# Math Help - integration

1. ## integration

integrate the following using trigonometric identities and/or substitutions:

﻿﻿﻿x^2 + 1 / sqrt (9 - x^2) dx

x^3 / sqrt (x^2 + 4) dx

(tan x)^5 * (sec x)^6 dx

(cos x)^5 / sqrt (sin x) dx

with the first two i started out confidently but ended up wasting half an hour trying o figure them out .... and the last two are just impossible ...

2. Originally Posted by razorfever
(tan x)^5 * (sec x)^6 dx
Note that $\int\tan^5x\sec^6x\,dx=\int\tan^5x\left(\sec^2x\ri ght)^2\sec^2x\,dx=\int\tan^5x\left(\tan^2x+1\right )^2\sec^2x\,dx$ $=\int\left(\tan^9x+2\tan^7x+\tan^5x\right)\sec^2x\ ,dx$

Now apply the substitution $z=\tan x$

(cos x)^5 / sqrt (sin x) dx
Note that $\int\frac{\cos^5x\,dx}{\sqrt{\sin x}}=\int\frac{\left(\cos^2x\right)^2\cos x\,dx}{\sqrt{\sin x}}=\int\frac{\left(1-\sin^2x\right)^2\cos x\,dx}{\sqrt{\sin x}}$

Not apply the substitution $z=\sin x$

Can you try these two now?

3. Hello, razorfever!

$\int \frac{x^2+1}{\sqrt{9-x^2}}\,dx$

Let: $x \:=\:3\sin\theta \quad\Rightarrow\quad dx \:=\:3\cos\theta\,d\theta \quad\Rightarrow\quad \sqrt{9=x^2} \:=\:3\cos\theta$

Substitute: . $\int \frac{9\sin^2\!\theta +1}{3\cos\theta}(3\cos\theta\,d\theta) \;=\;\int\left(9\sin^2\!\theta + 1\right)\,d\theta$

Can you finish it?

$\int \frac{x^3}{\sqrt{x^2 + 4}}\,dx$

Let: $x \:=\: 2\tan\theta \quad\Rightarrow\quad dx \:=\:2\sec^2\!\theta \quad\Rightarrow\quad \sqrt{x^2+4} \:=\:2\sec\theta$

Substitute: . $\int\frac{8\tan^3\!\theta}{2\sec\theta}\,\left(2\s ec^2\!\theta\,d\theta\right) \;=\;8\int\sec\theta\tan^3\!\theta\,d\theta$

. . $= \;8\int\tan^2\!\theta(\sec\theta\tan\theta\,d\thet a) \;=\;8\int(\sec^2\!\theta - 1)(\sec\theta\tan\theta\,d\theta)$

Now let: . $u \:=\:\sec\theta$

Got it?

4. For the first one, use trig sub: $x = 3\sin \theta \ \Rightarrow \ dx = 3\cos \theta$

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For the second one, use the sub: $u = x^2 + 4 \ \Rightarrow \frac{du}{2} = x \ dx$

and note that you can rearrange your sub to get: $x^2 = u - 4$

This will give you: $\int \frac{x^2 \cdot x \ dx}{\sqrt{x^2 + 4}} = \frac{1}{2}\int \frac{u-4}{u^{\frac{1}{2}}} \ du = \frac{1}{2}\int (u^\frac{1}{2} - 4u^{-\frac{1}{2}} ) du$

5. thanks