Evaluate the integral: I = integral cos2x cot³x dx

Evaluate the integral: $$ I = \int \cos2x\,\cot^3x\,dx. $$

Step 1 · Rewrite and split

Using $\cos2x=\cos^2x-\sin^2x$ and $\cot^3x=\cos^3x/\sin^3x$:

$$ \cos2x\cot^3x = \left(\cos^2x-\sin^2x\right)\frac{\cos^3x}{\sin^3x} = \frac{\cos^5x}{\sin^3x} - \frac{\cos^3x}{\sin x}. $$

So the integral splits as:

$$ I = \int \frac{\cos^5x}{\sin^3x}\,dx - \int \frac{\cos^3x}{\sin x}\,dx. $$

Step 2 · Evaluate the first piece

Write $\cos^5x=(1-\sin^2x)^2\cos x$, and let $u=\sin x$, $du=\cos x\,dx$:

$$ \int \frac{\cos^5x}{\sin^3x}\,dx = \int \frac{(1-u^2)^2}{u^3}\,du = \int \left(u^{-3}-2u^{-1}+u\right) du = -\frac{1}{2u^2} - 2\ln|u| + \frac{u^2}{2}. $$

Substituting back $u=\sin x$:

$$ \int \frac{\cos^5x}{\sin^3x}\,dx = -\frac{1}{2\sin^2x} - 2\ln|\sin x| + \frac{\sin^2x}{2}. $$

Step 3 · Evaluate the second piece

Write $\cos^3x=(1-\sin^2x)\cos x$, and let $u=\sin x$, $du=\cos x\,dx$:

$$ \int \frac{\cos^3x}{\sin x}\,dx = \int \frac{1-u^2}{u}\,du = \int \left(\frac1u - u\right) du = \ln|u| - \frac{u^2}{2}. $$

Substituting back $u=\sin x$:

$$ \int \frac{\cos^3x}{\sin x}\,dx = \ln|\sin x| - \frac{\sin^2x}{2}. $$

Step 4 · Combine the results

Subtracting the second piece from the first:

$$ I = \left(-\frac{1}{2\sin^2x} - 2\ln|\sin x| + \frac{\sin^2x}{2}\right) - \left(\ln|\sin x| - \frac{\sin^2x}{2}\right). $$

Simplifying:

$$ I = -\frac{1}{2\sin^2x} - 3\ln|\sin x| + \sin^2x + C. $$

Summary Table

Additional Notes

• The domain excludes $x=k\pi$ for integer $k$, where $\sin x=0$.
• A common pitfall in this problem is losing track of the $\sin^2x$ terms that arise from each piece — they partially cancel (the two $\tfrac{\sin^2x}{2}$ terms combine to $\sin^2x$ rather than canceling to zero, since the second piece is subtracted), and dropping them entirely leaves an incomplete antiderivative that fails to differentiate back to the original integrand.
• The result can be checked by differentiating $-\tfrac{1}{2\sin^2x}-3\ln|\sin x|+\sin^2x$: this gives $\dfrac{\cos x}{\sin^3x}-\dfrac{3\cos x}{\sin x}+2\sin x\cos x$, which simplifies to $\cos2x\cot^3x$ after combining over a common denominator.