1. ## Integration by substitution

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

2. 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

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

Can you take it from here ?

3. Merci

I can see where I was going wrong

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

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

5. Thanks again