evaluate the integral using the indicated trigonometric substitution.\n int \frac{3x^{3}}{sqrt{x^{2}+25}}…

Explanation:

Step1: Find $dx$

Given $x = 5\tan(\theta)$, then $dx=5\sec^{2}(\theta)d\theta$.

Step2: Substitute $x$ and $dx$ into the integral

Substitute $x = 5\tan(\theta)$ and $dx = 5\sec^{2}(\theta)d\theta$ into $\int\frac{3x^{3}}{\sqrt{x^{2}+25}}dx$. First, $x^{2}+25 = 25\tan^{2}(\theta)+25=25(\tan^{2}(\theta)+1)=25\sec^{2}(\theta)$, so $\sqrt{x^{2}+25}=5\sec(\theta)$. And $x^{3}=125\tan^{3}(\theta)$. The integral becomes $\int\frac{3\times125\tan^{3}(\theta)}{5\sec(\theta)}\times5\sec^{2}(\theta)d\theta=\int375\tan^{3}(\theta)\sec(\theta)d\theta$.

Step3: Rewrite $\tan^{3}(\theta)$

$\tan^{3}(\theta)=\tan^{2}(\theta)\tan(\theta)=(\sec^{2}(\theta)-1)\tan(\theta)$. So the integral is $\int375(\sec^{2}(\theta)-1)\tan(\theta)\sec(\theta)d\theta$. Let $u = \sec(\theta)$, then $du=\sec(\theta)\tan(\theta)d\theta$. The integral becomes $375\int(u^{2}-1)du$.

Step4: Integrate with respect to $u$

$375\int(u^{2}-1)du=375(\frac{u^{3}}{3}-u)+C = 125u^{3}-375u + C$.

Step5: Substitute back $u=\sec(\theta)$

Since $x = 5\tan(\theta)$, we know that $\sec(\theta)=\frac{\sqrt{x^{2}+25}}{5}$. So the result is $125\times(\frac{\sqrt{x^{2}+25}}{5})^{3}-375\times\frac{\sqrt{x^{2}+25}}{5}+C=\frac{1}{5}(x^{2}+25)^{\frac{3}{2}}-75\sqrt{x^{2}+25}+C$.

Answer:

$\frac{1}{5}(x^{2}+25)^{\frac{3}{2}}-75\sqrt{x^{2}+25}+C$