Two trig integrals

**symstar** · September 17th 2008, 04:56 PM

First:
$\int \sin^3x \cos^3x dx$

I used $u=\cos x \text{ }du=-\sin x dx$

My answer was : $=\tfrac{1}{6}cos^{6} x-\tfrac{1}{4}\cos^{4} x + C$

Apparently, the answer given was with $u=\sin x$ which gives the answer: $=\tfrac{1}{4}\sin^4x-\tfrac{1}{6}\sin^6x + C$

So, are these answers the same or was I just wrong?

Second:
$\int^1_{-1} \frac{(x+2)^2}{x^2+1}dx$
$x=\tan\theta \text{ }dx=\sec^2\theta d\theta$
$=\int^1_{-1} \tan^2\theta d\theta + 4\int^1_{-1} \tan\theta d\theta + 4\int^1_{-1} d\theta$
$=\int^1_{-1}\frac{\sin^2\theta}{\cos^2\theta} d\theta + 4*\ln{\left|\sec\theta\right| \Big|^1_{-1}} + 4*\theta \Big|^1_{-1}$
$=\int^1_{-1}\frac{1-cos^2\theta}{cos^2\theta}d\theta + 4*\ln{(\sqrt{x^2+1})\Big|^1_{-1}} + 4*\tan^{-1}x\Big|^1_{-1}$
$=\int^1_{-1}\sec^2\theta d\theta - \int^1_{-1} d\theta + 4(\ln{(\sqrt{2})}-\ln{(\sqrt{2})}) + 4(\tan^{-1}(1)-\tan^{-1}(-1))$
$=\tan\theta\Big|^1_{-1} - \theta \Big|^1_{-1} + 4(\tfrac{\pi}{4}-\tfrac{3\pi}{4})$
$=2-(\tfrac{\pi}{4}-\tfrac{3\pi}{4}) + 4(-\tfrac{\pi}{2})$
$=2-\tfrac{3\pi}{2}$

The answer given is: $=2+\tfrac{3\pi}{2}$

Where did the sign change come from or is the given answer wrong?

**skeeter** · September 17th 2008, 05:27 PM

for the 1st problem, you're ok ... the two expressions only differ by a constant.

$\int_{-1}^1 \frac{x^2 + 2x + 4}{x^2 + 1} dx =$

$\int_{-1}^1 \frac{x^2}{x^2+1} + \frac{2x}{x^2+1} + \frac{4}{x^2 + 1} dx =<br />$

$\int_{-1}^1 1 - \frac{1}{x^2+1} + \frac{2x}{x^2+1} + \frac{4}{x^2+1} \, dx =$

$\int_{-1}^1 1 + \frac{2x}{x^2+1} + \frac{3}{x^2+1} \, dx =$

$[x]_{-1}^1 + [\ln(x^2+1)]_{-1}^1 + 3[\arctan{x}]_{-1}^1 =$

$2 + \frac{3\pi}{2}$

**Chop Suey** · September 17th 2008, 05:29 PM

The first one is correct. As for the second one, I can't really check your work but perhaps you should re-evaluate it. Alternatively, if you expand the numerator and use polynomial division, you will get a much simpler integral to evaluate (and the answer would be the one given in the book).

**symstar** · September 17th 2008, 05:46 PM

Yeah, but where's the sport in doing it the easy way

.

**skeeter** · September 17th 2008, 05:50 PM

"sport" only counts when chasing members of the opposite sex.

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