Evaluate int [(sqrt(x) - 1/sqrt(x))^3] dx, etc.

[Sophie]

Hello

I am getting a little confused with a few of the integrals especially as they get more complex. I have to evalute the integral on the following 2 questions.

1.

\(\displaystyle \L\int \left( {\sqrt {\text{x}} {\text{ - }}\frac{{\text{1}}}
{{\sqrt {\text{x}} }}} \right)^{\text{3}} }
\\)

I multiplied it out to get

\(\displaystyle \L\
= \frac{{2x^{\frac{5}
{2}} }}
{5} - 2x^{\frac{3}
{2}} + 6x^{\frac{1}
{2}} + 2x^{\frac{{ - 1}}
{2}} + c
\\)

But I was wondering if it would have been better to use the substitution rule, if that is the case it did not seem very simple. If substitution is the better method could someone point me in the right direction.

Question 2.

\(\displaystyle \L\int_{\frac{{ - \pi }}
{4}}^{\frac{{3\pi }}
{4}} {\left| {\cos x} \right|} {\text{dx}}
\\)

I have attempted to split it into x >0 and x<0 and added these together, but I am not confidant in my method. If someone could check and guild me in the right direction if I have made an error.

\(\displaystyle \L\int_{\frac{{ - \pi }}
{4}}^{\frac{{3\pi }}
{4}} {\left| {\cos x} \right|} {\text{dx}} \hfill \\
\int_{\frac{{ - \pi }}
{4}}^{\frac{{3\pi }}
{4}} {\left| {\cos x} \right|} {\text{dx = }}\int_{\frac{{ - \pi }}
{4}}^{\frac{\pi }
{2}} {\cos x{\text{ }}} {\text{dx + }}\int_{\frac{\pi }
{2}}^{\frac{{3\pi }}
{4}} { - \cos x{\text{ }}} {\text{dx}} \hfill \\
{\text{ = }}\left[ {{\text{sinx}}} \right]_{\frac{{ - \pi }}
{4}}^{\frac{\pi }
{2}} + \left[ {{\text{sinx}}} \right]_{\frac{\pi }
{2}}^{\frac{{3\pi }}
{4}} \hfill \\
= \frac{{\sqrt 2 }}
{2} + \frac{{\sqrt 2 }}
{2} - \frac{{\sqrt 2 }}
{2} + 1 = \frac{{\sqrt 2 }}
{2} + 1 \hfill \\\)

Thanks very much for your time, Sophie

 

[morson]

Here's something that might make 1. a little simpler:

Let u^2 = x, hence dx = 2u*du

So the integral becomes:

\(\displaystyle \int (u - \frac{1}{u}\)^3\,2udu.\ = 2 \int (u^4 - 3u^2 + 3 - \frac{1}{u^2}\)\,du.\\)

 

[soroban]

Re: Evaluate the integral

Hello, Sophie!

2) \(\displaystyle \L\int_{\frac{{-\pi }}{4}}^{\;\;\;\frac{{3\pi }}{4}} \left|\cos x\right|\,dx\)

I have attempted to split it into y >0 and y<0 and added these together.

\(\displaystyle \L\int_{\frac{{ - \pi }}{4}}^{\;\;\;\frac{3\pi }{4}} {\left| {\cos x} \right|} {\text{dx}} \;=\;

\int_{\frac{-\pi}{4}}^{\;\;\;\frac{\pi }{2}} \cos x\,dx\:-\:\int_{\frac{\pi }{2}}^{\;\;\;\frac{3\pi }{4}} \cos x\,dx\)

. . \(\displaystyle \L=\;\left[\sin x\right]\)\(\displaystyle \,\begin{array}{c}\frac{\pi}{2}\\ \\ -\frac{\pi}{4}\end{array}\L\:-\:\left[\sin x\right]\)\(\displaystyle \begin{array}{c}\frac{3\pi}{4}\\ \\ \frac{\pi}{2}\end{array}\) . . . Good!

Now be careful . . .

\(\displaystyle \L\left[\sin\left(\frac{\pi}{2}\right)\,-\,\sin\left(-\frac{\pi}{4}\right)\right]\:-\:\left[\sin\left(\frac{3\pi}{4}\right)\,-\,\sin\left(\frac{\pi}{2}\right)\right]\)

. . \(\displaystyle \L=\;\left[1\,-\,\left(-\frac{\sqrt{2}}{2}\right)\right]\:-\:\left[\frac{\sqrt{2}}{2}\,-\,1\right]\)

. . \(\displaystyle \L=\;1\,+\,\frac{\sqrt{2}}{2}\,-\,\frac{\sqrt{2}}{2}\,+\,1\)

. . \(\displaystyle \L=\;2\)

 

[Sophie]

Thanks for the input once again, Sophie