New Notes Blog: `int 1/(cos(theta) -1) d theta` Find the indefinite integral

Sunday, 4 October 2015

$\displaystyle\int\frac{{1}}{{{\cos{{\left(\theta\right)}}}-{1}}}{d}\theta$ Find the indefinite integral

Explanation for $\displaystyle{\cos{{\left({x}\right)}}}=\frac{{{1}-{{u}}^{{2}}}}{{{1}+{{u}}^{{2}}}}$

before that , we know

$\displaystyle{\cos{{\left({2}{x}\right)}}}={{\cos}}^{{2}}{\left({x}\right)}-{{\sin}}^{{2}}{\left({x}\right)}$

as $\displaystyle{{\cos}}^{{2}}{\left({x}\right)}$ can be written as $\displaystyle\frac{{1}}{{{{\sec}}^{{2}}{\left({x}\right)}}}$

and we can show $\displaystyle{{\sin}}^{{2}}{\left({x}\right)}=\frac{{\frac{{{{\sin}}^{{2}}{\left({x}\right)}}}{{{{\cos}}^{{2}}{\left({x}\right)}}}}}{{\frac{{1}}{{{{\cos}}^{{2}}{\left({x}\right)}}}}}$

= $\displaystyle{{\tan}}^{{2}}\frac{{{x}}}{{{\sec}}^{{2}}}{x}$

so now ,

$\displaystyle{\cos{{\left({2}{x}\right)}}}={{\cos}}^{{2}}{\left({x}\right)}-{{\sin}}^{{2}}{\left({x}\right)}$

= $\displaystyle{\left(\frac{{1}}{{{\sec}}^{{2}}}{\left({x}\right)}\right)}-{\left({{\tan}}^{{2}}\frac{{{x}}}{{{\sec}}^{{2}}}{\left({x}\right)}\right)}$

= $\displaystyle\frac{{{1}-{{\tan}}^{{2}}{\left({x}\right)}}}{{{{\sec}}^{{2}}{\left({x}\right)}}}$

but $\displaystyle{{\sec}}^{{2}}{\left({x}\right)}={1}+{{\tan}}^{{2}}{\left({x}\right)}$ ,as its an identity

so,

= $\displaystyle\frac{{{1}-{{\tan}}^{{2}}{\left({x}\right)}}}{{{{\sec}}^{{2}}{\left({x}\right)}}}$

= $\displaystyle\frac{{{1}-{{\tan}}^{{2}}{\left({x}\right)}}}{{{1}+{\left({{\tan}}^{{2}}{\left({x}\right)}\right)}}}$

so ,

$\displaystyle{\cos{{\left({2}{x}\right)}}}=\frac{{{1}-{{\tan}}^{{2}}{\left({x}\right)}}}{{{1}+{\left({{\tan}}^{{2}}{\left({x}\right)}\right)}}}$

so,

then

$\displaystyle{\cos{{\left({x}\right)}}}=\frac{{{1}-{{\tan}}^{{2}}{\left(\frac{{x}}{{2}}\right)}}}{{{1}+{\left({{\tan}}^{{2}}{\left(\frac{{x}}{{2}}\right)}\right)}}}$

as before we told to assume that $\displaystyle{u}={\tan{{\left(\frac{{x}}{{2}}\right)}}},$

so,

$\displaystyle{\cos{{\left({x}\right)}}}=\frac{{{1}-{{u}}^{{2}}}}{{{1}+{{u}}^{{2}}}}$

Hope this helps to understand better

New Notes Blog

Sunday, 4 October 2015

$\displaystyle\int\frac{{1}}{{{\cos{{\left(\theta\right)}}}-{1}}}{d}\theta$ Find the indefinite integral

No comments:

Post a Comment

In "By the Waters of Babylon," under the leadership of John, what do you think the Hill People will do with their society?

Sunday, 4 October 2015

Find the indefinite integral

No comments:

Post a Comment

In &quot;By the Waters of Babylon,&quot; under the leadership of John, what do you think the Hill People will do with their society?

$\displaystyle\int\frac{{1}}{{{\cos{{\left(\theta\right)}}}-{1}}}{d}\theta$ Find the indefinite integral

In "By the Waters of Babylon," under the leadership of John, what do you think the Hill People will do with their society?