Trigonometric Substitution

[Ryo]

Hello,

First post here, I'm experienced with use of forums and have read the rules, so here goes:

I need some major help with Trigonometric substitution in my Calc 2 class. I just don't seem to ever get it right, no matter how hard I try. I learned Integration by Parts easily enough but I somehow still managed to bomb my quiz for that. So I'm pretty worried about passing this one with better than a D, since I need Calc 2 to move on in my Computer Engineering Degree.

Ok, so an example problem off webassign:

Indefinite Integral of: [5*sin^3 (x^1/2)] / (x^1/2)

Some stuff i have tried : I know I can't use integration by parts or it wouldn't benefit as neither function goes to zero. U substitution looked promising but I either messed it up or something because it didn't help, at all.

When explaining stuff to me, it would help if you explained each step, I am not stupid, but I don't always get implied meanings or steps.

 

[MarkFL]

We are given to evaluate:

\(\displaystyle \displaystyle 5\int\frac{\sin^3(\sqrt{x})}{\sqrt{x}}\,dx\)

Let's first try a u-substitution:

\(\displaystyle \displaystyle u=\sqrt{x}\,\therefore\,du=\frac{1}{2\sqrt{x}}\,dx\) and we have:

\(\displaystyle \displaystyle 10\int\sin^3(u)\,du\)

Next, let's consider the power-reduction identity:

\(\displaystyle \sin^3(\theta)=\dfrac{3\sin(\theta)-\sin(3\theta)}{4}\)

and now we have:

\(\displaystyle \displaystyle \frac{5}{2}\int3\sin(u)-\sin(3u)\,du\)

Can you proceed from here?

 

[soroban]

Hello, Ryo!

A somewhat different finish to the problem . . .

\(\displaystyle \displaystyle \int \frac{5\sin^3(\sqrt{x})}{\sqrt{x}} \,dx\)

We have: .\(\displaystyle \displaystyle 5\int \sin^3(x^{\frac{1}{2}})\,(x^{\text{-}\frac{1}{2}}dx) \)

Let \(\displaystyle u = x^{\frac{1}{2}} \quad\Rightarrow\quad du = \frac{1}{2}x^{\text{-}\frac{1}{2}}dx \quad\Rightarrow\quad x^{\text{-}\frac{1}{2}}dx =2\,du \)

Substitute: .\(\displaystyle \displaystyle 5\int(\sin^3\!u)(2\,du) \;=\;10\int \sin^3\!u\,du\)

. . . . . . \(\displaystyle \displaystyle =\;10\int \sin^2\!u(\sin u\,du) \;=\;10\int(1-\cos^2\!u)(\sin u\,du)\)


Let \(\displaystyle z = \cos u \quad\Rightarrow\quad dz = -\sin u\,du \quad\Rightarrow\quad \sin u\,du = -dz\)

Substitute: .\(\displaystyle \displaystyle10\int(1 - z^2)(-dz) \;=\;10\int(z^2-1)\,dz\)

. . . . . . . \(\displaystyle =\;10\left(\frac{1}{3}z^3 - z\right) + C \;=\;\frac{10}{3}z(z^2-3)+C\)


Back-substitute: .\(\displaystyle \frac{10}{3}\cos u(\cos^2\!u - 3) + C\)


Back-substitute: .\(\displaystyle \frac{10}{3}\cos\!\sqrt{x}\, \left(\cos^2\!\!\sqrt{x} - 3\right)+C\)

 

[Ryo]

^{Thanks for the help guys! I got it correct, Soroban was immensely helpful as that is the approach I have in my notes.}