
Two trig integrals
First:
$\displaystyle \int \sin^3x \cos^3x dx$
I used $\displaystyle u=\cos x \text{ }du=\sin x dx$
My answer was : $\displaystyle =\tfrac{1}{6}cos^{6} x\tfrac{1}{4}\cos^{4} x + C$
Apparently, the answer given was with $\displaystyle u=\sin x$ which gives the answer: $\displaystyle =\tfrac{1}{4}\sin^4x\tfrac{1}{6}\sin^6x + C$
So, are these answers the same or was I just wrong?
Second:
$\displaystyle \int^1_{1} \frac{(x+2)^2}{x^2+1}dx$
$\displaystyle x=\tan\theta \text{ }dx=\sec^2\theta d\theta$
$\displaystyle =\int^1_{1} \tan^2\theta d\theta + 4\int^1_{1} \tan\theta d\theta + 4\int^1_{1} d\theta$
$\displaystyle =\int^1_{1}\frac{\sin^2\theta}{\cos^2\theta} d\theta + 4*\ln{\left\sec\theta\right \Big^1_{1}} + 4*\theta \Big^1_{1}$
$\displaystyle =\int^1_{1}\frac{1cos^2\theta}{cos^2\theta}d\theta + 4*\ln{(\sqrt{x^2+1})\Big^1_{1}} + 4*\tan^{1}x\Big^1_{1}$
$\displaystyle =\int^1_{1}\sec^2\theta d\theta  \int^1_{1} d\theta + 4(\ln{(\sqrt{2})}\ln{(\sqrt{2})}) + 4(\tan^{1}(1)\tan^{1}(1))$
$\displaystyle =\tan\theta\Big^1_{1}  \theta \Big^1_{1} + 4(\tfrac{\pi}{4}\tfrac{3\pi}{4})$
$\displaystyle =2(\tfrac{\pi}{4}\tfrac{3\pi}{4}) + 4(\tfrac{\pi}{2})$
$\displaystyle =2\tfrac{3\pi}{2}$
The answer given is: $\displaystyle =2+\tfrac{3\pi}{2}$
Where did the sign change come from or is the given answer wrong?

for the 1st problem, you're ok ... the two expressions only differ by a constant.
$\displaystyle \int_{1}^1 \frac{x^2 + 2x + 4}{x^2 + 1} dx =$
$\displaystyle \int_{1}^1 \frac{x^2}{x^2+1} + \frac{2x}{x^2+1} + \frac{4}{x^2 + 1} dx =
$
$\displaystyle \int_{1}^1 1  \frac{1}{x^2+1} + \frac{2x}{x^2+1} + \frac{4}{x^2+1} \, dx =$
$\displaystyle \int_{1}^1 1 + \frac{2x}{x^2+1} + \frac{3}{x^2+1} \, dx =$
$\displaystyle [x]_{1}^1 + [\ln(x^2+1)]_{1}^1 + 3[\arctan{x}]_{1}^1 =$
$\displaystyle 2 + \frac{3\pi}{2}$

The first one is correct. As for the second one, I can't really check your work but perhaps you should reevaluate 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).

Yeah, but where's the sport in doing it the easy way (Smirk).

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