\ln\left|\frac{1+\sin x}{1-\sin x}\right| + C. $$

Summary Table

Additional Notes

• The domain excludes $x=k\pi$ and $x=\tfrac\pi2+k\pi$ for integer $k$, where $\sin x=0$ or $\cos x=0$.
• Because $u^2(1-u^2)$ has a repeated linear factor at $u=0$ and two distinct simple factors at $u=\pm1$, the decomposition needs four terms in general ($A/u$, $B/u^2$, $C/(1-u)$, $D/(1+u)$) — it is only after solving that the $A/u$ term vanishes here.
• The result can be checked by differentiating $-\tfrac{1}{\sin x}+\tfrac12\ln\left|\tfrac{1+\sin x}{1-\sin x}\right|$ and confirming it returns $\dfrac{1}{\sin^2x\cos x}$.