1. ## definite intergral

Find the exact value of $\int^{\frac{\pi}{2}}_{\frac{\pi}{6}}\frac{cot^3x}{ cosecx}dx$

2. ## Re: definite intergral

$\int^{\frac{\pi}{2}}_{\frac{\pi}{6}}\frac{cot^3x}{ cosecx}dx=\int^{\frac{\pi}{2}}_{\frac{\pi}{6}}\fra c{sinx}{tanx(sec^2x-1)}dx$

$=\int^{\frac{\pi}{2}}_{\frac{\pi}{6}}\frac{1}{\fra c{sec^2x-1}{cosx}}dx$

$=\int^{\frac{\pi}{2}}_{\frac{\pi}{6}}\frac{1}{\fra c{1}{cos^3x}-secx}dx$

3. ## Re: definite intergral

\displaystyle \begin{align*} \int_{\frac{\pi}{6}}^{\frac{\pi}{2}}{\frac{\cot^3{ x}}{\csc{x}}\,dx} &= \int_{\frac{\pi}{6}}^{\frac{\pi}{2}}{\frac{\frac{\ cos^3{x}}{\sin^3{x}}}{\frac{1}{\sin{x}}}\,dx} \\ &= \int_{\frac{\pi}{6}}^{\frac{\pi}{2}}{\frac{\cos^3{ x}}{\sin^2{x}}\,dx} \\ &= \int_{\frac{\pi}{6}}^{\frac{\pi}{2}}{\frac{\cos{x} (1 - \sin^2{x})}{\sin^2{x}}\,dx} \end{align*}

Now make the substitution $\displaystyle u = \sin{x} \implies du = \cos{x}\,dx$. Also note that when $\displaystyle x = \frac{\pi}{6}, u = \frac{1}{2}$ and when $\displaystyle x = \frac{\pi}{2}, u = 1$. The integral becomes

\displaystyle \begin{align*} \int_{\frac{1}{2}}^1{\frac{1 - u^2}{u^2}\,du} &= \int_{\frac{1}{2}}^1{u^{-2} - 1\,du} \\ &= \left[\frac{u^{-1}}{-1} - u\right]_{\frac{1}{2}}^1 \\ &= \left[-\frac{1}{u} - u\right]_{\frac{1}{2}}^1 \\ &= \left(-\frac{1}{1} - 1\right) - \left(- \frac{1}{\frac{1}{2}} - \frac{1}{2}\right) \\ &= \left(-1 - 1\right) - \left(-2 - \frac{1}{2}\right) \\ &= -2 - \left(-\frac{5}{2}\right) \\ &= \frac{1}{2} \end{align*}

4. ## Re: definite intergral

Originally Posted by Prove It
\displaystyle \begin{align*} \int_{\frac{\pi}{6}}^{\frac{\pi}{2}}{\frac{\cot^3{ x}}{\csc{x}}\,dx} &= \int_{\frac{\pi}{6}}^{\frac{\pi}{2}}{\frac{\frac{\ cos^3{x}}{\sin^3{x}}}{\frac{1}{\sin{x}}}\,dx} \\ &= \int_{\frac{\pi}{6}}^{\frac{\pi}{2}}{\frac{\cos^3{ x}}{\sin^2{x}}\,dx} \\ &= \int_{\frac{\pi}{6}}^{\frac{\pi}{2}}{\frac{\cos{x} (1 - \sin^2{x})}{\sin^2{x}}\,dx} \end{align*}

Now make the substitution $\displaystyle u = \sin{x} \implies du = \cos{x}\,dx$. Also note that when $\displaystyle x = \frac{\pi}{6}, u = \frac{1}{2}$ and when $\displaystyle x = \frac{\pi}{2}, u = 1$. The integral becomes

\displaystyle \begin{align*} \int_{\frac{1}{2}}^1{\frac{1 - u^2}{u^2}\,du} &= \int_{\frac{1}{2}}^1{u^{-2} - 1\,du} \\ &= \left[\frac{u^{-1}}{-1} - u\right]_{\frac{1}{2}}^1 \\ &= \left[-\frac{1}{u} - u\right]_{\frac{1}{2}}^1 \\ &= \left(-\frac{1}{1} - 1\right) - \left(- \frac{1}{\frac{1}{2}} - \frac{1}{2}\right) \\ &= \left(-1 - 1\right) - \left(-2 - \frac{1}{2}\right) \\ &= -2 - \left(-\frac{5}{2}\right) \\ &= \frac{1}{2} \end{align*}
Thank you for your help but there is something i need to clarify. I remember my teacher telling me that all substitution questions comes with a given substitute. This question, however, does not come with one. How then did you realise that this question requires a substitution? And would it be appropriate to do substitution for a question that doesn't come with a given substitute?

5. ## Re: definite intergral

Originally Posted by Punch
Thank you for your help but there is something i need to clarify. I remember my teacher telling me that all substitution questions comes with a given substitute. This question, however, does not come with one. How then did you realise that this question requires a substitution? And would it be appropriate to do substitution for a question that doesn't come with a given substitute?
The point of substitution is to be able to find an "inner function" multiplied by the inner function's derivative. With trigonometric functions, I tend to convert them to sines and cosines if possible, because they are each others derivatives (up to a constant multiple) and it is easy to use identities to get the function to a function of sine or a function of cosine only (except for the multiple of the other to be used as the derivative). In this case, I had to convert it to a function of sine (the "inner function"), multiplied by a single cosine (the derivative of the inner function).