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  1. #1
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    definite intergral

    Find the exact value of \int^{\frac{\pi}{2}}_{\frac{\pi}{6}}\frac{cot^3x}{  cosecx}dx
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    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
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  3. #3
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    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*}
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    Re: definite intergral

    Quote Originally Posted by Prove It View Post
    \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?
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    Re: definite intergral

    Quote Originally Posted by Punch View Post
    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).
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