Question

evaluate the integral.
int\tan12xsec^{2}12x dx
int\tan12xsec^{2}12x dx=square

Explanation:

Step1: Use substitution

Let $u = \tan12x$. Then $du=12\sec^{2}12x dx$, so $\sec^{2}12x dx=\frac{1}{12}du$.

Step2: Rewrite the integral

The original integral $\int\tan12x\sec^{2}12x dx$ becomes $\int u\times\frac{1}{12}du=\frac{1}{12}\int udu$.

Step3: Integrate with respect to u

We know that $\int udu=\frac{u^{2}}{2}+C$. So $\frac{1}{12}\int udu=\frac{1}{12}\times\frac{u^{2}}{2}+C=\frac{u^{2}}{24}+C$.

Step4: Substitute back u

Substituting $u = \tan12x$ back, we get $\frac{\tan^{2}12x}{24}+C$.

Answer:

$\frac{\tan^{2}12x}{24}+C$