# How do you evaluate the integral int cscx-sinx from [pi/4, pi/2]?

Oct 29, 2017

$\ln \left(1 + {2}^{\frac{1}{2}}\right) - {2}^{- \frac{1}{2}}$

#### Explanation:

A possible way of computing this integral ${\int}_{\frac{\pi}{4}}^{\frac{\pi}{2}} \csc x - \sin x \mathrm{dx}$, is by simplifying to give; ${\int}_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cos x \cot x \mathrm{dx}$ But a more interesting method is via individually finding $\int \csc x \mathrm{dx}$

let $u$ = $\cot x$
then $\mathrm{du} = - {\csc}^{2} x \mathrm{dx}$

Via quotient rule;
$\cot x$ = $\cos \frac{x}{\sin} x$
hence $\frac{d}{\mathrm{dx}} \left(\cot x\right)$ = $\frac{\left(\sin x\right) \left(- \sin x\right) - \left(\cos x\right) \left(\cos x\right)}{\sin} ^ 2 x$
Hence = $- {\csc}^{2} x$

Hence $\frac{- \mathrm{du}}{\csc} x = \csc x \mathrm{dx}$

Hence $\int \csc x \mathrm{dx}$ becomes $- \int \frac{\mathrm{du}}{\csc} x$

Considering $1 + {\cot}^{2} x = {\csc}^{2} x$

Hence if $u$ = $\cot x$ then $\csc x$ = ${\left(1 + {u}^{2}\right)}^{\frac{1}{2}}$

Hence $- \int \frac{\mathrm{du}}{\csc} x$ becomes $- \int \frac{\mathrm{du}}{1 + {u}^{2}} ^ \left(\frac{1}{2}\right)$

Then make a new substitution of $u$ = $\sinh \theta$
Hence $\mathrm{du}$ = $\cosh \theta d \theta$

Now by cosindering ${\cosh}^{2} \theta - {\sinh}^{2} \theta = 1$

$- \int \frac{\mathrm{du}}{1 + {u}^{2}} ^ \left(\frac{1}{2}\right)$ becomes $- \int d \theta$

= $- \theta + c$

As $u$ = $\sinh \theta$ then $\theta$ = $\arcsin h \left(u\right)$

Hence $- \arcsin h \left(u\right) + c$

Hence as $u$ = $\cot \theta$

Hence $\int \csc x \mathrm{dx}$ = $c - {\sinh}^{-} 1 \left(\cot x\right)$

Hence $\int \csc x - \sin x \mathrm{dx}$ = $\cos x - {\sinh}^{-} 1 \left(\cot x\right) + c$

Hence ${\int}_{\frac{\pi}{4}}^{\frac{\pi}{2}} \csc x - \sin x \mathrm{dx}$ = | $\cos x - {\sinh}^{-} 1 \left(\cot x\right)$ | from $\frac{\pi}{4}$ to $\frac{\pi}{2}$

So hence by using ${\sinh}^{-} 1 \left(x\right) = \ln \left(x + {\left(1 + {x}^{2}\right)}^{\frac{1}{2}}\right)$, We can simply evaluate to give;

$\ln \left(1 + {2}^{\frac{1}{2}}\right) - {2}^{- \frac{1}{2}}$