evaluate $int_{0}^{pi/33}x\tan^{2}(11x)dx$.\n$int_{0}^{pi/33}x\tan^{2}(11x)dx = square$ (type an exact…

Explanation:

Step1: Use trigonometric identity

Recall that $\tan^{2}\theta=\sec^{2}\theta - 1$. So, $x\tan^{2}(11x)=x(\sec^{2}(11x)-1)=x\sec^{2}(11x)-x$. Then $\int_{0}^{\frac{\pi}{33}}x\tan^{2}(11x)dx=\int_{0}^{\frac{\pi}{33}}(x\sec^{2}(11x)-x)dx=\int_{0}^{\frac{\pi}{33}}x\sec^{2}(11x)dx-\int_{0}^{\frac{\pi}{33}}x dx$.

Step2: Evaluate $\int_{0}^{\frac{\pi}{33}}x dx$

Use the power - rule for integration $\int x^n dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$). Here $n = 1$, so $\int_{0}^{\frac{\pi}{33}}x dx=\left[\frac{x^{2}}{2}\right]_{0}^{\frac{\pi}{33}}=\frac{(\frac{\pi}{33})^{2}}{2}-0=\frac{\pi^{2}}{2\times33^{2}}$.

Step3: Evaluate $\int_{0}^{\frac{\pi}{33}}x\sec^{2}(11x)dx$ using integration by parts

The integration - by - parts formula is $\int_{a}^{b}u dv=uv|{a}^{b}-\int{a}^{b}v du$. Let $u = x$, $dv=\sec^{2}(11x)dx$. Then $du=dx$, and $v=\frac{1}{11}\tan(11x)$. $\int_{0}^{\frac{\pi}{33}}x\sec^{2}(11x)dx=\left[\frac{x}{11}\tan(11x)\right]{0}^{\frac{\pi}{33}}-\frac{1}{11}\int{0}^{\frac{\pi}{33}}\tan(11x)dx$. First, $\left[\frac{x}{11}\tan(11x)\right]{0}^{\frac{\pi}{33}}=\frac{\frac{\pi}{33}}{11}\tan(11\times\frac{\pi}{33})-0=\frac{\pi}{33\times11}\tan(\frac{\pi}{3})=\frac{\pi\sqrt{3}}{33\times11}$. Second, $\int\tan(11x)dx=-\frac{1}{11}\ln|\cos(11x)|+C$. So, $\frac{1}{11}\int{0}^{\frac{\pi}{33}}\tan(11x)dx=\frac{1}{11}\left[-\frac{1}{11}\ln|\cos(11x)|\right]_{0}^{\frac{\pi}{33}}=\frac{-1}{121}(\ln|\cos(\frac{\pi}{3})|-\ln|\cos(0)|)=\frac{-1}{121}(\ln\frac{1}{2}-\ln1)=\frac{\ln2}{121}$.

Step4: Combine the results

$\int_{0}^{\frac{\pi}{33}}x\tan^{2}(11x)dx=\frac{\pi\sqrt{3}}{33\times11}-\frac{\ln2}{121}-\frac{\pi^{2}}{2\times33^{2}}$.

Answer:

$\frac{\pi\sqrt{3}}{363}-\frac{\ln2}{121}-\frac{\pi^{2}}{2178}$