Thread: Integration by substitution

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

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

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

Since $\displaystyle u\geq0,\, \sqrt{u^2}=|u|=u$ so
\displaystyle \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)
$\displaystyle u=\cos t$ so $\displaystyle \frac{\mathrm{d}u}{\mathrm{d}t}=-\sin t \implies \mathrm{d}u=-\sin(t)\,\mathrm{d}t$.

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

5. Thanks again