[SPLIT] int [cos^2(x) - sin^2(x)] dx, int [tan^4(x)....

[warwick]

57. Find the volume of the solid that results when the region enclosed by y = cos x, y = sin x, x = 0, and x = pi/4 is revolved about the x-axis.

\(\displaystyle \L\\\int{cos^{2}(x)-sin^{2}(x)}dx\)

\(\displaystyle \L\\\int{[(1/2)(1+cos(2x)] - [(1/2)(1-cos(2x)]}dx\)

(1/2)\(\displaystyle \L\\\int{cos(2x)}dx\)

(1/2)sin (2x)

1/2

35. \(\displaystyle \L\\\int{tan^{4}(x)sec(x)}dx\)

 

[galactus]

57. Find the volume of the solid that results when the region enclosed by y = cos x, y = sin x, x = 0, and x = pi/4 is revolved about the x-axis.

\(\displaystyle \L\\\int{cos^{2}(x)-sin^{2}(x)}dx\)

\(\displaystyle \L\\\int{[(1/2)(1+cos(2x)] - [(1/2)(1-cos(2x)]}dx\)

(1/2)\(\displaystyle \L\\\int{cos(2x)}dx\)

(1/2)sin (2x)

1/2

What happened to Pi?. \(\displaystyle \L\\{\pi}\int_{0}^{\frac{\pi}{2}}[cos^{2}(x)-sin^{2}(x)]dx\)

35. \(\displaystyle \L\\\int{tan^{4}(x)sec(x)}dx\)

Rewrite as \(\displaystyle \L\\sec(x)(sec^{2}(x)-1)^{2}\)

 

[warwick]

Haha. That's it, man. I forgot pi in the volume equation. I was very, very tired when I attempting that problem last night.

I thought about rewriting #35 like that, but I don't think I ever got around to doing it on paper. I'll give it a shot. Thanks.

 

[galactus]

Be careful with the first one. You may want to use |cos(x)-sin(x)|.

Otherwise, you'll get 0 as an answer.

 

[warwick]

Be careful with the first one. You may want to use |cos(x)-sin(x)|.

Otherwise, you'll get 0 as an answer.

Your limits of integration are incorrect. It's 0 to pi/4.

 

[galactus]

Sorry, Warwick.

Anyway, write as:

\(\displaystyle \L\\{\pi}\left[\int_{0}^{\frac{\pi}{4}}cos^{2}(x)-\int_{0}^{\frac{\pi}{4}}sin^{2}(x)\right]dx\)

Now, use the identities:

\(\displaystyle \L\\cos^{2}(x)=\frac{1}{2}\left(cos(2x)+1\right)\)

and

\(\displaystyle \L\\sin^{2}(x)=\frac{1}{2}\left(1-cos(2x)\right)\)

 

[warwick]

Sorry, Warwick.

Anyway, write as:

\(\displaystyle \L\\{\pi}\left[\int_{0}^{\frac{\pi}{4}}cos^{2}(x)-\int_{0}^{\frac{\pi}{4}}sin^{2}(x)\right]dx\)

Now, use the identities:

\(\displaystyle \L\\cos^{2}(x)=\frac{1}{2}\left(cos(2x)+1\right)\)

and

\(\displaystyle \L\\sin^{2}(x)=\frac{1}{2}\left(1-cos(2x)\right)\)

Yeah, I worked that one out and got it correct. The Cal book has the odd-problem answers in the back, but I don't have a solutions manual. I should get one.