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TrueTears · « **Reply #75 on:** January 02, 2009, 06:50:50 pm »

and
1. $\int sec^4(x) dx$

ed_saifa · « **Reply #76 on:** January 02, 2009, 07:21:12 pm »

$\int sec^4(x)dx$

$\int sec^4(x)dx$ = $\int sec^2(x).sec^2(x)$

$sec^2(x)= 1 + tan^2(x)$

$\int sec^2(x)(1 + tan^2(x))dx$

Let $u=tan(x)$

$\frac{du}{dx}=sec^2(x)$

$dx=\frac{du}{sec^2(x)}$

$\int sec^2(x)(1 + tan^2(x))dx$ = $\int 1 + u^2 du$

= $u + \frac{u^3}{3} + C$

= $tan(x) + \frac{tan^3(x)}{3} + C$

TrueTears · « **Reply #77 on:** January 02, 2009, 07:33:33 pm »

thanks again ed_saifa ^^, just realised it before u posted xD

just some other one ><

1. Let $U_{n} = \int^{\frac{\pi}{4}}}_{0} tan^n(x) dx$ where $n \in Z$ and $n>1$

Express $U_{n} + U_{n-2}$ in terms of $n$

/0 · « **Reply #78 on:** January 02, 2009, 07:46:22 pm »

Quote from: TrueTears on January 02, 2009, 06:50:50 pm

and
1. $\int sec^4(x) dx$

2. Let $U_{n} = \int^{\frac{\pi}{4}}}_{0} tan^n(x) dx$ where $n \in Z$ and $n>1$

Express $U_{n} + U_{n-2}$ in terms of $n$

$U_n+U_{n-2} = \int_0^{\frac{\pi}{4}} \tan^n(x)+\tan^{n-2}(x)\,dx$

$=\int_0^\frac{\pi}{4} \tan^{n-2}(x)(1+\tan^2(x))\,dx$

$=\int_0^\frac{\pi}{4} \tan^{n-2}(x)\sec^2(x)\,dx$

Let $u=\tan(x)\Rightarrow \frac{du}{dx} = \sec^2(x)$

$dx = \frac{du}{\sec^2(x)}$

$\Rightarrow U_n+U_{n-2} = \int_0^{1} tan^{n-2}(x) du=\int_0^1 u^{n-2}du$

$\therefore U_n+U_{n-2} = \left[ \frac{u^{n-1}}{n-1}\right]_0^1 = \frac{1}{n-1}$

TrueTears · « **Reply #79 on:** January 02, 2009, 07:52:19 pm »

thanks /0 and also

Let $\alpha = \frac{\pi}{2} - \theta$ . Show that $\int^{\frac{\pi}{2}}}_{0} \frac{d\theta}{1+tan\theta} = \int^{\frac{\pi}{2}}}_{0} \frac{d\alpha}{1+cot\alpha}$

I keep getting $-\int^{\frac{\pi}{2}}}_{0} \frac{d\alpha}{1+cot\alpha}$ for Q2, this is my working:

$\alpha = \frac{\pi}{2} - \theta$

$\implies \theta = \frac{\pi}{2} - \alpha$

$\implies \frac{d\alpha}{d\theta} = - 1$

$\implies d\alpha = - d\theta$

$\implies -\int^{\frac{\pi}{2}}}_{0} \frac{d\alpha}{1+tan(\frac{\pi}{2} - \alpha)} = -\int^{\frac{\pi}{2}}}_{0} \frac{d\alpha}{1+cot\alpha}$

Can anyone check if anything is wrong?

/0 · « **Reply #80 on:** January 02, 2009, 07:56:49 pm »

When you change $d\theta$ to $d\alpha$ you also have to change the terminals on the integral to:

$-\int_{\frac{\pi}{2}-0}^{\frac{\pi}{2}-\frac{\pi}{2}} \frac{d\alpha}{1+\cot{\alpha}}$

This will cancel out the negative sign.

(since $\alpha = \frac{\pi}{2} - \theta$ )

TrueTears · « **Reply #81 on:** January 02, 2009, 08:11:48 pm »

how u get the $-\int_{\frac{\pi}{2}-0}^{\frac{\pi}{2}-\frac{\pi}{2}}$ ?

/0 · « **Reply #82 on:** January 02, 2009, 08:19:26 pm »

The $\frac{\pi}{2}$ and $0$ terminals are valid only with respect to $\theta$ . If you transform $\theta$ into another variable, you also have to transform the terminals to fit the new variable. Transforming the terminals according to $\alpha = \frac{\pi}{2} - \theta$ assign new terminals to fit the new variable and ensures the integral is still being taken over the same area. Page 270 has some information on it.

TrueTears · « **Reply #83 on:** January 02, 2009, 08:39:01 pm »

thanks /0. xD

TrueTears · « **Reply #84 on:** January 02, 2009, 08:45:58 pm »

If $F'(x) = f(x)$ , then an antiderivative of $3f(3-2x)$ is?

/0 · « **Reply #85 on:** January 02, 2009, 08:54:48 pm »

$\int 3f(3-2x)dx = 3 \int f(3-2x)dx$

$u = 3-2x \Rightarrow \frac{du}{dx} = -2$

$\Rightarrow 3 \int f(3-2x)dx = 3 \int f(u)\frac{du}{-2}=\frac{-3}{2} \int f(u)du = \frac{-3}{2}F(u)+C=-\frac{3}{2}F(3-2x)+C$

$\therefore$ An antiderivative is $-\frac{3}{2}F(3-2x)$

TrueTears · « **Reply #86 on:** January 02, 2009, 08:58:06 pm »

yeap thx. Hey can u take a look at Q 27 b) and Q 28 c) on pg 281, /0 xD

brendan · « **Reply #87 on:** January 02, 2009, 09:00:03 pm »

Quote from: ed_saifa on January 02, 2009, 07:21:12 pm

Let $u=tan(x)$

$\frac{du}{dx}=tan(x)$

thats not right...

$\frac{d(tan(x))}{dx}=sec^2(x)$

/0 · « **Reply #88 on:** January 02, 2009, 09:52:49 pm »

Quote from: TrueTears on January 02, 2009, 08:58:06 pm

yeap thx. Hey can u take a look at Q 27 b) and Q 28 c) on pg 281, /0 xD

lol

27b)

You have $U_n+U_{n-2}$ . If you find $U_2=\int_0^\frac{\pi}{4} \tan^2(x)dx$ then you can find $U_4$ , then $U_6$ .

28c)

$\int_0^\frac{\pi}{2} \frac{d\theta}{1+\tan{\theta}}+\int_0^\frac{\pi}{2} \frac{d\theta}{1+\cot{\theta}}=\int_0^\frac{\pi}{2} \left(\frac{1}{1+\tan\theta}+\frac{1}{1+\cot\theta}\right)d\theta=\int_0^\frac{\pi}{2}d\theta=\frac{\pi}{2}$

Let $\int_0^\frac{\pi}{2} \frac{d\theta}{1+\tan\theta} = \int_0^\frac{\pi}{2} \frac{d\phi}{1+\cot\phi}=x$

If $\int_0^\frac{\pi}{2} \frac{d\phi}{1+\cot\phi}=x$ ,

we can say... for instance, that $\int_0^\frac{\pi}{2} \frac{da}{1+\cot{a}}=x$ or $\int_0^\frac{\pi}{2} \frac{dp}{1+\cot{p}}=x$ , simply by changing variables.

Hence it follows that $\int_0^\frac{\pi}{2} \frac{d\theta}{1+\cot{\theta}}=x$ .

Therefore, $\int_0^\frac{\pi}{2} \frac{d\theta}{1+\tan{\theta}}=\int_0^\frac{\pi}{2} \frac{d\theta}{1+\cot{\theta}}$

So... $\int_0^\frac{\pi}{2} \frac{d\theta}{1+\tan{\theta}}=\int_0^\frac{\pi}{2} \frac{d\theta}{1+\cot{\theta}}= \frac{\pi}{4}$

ed_saifa · « **Reply #89 on:** January 02, 2009, 10:30:45 pm »

Quote from: Brendan on January 02, 2009, 09:00:03 pm

Quote from: ed_saifa on January 02, 2009, 07:21:12 pm
Let $u=tan(x)$

$\frac{du}{dx}=tan(x)$

thats not right...

$\frac{d(tan(x))}{dx}=sec^2(x)$

Whoops, I'll change it.

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