Trig substitution help (easy one )

[RadiationX]

Use trig substitution to find \int_{0}^{5} \frac{dt}{25 + x^2}dt

I can solve it to here \int_{0}^{\frac{\pi}{4}}\frac{25sec^2\theta}{(25 + tan^2\theta)^2}

and from this point i can factor the denominator into {625(1+ \tan^2\theta)}^2

which becomes 625\sec^4\theta

now i have the integral \int_{0}^{\frac{\pi}{4}}\frac{25sec^2\theta}{625\sec^4\theta}

this now reduces to \int_{0}^{\frac{\pi}{4}}\frac{cos^2\theta}{25}

and at this point i can use a power reducing formula to get rid of the \cos^2\theta

assuming that the last integral is correct and that i use the power reducing formula to reduce \cos^2\theta correctly, what am i doing wrong?
i have a TI-89 graphing calculator, and when i integrat this problem on it i get a different answer than when i do it by hand. where is my mistake?




this post is incorrect look further down for the correction.
i'm really sorry about this.

 

[Nylex]

You can use the double angle formula for cos 2x to get rid of cos^2 x. cos 2x = 2cos^2 x - 1.

 

[codyg1985]

As the integral stands now, you could use another formula to find the answer, but in order to put the integral into a form where you can use a trig substution you have to play around with it:

25 + t^2 = (\sqrt{25 + t^2})^2

So the integral becomes:

\int_{0}^{5}\frac{dt}{(\sqrt{25 + t^2})^2}

Make the substutions:
t = 5 \tan{\theta}
dt = 5 \sec^2{\theta} d\theta

and work it from there. :)

 

[Data]

This is actually a standard integral. Anyways, you seem to have made a few errors. Here we want

\int_0^5 \frac{dt}{25 + t^2}

let t = 5\tan{\theta} \Longrightarrow dt = 5 \sec^2{\theta} and note that t=5 \Longrightarrow 5 = 5\tan{\theta} \Longrightarrow \theta = \pi / 4. Thus we just have

\int_0^{\frac{\pi}{4}} \frac{5\sec^2 \theta}{25 + (5 \tan \theta)^2} \ d\theta

= \frac{1}{5}\int_0^{\frac{\pi}{4}} \frac{\sec^2 \theta}{1 + \tan^2 \theta} \ d\theta

now, review your trig identities and see if you can find an easy way to do this

The standard integral is

\int \frac{dx}{a^2+x^2} = \frac{1}{a}\arctan\left(\frac{x}{a}\right) \ + \ C,

by the way.

 

[RadiationX]

Use trig substitution to find \int_{0}^{5} \frac{dt}{25 + x^2}dt

I can solve it to here \int_{0}^{\frac{\pi}{4}}\frac{25sec^2\theta}{(25 + tan^2\theta)^2}

and from this point i can factor the denominator into {625(1+ \tan^2\theta)}^2

which becomes 625\sec^4\theta

now i have the integral \int_{0}^{\frac{\pi}{4}}\frac{25sec^2\theta}{625\sec^4\theta}

this now reduces to \int_{0}^{\frac{\pi}{4}}\frac{cos^2\theta}{25}

and at this point i can use a power reducing formula to get rid of the \cos^2\theta

assuming that the last integral is correct and that i use the power reducing formula to reduce \cos^2\theta correctly, what am i doing wrong?
i have a TI-89 graphing calculator, and when i integrat this problem on it i get a different answer than when i do it by hand. where is my mistake?

I made a huge mistake in my original post! the first integral is

\int_{0}^{5} \frac{dt}{(25 + x^2)^2}dt

and what follows is what i get. I'm really sorry about posting this problem incorrectly.

 

[RadiationX]

i'm putting this back to the top because i screwed it up the first time.

 

[Data]

No problem

Anyways, in that case your work is almost right. We want

\int_0^5 \frac{dx}{(25+x^2)^2}

substitute x = 5\tan \theta \Longrightarrow dx = 5\sec^2 \theta \ d\theta and so x=5 \Longrightarrow \theta = \pi / 4. Thus our integral is the same as

\int_0^{\frac{\pi}{4}} \frac{5\sec^2 \theta}{(25 + 25\tan^2 \theta)^2} \ d\theta

= \frac{1}{125}\int_0^{\frac{\pi}{4}} \frac{\sec^2 \theta}{\sec^4 \theta} \ d\theta

= \frac{1}{125}\int_0^{\frac{\pi}{4}} \cos^2 \theta \ d\theta

 

[RadiationX]

No problem

Anyways, in that case your work is almost right. We want

\int_0^5 \frac{dx}{(25+x^2)^2}

substitute x = 5\tan \theta \Longrightarrow dx = 5\sec^2 \theta \ d\theta and so x=5 \Longrightarrow \theta = \pi / 4. Thus our integral is the same as

\int_0^{\frac{\pi}{4}} \frac{5\sec^2 \theta}{(25 + 25\tan^2 \theta)^2} \ d\theta

= \frac{1}{125}\int_0^{\frac{\pi}{4}} \frac{\sec^2 \theta}{\sec^4 \theta} \ d\theta

= \frac{1}{125}\int_0^{\frac{\pi}{4}} \cos^2 \theta \ d\theta

i worked this down to exactly the same thing that you did but when i use my
TI-89 graphing calculator to integrate this i get a different answer than what i get by hand.

 

[Data]

\frac{1}{125}\int_0^{\frac{\pi}{4}} \cos^2 \theta d\theta

= \frac{1}{125} \int_0^{\frac{\pi}{4}} \frac{1}{2} \left(1 + \cos (2\theta)) \ d\theta

= \frac{1}{250}\left[\theta + \frac{\sin (2\theta)}{2} \right]_0^{\frac{\pi}{4}}

= \frac{1}{250}\left(\frac{\pi}{4} + \frac{1}{2}\right).

 

[RadiationX]

now i see my mistake. thankyou Data