Stuck on a few integration problems

[Guest]

I am stuck on a few integration problems:

1.) integral from -1 to 1 ln[x+2] dx

2.) integrate (sin[x]^4)(cos[x]^3) dx

3.) integrate dx/sqrt[(x^2)-(6x)+10]

Basically, I checked for the solutions on quickmath.com and they do not match mine. But here is how I did them:

1.) Using U-Substitution, set u = ln[x+2] and (x+2) du = dx. After doing it out, I ended up with the equation (x+2)*ln[x+2]. After plugging in the 1 and -1, I got (3ln[3] - 0)

2.) I took out a factor of cos[x] and set cos[x]^2 = [1-(sin[x]^2)]. After doing this, I used u-substitution and set u=sin[x] and du=cos[x] dx. After doing it out, I got [(1/5)(sin[x]^5)] - [(1/7)(sin[x]^7) + C

3.) For this one, I set the denominator equal to sqrt[(x^2)-(6x)+9+10-9] and factored the [(x^2)-(6x)+9] into (x-3)^2.

I then set u=x-3, and du=dx.

I plugged in u, and then did another u-sub with u=tan[theta], du=sec[theta]^2 d[theta].

After working this out, I got the equation: integral of sqrt[sec[theta]^2] d[theta].


Can someone please tell me what I did wrong with the problems above?

 

[soroban]

Hello, bubblebrains!

\(\displaystyle 1)\L\;\int^{\;\;\;1}_{-1}\ln(x\,+\,2)\;dx\)

\(\displaystyle 2)\L\;\int \sin^4x\cos^3x\,dx\)

\(\displaystyle 3)\L\;\int \frac{dx}{\sqrt{x^2\,-\.6x\,+\,10}\)


2) I took out a factor of $\displaystyle \cos x$ and set $\displaystyle \cos^2x\:=\:1\,-\,\sin^2 x$
After doing this, I used u-substitution and set $\displaystyle u\,=\,\sin x$ and $\displaystyle du\,=\,\cos x\,dx$
After doing it out, I got:$\displaystyle \,\frac{1}{5}\sin^5x\,-\,\frac{1}{7}\sin^7x\,+\,C\;$ . . . This is correct!

#1 requires integration by parts . . .

Let: $\displaystyle u\,=\,\ln(x+2)\;\;dv\,=\,dx$
Then: $\displaystyle du\,=\,\frac{dx}{x+2}\;\;\;v\,=\,x$

Then we have: \(\displaystyle \,x\cdot\ln(x+2)\,-\,\L\int\frac{x}{x+2}\,dx\)\(\displaystyle \;=\;x\cdot\ln(x+2)\,-\,\L\int\left(1\,-\,\frac{2}{x+2}\right)\,dx\)

$\displaystyle \;\;\;=\;x\cdot\ln(x+2)\,-\,x\,+\,2\ln(x+2)$


Evaluate: $\displaystyle \,[1\cdot\ln 3\,-\,1\,+\,2\cdot\ln 3]\,- \,[-1\cdot\ln 1\,-\,(-1)\,+\,2\cdot\ln 1]$

$\displaystyle \;\;\;= \;(\ln 3\,-\,1\,+\,2\cdot\ln3)\,-\,(0\,+\,1\,+\,0)\;=\;3\cdot\ln3\,-\,2$

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In #3, the denominator has: \(\displaystyle \,x^2\,-\,6x\,+\,10\:=\:x^2\,-\,6x\,+\,9\,+\,1\:=\x-3)^2\,+\,1\)

So we have: \(\displaystyle \L\,\int\frac{dx}{\sqrt{(x-3)^2\,+\,1}}\)

Let $\displaystyle x-3\,=\,\tan\theta\;\;\Rightarrow\;\;dx\,=\,\sec^2\theta\,d\theta$

Substitute: \(\displaystyle \L\,\int\frac{\sec^2\theta\,d\theta}{\sec\theta} \;= \;\int\sec\theta\,d\theta\) . . . and so on.

 

[Guest]

Great... thanks a lot for the quick reply!