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:

$\displaystyle \int^1_{-1} \frac{(x+2)^2}{x^2+1}dx$

$\displaystyle x=\tan\theta \text{ }dx=\sec^2\theta d\theta$

$\displaystyle =\int^1_{-1} \tan^2\theta d\theta + 4\int^1_{-1} \tan\theta d\theta + 4\int^1_{-1} d\theta$

$\displaystyle =\int^1_{-1}\frac{\sin^2\theta}{\cos^2\theta} d\theta + 4*\ln{\left|\sec\theta\right| \Big|^1_{-1}} + 4*\theta \Big|^1_{-1}$

$\displaystyle =\int^1_{-1}\frac{1-cos^2\theta}{cos^2\theta}d\theta + 4*\ln{(\sqrt{x^2+1})\Big|^1_{-1}} + 4*\tan^{-1}x\Big|^1_{-1}$

$\displaystyle =\int^1_{-1}\sec^2\theta d\theta - \int^1_{-1} d\theta + 4(\ln{(\sqrt{2})}-\ln{(\sqrt{2})}) + 4(\tan^{-1}(1)-\tan^{-1}(-1))$

$\displaystyle =\tan\theta\Big|^1_{-1} - \theta \Big|^1_{-1} + 4(\tfrac{\pi}{4}-\tfrac{3\pi}{4})$

$\displaystyle =2-(\tfrac{\pi}{4}-\tfrac{3\pi}{4}) + 4(-\tfrac{\pi}{2})$

$\displaystyle =2-\tfrac{3\pi}{2}$

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

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