question 9 1 point use substitution to evaluate the definite integral given below. ∫₀^(√(π/12)) -5x…

Explanation:

Step1: Set the substitution

Let $u = 3x^{2}$. Then $du=6x dx$, and $x dx=\frac{1}{6}du$. When $x = 0$, $u=3\times0^{2}=0$. When $x=\sqrt{\frac{\pi}{12}}$, $u = 3\times(\sqrt{\frac{\pi}{12}})^{2}=\frac{\pi}{4}$. The integral $\int_{0}^{\sqrt{\frac{\pi}{12}}}- 5x\tan^{2}(3x^{2})\sec^{2}(3x^{2})dx$ becomes $\int_{0}^{\frac{\pi}{4}}-5\times\frac{1}{6}\tan^{2}(u)\sec^{2}(u)du=-\frac{5}{6}\int_{0}^{\frac{\pi}{4}}\tan^{2}(u)\sec^{2}(u)du$.

Step2: Use another substitution

Let $t=\tan(u)$. Then $dt=\sec^{2}(u)du$. When $u = 0$, $t=\tan(0)=0$. When $u=\frac{\pi}{4}$, $t=\tan(\frac{\pi}{4}) = 1$. The integral $-\frac{5}{6}\int_{0}^{\frac{\pi}{4}}\tan^{2}(u)\sec^{2}(u)du$ becomes $-\frac{5}{6}\int_{0}^{1}t^{2}dt$.

Step3: Evaluate the integral

We know that $\int t^{n}dt=\frac{t^{n + 1}}{n+1}+C$ for $n\neq - 1$. So, $-\frac{5}{6}\int_{0}^{1}t^{2}dt=-\frac{5}{6}\left[\frac{t^{3}}{3}\right]_{0}^{1}$. $=-\frac{5}{6}\times\left(\frac{1^{3}}{3}-\frac{0^{3}}{3}\right)=-\frac{5}{18}$.

Answer:

$-\frac{5}{18}$