Question

\int _{-1}^{-\frac{1}{2}}t^{-2}sin^2\left(1+\frac{1}{t}\right)\:dt

Answer 1

$\boxed{\begin{array}{l}\displaystyle\sf\int_{-1}^{-\frac{1}{2}}t^{-2} sen^2\bigg(1+\dfrac{1}{t}\bigg)~dt=-\int_{-1}^{-\frac{1}{2}}-t^{-2}sen^2\bigg(1+t^{-1}\bigg)dt\\\underline{\boldsymbol{fac_{\!\!,}a}}\\\sf u=1+t^{-1}\\\sf quando~t=-1\implies u=0\\\sf quando~t=-\dfrac{1}{2}\implies u=-1\\\sf du=-t^{-2}dt\\\underline{\rm substituindo~temos:}\end{array}}$

$\boxed{\begin{array}{l}\displaystyle\sf-\int_{-1}^{-\frac{1}{2}}-t^{-2} sen^2\bigg(1+\dfrac{1}{t}\bigg)dt=-\int_0^{-1}sen^2(u)du=\int_{-1}^0sen ^2(u)~du\\\sf lembrando~que~sen^2(u)=\dfrac{1}{2}\cdot[1-cos(2u)]~temos:\\\displaystyle\sf\dfrac{1}{2}\int_{-1}^0 [1-cos(2u)]du=\dfrac{1}{2}\bigg[u-\dfrac{1}{2}sen(2u)\bigg]_{-1}^0\\\sf \dfrac{1}{2}\bigg[0-\dfrac{1}{2}sen(2\cdot0)-\bigg(-1-\dfrac{1}{2}sen(2\cdot(-1))\bigg)\bigg]\end{array}}$

$\boxed{\begin{array}{l}\sf\dfrac{1}{2}\bigg[0-\bigg(-1+\dfrac{1}{2}sen(2)\bigg)\bigg]=\dfrac{1}{2}\bigg[1-\dfrac{1}{2}sen(2)\bigg]\\\sf\dfrac{1}{2}-\dfrac{1}{4}sen(2)\end{array}}$

$\Large\boxed{\begin{array}{l}\displaystyle\sf\int_{-1}^{0}t^{-2}sen^2\bigg(1+\dfrac{1}{t}\bigg)dt=\dfrac{1}{2}-\dfrac{1}{4}sen(2)\end{array}}$