Question

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1. - / 2.08 points

evaluate the integral. (remember the constant of integration.)
int\tan^{2}(x)cos^{3}(x)dx

Explanation:

Step1: Rewrite $\tan^{2}(x)$

Since $\tan(x)=\frac{\sin(x)}{\cos(x)}$, then $\tan^{2}(x)=\frac{\sin^{2}(x)}{\cos^{2}(x)}$. So the integral $\int\tan^{2}(x)\cos^{3}(x)dx=\int\frac{\sin^{2}(x)}{\cos^{2}(x)}\cdot\cos^{3}(x)dx=\int\sin^{2}(x)\cos(x)dx$.

Step2: Use substitution

Let $u = \sin(x)$, then $du=\cos(x)dx$. The integral $\int\sin^{2}(x)\cos(x)dx$ becomes $\int u^{2}du$.

Step3: Integrate $u^{2}$

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

Step4: Substitute back $u=\sin(x)$

Substituting $u = \sin(x)$ back into the result, we get $\frac{\sin^{3}(x)}{3}+C$.

Answer:

$\frac{\sin^{3}(x)}{3}+C$