Integrating Trigonometric Functions: Need Help!

[Jeebus]

I need help with these problems:

(1) STan^-1(y)dy

and

(2) Ssin^-1(x)dx

(S=integral)

I can't seem to figure out the trigonometric formula for integration. A little help or hint would be nice. Thanks!

 

[arildno]

Are you talking about arctan and arcsine (the inverse functions)?
If so, use integration by parts in both cases, with 1 as the function you integrate.

 

[TALewis]

Do you have to use a trigonometric method to integrate those? I think they could both be readily solved with integration by parts.

Edit: arildno beat me to it :)

If you need help with integration by parts, let us know. It works really well in this case.

 

[Jeebus]

Um, arildno or TALewis can you show me how to do it using integration by parts. I just learned this relatively recently and haven't got the core concepts down. If you could show me, it would be much appreciated.

Thanks!

 

[Gokul43201]

Look at :
http://www.sosmath.com/calculus/integration/byparts/byparts.html

and check out example 4, it's close to what you want, except you make du = 1*dx

 

[franznietzsche]

Aarr, help

Aasrr, i was born on a pirate ship! ok now that i got that out of my system...

i'm assuming for the problem you mean:

1 )$$\int arctan(y)$$ $$dy$$

and

2)$$\int arcsin(x)$$ $$dx$$

(thats the inverse tangent and inverse sine functions respectively). In that case you must perfomr a substitution before using integration by parts:

1)

$$\theta = arctan(y)$$

$$tan(\theta) = y$$

$$dy = sec^2(\theta)$$ $$d\theta$$

So the integral becomes:

$$\int \theta*sec^(\theta)$$ $$d\theta$$

which using integration by parts gives:

$$\int u$$ $$dv = u*v-\int v$$ $$du$$

with

$$u = \theta$$
$$dv = sec^2(\theta)$$

so

$$\int \theta*sec^2(\theta)$$ $$d\theta = \theta*tan(\theta)-\int tan(\theta)$$ $$d\theta$$

$$= \theta*tan(\theta) + ln|cos \theta|$$

So substituting for theta we get

$$\int arctan (y)$$ $$dy = yarctan(y) + ln|cos(arctan y)|$$
$$=y*arctan(y) + ln|\frac{1}{\sqrt{1+y^2}}|$$
$$=y*arctan(y) - \frac{1}{2}ln|y^2+1|$$
$$=y*arctan(y) - \frac{1}{2}ln(y^2 + 1)$$

U can use a similar substitution in the second problem (simply construct a right triangle, substitute theta in, integrate, and the substitute x back in.

Note: LaTeX still hates me...eventually i'll get it to look right...

 

[TALewis]

franznietzsche, I am curious why you made that substitution. I approach the problem this way. We want to evaluate:

$$\int \tan^{-1}y\,dy = I$$

I use capital I to denote the integral to be evaluated.

I choose the following directly from the integral:

$$\begin{align*} u &= \tan^{-1}y & du &= \frac{dy}{1+y^2}\\ dv &= dy & v &= y \end{align}$$

Integration by parts is given by:

$$\int u\,dv = uv-\int v\,du$$

So I have:

$$\begin{align*} I&=y\tan^{-1}y - \int \frac{y\,dy}{1+y^2}\\ \therefore \int \tan^{-1}y\,dy&=y\tan^{-1}y - \frac{1}{2}\ln(1+y^2) \end{align}$$

 

[franznietzsche]

franznietzsche, I am curious why you made that substitution. I approach the problem this way. We want to evaluate:

$$\int \tan^{-1}y\,dy = I$$

I use capital I to denote the integral to be evaluated.

I choose the following directly from the integral:

$$\begin{align*} u &= \tan^{-1}y & du &= \frac{dy}{1+y^2}\\ dv &= dy & v &= y \end{align}$$

Integration by parts is given by:

$$\int u\,dv = uv-\int v\,du$$

So I have:

$$\begin{align*} I&=y\tan^{-1}y - \int \frac{y\,dy}{1+y^2}\\ \therefore \int \tan^{-1}y\,dy&=y\tan^{-1}y - \frac{1}{2}\ln(1+y^2) \end{align}$$

I suppose some combination of liking trig substitutions inherently and not wanting to differetiate $$arctan y$$ simply because i didn't know it offhand and didn't want to look it up.

 

[TALewis]

Hm, I agree. If you don't have the derivative of arctan memorized, you have to look it up. And while you're looking things up, you might as well just use an integral table anyway.

 

[franznietzsche]

Hm, I agree. If you don't have the derivative of arctan memorized, you have to look it up. And while you're looking things up, you might as well just use an integral table anyway.

yup, exactly.