Integration by substitution

**Beard** · December 23rd 2008, 01:27 AM

I'm stuck on several questions that are driving me round the bend. If you can help with any it would be greatly appreciated.

a)Integral of: 1/sqrtx(2+sqrtx).dx = ln 16, between the values of 36 and 0 and where the substitution is x=u^2

or

b)Integral of: cos^2(t)sin(t).dt, between the values of pi/2 and pi/3 where u = cos(t)

sorry if it is hard to understand but I am new to this and don't understand how to insert the proper maths lingo and symbols.

Thanks Beard

**flyingsquirrel** · December 23rd 2008, 01:51 AM

Hi,

Originally Posted by Beard

a)Integral of: 1/sqrtx(2+sqrtx).dx = ln 16, between the values of 36 and 0 and where the substitution is x=u^2

$I=\int_0^{36}\frac{1}{\sqrt{x}\left(2+\sqrt{x}\rig ht)}\,\mathrm{d}x$

Let $x=u^2$ . We have $\frac{\mathrm{d}x}{\mathrm{d}u}=2u$ hence $\mathrm{d}x=2u\,\mathrm{d}u$ .

$I=\int_{\sqrt{0}}^{\sqrt{36}} \frac{1}{\sqrt{u^2}\left(2+\sqrt{u^2}\right)}\cdot 2u\,\mathrm{d}u$

Since $u\geq0,\, \sqrt{u^2}=|u|=u$ so

$<br /> \begin{aligned}<br /> I&=2\int_0^6 \frac{1}{u\left(2+u\right)}\cdot u\,\mathrm{d}u\\<br /> &=2\int_0^6 \frac{1}{2+u}\,\mathrm{d}u\\<br /> \end{aligned}$

Can you take it from here ?

**Beard** · December 23rd 2008, 02:01 AM

Merci

I can see where I was going wrong

**flyingsquirrel** · December 23rd 2008, 04:02 AM

Originally Posted by Beard

Merci

De rien.

Originally Posted by Beard

b)Integral of: cos^2(t)sin(t).dt, between the values of pi/2 and pi/3 where u = cos(t)

$u=\cos t$ so $\frac{\mathrm{d}u}{\mathrm{d}t}=-\sin t \implies \mathrm{d}u=-\sin(t)\,\mathrm{d}t$ .

$\int_{\pi/2}^{\pi/3}\underbrace{\cos^2(t)}_{u^2}\underbrace{\sin(t)\ ,\mathrm{d}t}_{-\mathrm{d}u}=\int_{\cos(\pi/3)}^{\cos(\pi/2)}u^2\mathrm{d}u=\ldots$

**Beard** · December 23rd 2008, 04:56 AM

Thanks again

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