Suggested languages for you:

Americas

Europe

Q30E

Expert-verified
Found in: Page 326

### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

# Evaluate the Equation$$\int\limits_{\frac{\pi }{4}}^{\frac{\pi }{2}} {co{t^3}x\,dx}$$

The answer is $$\frac{1}{2} - \frac{{ln\left( 2 \right)}}{2}$$

First, convert $$co{t^3}x$$ in $$cosec\,x$$ using trigonometric identities and then substituting $$cosec\,x$$ with $$u$$, solve the definite integral.

See the step by step solution

## Step 1: Converting $$co{t^3}x$$ in $$cosec\,x$$ using the trigonometric identity.

\begin{aligned}{l}\int\limits_{\frac{\pi }{4}}^{\frac{\pi }{2}} {co{t^3}x\,dx} \\ &= \int\limits_{\frac{\pi }{4}}^{\frac{\pi }{2}} {cot\,x\left( {co{t^2}x} \right)\,dx} \\ &= \int\limits_{\frac{\pi }{4}}^{\frac{\pi }{2}} {cot\,x\left( {cose{c^2}x - 1} \right)\,dx} \end{aligned}

## Step 2: Putting $$cosec\,x = u$$

Let $$cosec\,x = u$$

\begin{aligned}{l} \Rightarrow - cot x cosec x dx &= du\\ \Rightarrow cot x dx &= - \frac{{du}}{{cosec\,x}}\\ \Rightarrow cotx\,dx &= - \frac{{du}}{u}\\ \Rightarrow - \frac{{du}}{u} &= cot\,x\,dx\,\,.....(1)\end{aligned}

So,

\begin{aligned}{l} &= \int\limits_{\frac{\pi }{4}}^{\frac{\pi }{2}} {cot\,x\left( {cose{c^2}x - 1} \right)\,dx} \\ &= \int\limits_{\frac{\pi }{4}}^{\frac{\pi }{2}} {\left( {cose{c^2}x - 1} \right)cot\,x\,dx} \\ &= \int\limits_{\frac{\pi }{4}}^{\frac{\pi }{2}} {\left( {{u^2} - 1} \right)\left( { - \frac{{du}}{u}} \right)\,\,\,\,\left( {From\,\,(i)} \right)} \\ &= \int\limits_{\frac{\pi }{4}}^{\frac{\pi }{2}} {\frac{{ - \left( {{u^2} - 1} \right)}}{u}} \,du\\ &= \int\limits_{\frac{\pi }{4}}^{\frac{\pi }{2}} {\left( { - u + \frac{1}{u}} \right)} \,du\\ &= \int\limits_{\frac{\pi }{4}}^{\frac{\pi }{2}} {\left( {\frac{1}{u} - u} \right)} \,du\end{aligned}

## Step 3: Applying linearity and solving the integral.

\begin{aligned}{l} &= \int\limits_{\frac{\pi }{4}}^{\frac{\pi }{2}} {\left( {\frac{1}{u} - u} \right)} \,du\\ &= \int\limits_{\frac{\pi }{4}}^{\frac{\pi }{2}} {\left( {\frac{1}{u}} \right)} \,du - \int\limits_{\frac{\pi }{4}}^{\frac{\pi }{2}} {u\,} du\\ &= \left( {ln\left| u \right|} \right)_{\frac{\pi }{4}}^{\frac{\pi }{2}} - \left( {\frac{{{u^2}}}{2}} \right)_{\frac{\pi }{4}}^{\frac{\pi }{2}}\end{aligned}

## Step 4: Putting back $$u = cosec\,x$$ and putting the limit.

$$= \left( {ln\left| u \right|} \right)_{\frac{\pi }{4}}^{\frac{\pi }{2}} - \left( {\frac{{{u^2}}}{2}} \right)_{\frac{\pi }{4}}^{\frac{\pi }{2}}$$

$$= \left( {ln\left| {cosec\,x} \right|} \right)_{\frac{\pi }{4}}^{\frac{\pi }{2}} - \left( {\frac{{cose{c^2}x}}{2}} \right)_{\frac{\pi }{4}}^{\frac{\pi }{2}}$$

$$= \left( {ln\left| {cosec\,\left( {\frac{\pi }{2}} \right)} \right| - ln\left| {cosec\,\left( {\frac{\pi }{4}} \right)} \right|} \right) - \left( {\frac{{cose{c^2}\left( {\frac{\pi }{2}} \right)}}{2} - \frac{{cose{c^2}\left( {\frac{\pi }{4}} \right)}}{2}} \right)$$

$$= \left( {ln\left| 1 \right| - ln\left| {\sqrt 2 } \right|} \right) - \left( {\frac{{{1^2}}}{2} - \frac{{{{\left( {\sqrt 2 } \right)}^2}}}{2}} \right)$$

$$= 0 - ln\left| {\sqrt 2 } \right| - \left( {\frac{1}{2} - \frac{2}{2}} \right)$$

$$= - ln\left| {\sqrt 2 } \right| - \left( { - \frac{1}{2}} \right)$$

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

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

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

Hence, $$\int\limits_{\frac{\pi }{4}}^{\frac{\pi }{2}} {co{t^3}x\,dx} = \frac{1}{2} - \frac{{ln\left( 2 \right)}}{2}$$