\(∫_0^∞\frac{1}{1-sin(x)}dx\)

it's divergent, but i'm not sure how to actually integrate that thing inside

Guest Jun 24, 2020

#1**+3 **

**calc- integration**

\(\int \dfrac{1}{1-\sin(x)} \ dx\)

\(\begin{array}{|rclcrcl|} \hline && \mathbf{\int \dfrac{1}{1-\sin(x)} dx } \\\\ &=& \int \dfrac{1}{\Big(1-\sin(x)\Big)}*\dfrac{\Big(1+\sin(x)\Big)}{\Big(1+\sin(x)\Big)} dx \\\\ &=& \int \dfrac{1+\sin(x)}{1-\sin^2(x)} dx \quad | \quad 1-\sin^2(x)=\cos^2(x) \\\\ &=& \int \dfrac{1+\sin(x)}{ \cos^2(x)} dx \\\\ &=& \int \left( \dfrac{1}{ \cos^2(x)}+\dfrac{\sin(x)}{ \cos^2(x)}\right) dx \\\\ &=& \mathbf{\int \dfrac{1}{ \cos^2(x)} dx +\int \dfrac{\sin(x)}{ \cos^2(x)} dx } \\ \hline \end{array} \)

\(\begin{array}{|rclcrcl|} \hline && \mathbf{ \int \dfrac{1}{ \cos^2(x)} dx } \\\\ && \text{substitute:} & u&=& \tan(x) \\\\ && & u&=& \dfrac{\sin(x)}{\cos(x)} \\\\ && & du&=& \dfrac{\cos(x)\cos(x)-\sin(x)(-\sin(x)}{\cos^2(x)}\ dx \\\\ && & du&=& \dfrac{\cos^2(x)+\sin^2(x)}{\cos^2(x)} \ dx\\\\ && & du&=& \dfrac{1}{\cos^2(x)} \ dx \\\\ && & dx&=& \cos^2(x) \ du \\\\ &=& \int \dfrac{1}{ \cos^2(x)} \cos^2(x) \ du \\\\ &=& \int \ du \\\\ &=& u \\\\ \mathbf{ \int \dfrac{1}{ \cos^2(x)} dx }&=& \mathbf{ \tan(x) }+c \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline && \mathbf{\int \dfrac{\sin(x)}{ \cos^2(x)} dx } \\\\ &=& -\int \dfrac{-\sin(x)}{ \cos^2(x)} dx \\\\ && \text{substitute:} & u&=& \cos(x) \\\\ && & du&=& -\sin(x) \ dx \\\\ &=& -\int \dfrac{1}{u^2} du \\\\ &=& -\int u^{-2} du \\\\ &=& - \dfrac{u^{-2+1}}{-2+1} \\\\ &=& - \dfrac{u^{-1}}{-1} \\\\ &=& u^{-1} \\\\ &=& \dfrac{1}{u} \\\\ \mathbf{\int \dfrac{\sin(x)}{ \cos^2(x)} dx }&=& \mathbf{ \dfrac{1}{\cos(x)} +c} \\ \hline \end{array} \)

\(\begin{array}{|rclcrcl|} \hline && \mathbf{\int \dfrac{1}{1-\sin(x)} dx } \\\\ &=& \mathbf{\int \dfrac{1}{ \cos^2(x)} dx +\int \dfrac{\sin(x)}{ \cos^2(x)} dx } \\\\ &=& \mathbf{\tan(x) + \dfrac{1}{\cos(x)} +c} \\ \hline \end{array}\)

heureka Jun 25, 2020