1) $int_{\frac{pi}{2}}^{pi} ysin ydy$ 2) $int_{0}^{\frac{pi}{4}} tsec^{2}t dt$

Explanation:

Step1: Solve $\int_{\frac{\pi}{2}}^{\pi}y\sin ydy$ using integration - by - parts

Let $u = y$, $dv=\sin ydy$. Then $du = dy$, $v=-\cos y$. By the integration - by - parts formula $\int_{a}^{b}u;dv=uv|{a}^{b}-\int{a}^{b}v;du$, we have: [ \begin{align*} \int_{\frac{\pi}{2}}^{\pi}y\sin ydy&=-y\cos y|{\frac{\pi}{2}}^{\pi}+\int{\frac{\pi}{2}}^{\pi}\cos ydy\ \end{align*} ]

Step2: Evaluate $-y\cos y|{\frac{\pi}{2}}^{\pi}$ and $\int{\frac{\pi}{2}}^{\pi}\cos ydy$

First, evaluate $-y\cos y|{\frac{\pi}{2}}^{\pi}$: [ \begin{align*} -y\cos y|{\frac{\pi}{2}}^{\pi}&=-(\pi\cos\pi-\frac{\pi}{2}\cos\frac{\pi}{2})\ &=-(\pi\times(- 1)-\frac{\pi}{2}\times0)\ &=\pi \end{align*} ] Second, evaluate $\int_{\frac{\pi}{2}}^{\pi}\cos ydy=\sin y|{\frac{\pi}{2}}^{\pi}=\sin\pi-\sin\frac{\pi}{2}=0 - 1=-1$. So, $\int{\frac{\pi}{2}}^{\pi}y\sin ydy=\pi-1$.

Step3: Solve $\int_{0}^{\frac{\pi}{4}}t\sec^{2}t;dt$ using integration - by - parts

Let $u = t$, $dv=\sec^{2}t;dt$. Then $du = dt$, $v=\tan t$. By the integration - by - parts formula $\int_{a}^{b}u;dv=uv|{a}^{b}-\int{a}^{b}v;du$, we have: [ \begin{align*} \int_{0}^{\frac{\pi}{4}}t\sec^{2}t;dt&=t\tan t|{0}^{\frac{\pi}{4}}-\int{0}^{\frac{\pi}{4}}\tan t;dt \end{align*} ]

Step4: Evaluate $t\tan t|{0}^{\frac{\pi}{4}}$ and $\int{0}^{\frac{\pi}{4}}\tan t;dt$

First, evaluate $t\tan t|{0}^{\frac{\pi}{4}}$: [ \begin{align*} t\tan t|{0}^{\frac{\pi}{4}}&=\frac{\pi}{4}\tan\frac{\pi}{4}-0\times\tan0\ &=\frac{\pi}{4} \end{align*} ] Second, evaluate $\int_{0}^{\frac{\pi}{4}}\tan t;dt=-\ln|\cos t|{0}^{\frac{\pi}{4}}=-\ln\cos\frac{\pi}{4}+\ln\cos0=-\ln\frac{\sqrt{2}}{2}+0=\frac{1}{2}\ln2$. So, $\int{0}^{\frac{\pi}{4}}t\sec^{2}t;dt=\frac{\pi}{4}-\frac{1}{2}\ln2$.

Answer:

1. $\int_{\frac{\pi}{2}}^{\pi}y\sin ydy=\pi - 1$
2. $\int_{0}^{\frac{\pi}{4}}t\sec^{2}t;dt=\frac{\pi}{4}-\frac{1}{2}\ln2$