Question

use any method to evaluate $int\frac{cos^{3}12x}{sin^{4}12x}dx$.
$int\frac{cos^{3}12x}{sin^{4}12x}dx = square$

Explanation:

Step1: Rewrite the integrand

We know that $\cos^{3}12x=\cos12x\cdot\cos^{2}12x=\cos12x(1 - \sin^{2}12x)$. So the integral $\int\frac{\cos^{3}12x}{\sin^{4}12x}dx=\int\frac{\cos12x(1 - \sin^{2}12x)}{\sin^{4}12x}dx$.

Step2: Use substitution

Let $u = \sin12x$, then $du=12\cos12x dx$, and $\cos12x dx=\frac{1}{12}du$. The integral becomes $\frac{1}{12}\int\frac{1 - u^{2}}{u^{4}}du=\frac{1}{12}\int(u^{- 4}-u^{-2})du$.

Step3: Integrate term - by - term

Using the power - rule for integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n
eq - 1$), we have $\frac{1}{12}(\frac{u^{-4 + 1}}{-4 + 1}-\frac{u^{-2+1}}{-2 + 1})+C=\frac{1}{12}(\frac{u^{-3}}{-3}-\frac{u^{-1}}{-1})+C$.

Step4: Simplify the result

$\frac{1}{12}(-\frac{1}{3u^{3}}+\frac{1}{u})+C=\frac{1}{12u}(1-\frac{1}{3u^{2}})+C$.

Step5: Substitute back $u=\sin12x$

The final result is $\frac{1}{12\sin12x}(1-\frac{1}{3\sin^{2}12x})+C=\frac{1}{12\sin12x}-\frac{1}{36\sin^{3}12x}+C$.

Answer:

$\frac{1}{12\sin12x}-\frac{1}{36\sin^{3}12x}+C$