Difference in my and wolfram's integration.

Question

Calculate $$\int \frac{\sin ^3(x)+1}{\cos (x)+1} \, dx$$

Let $$u = \tan(x/2)$$ $\int \frac{\sin ^3(x)+1}{\cos (x)+1} \, dx = \int \frac{2\left(\frac{8u^3}{(u^2+1)^3}+1 \right)}{(u^2+1)\left( \frac{1-u^2}{u^2+1} + 1 \right)} \, du = \int \frac{8u^3}{(u^2+1)^3 } + 1 \, du$ $$ s := u^2 \wedge ds = 2u \, du $$ $ u + \int \frac{8su}{(s+1)^3 } (2u)^{-1} \, ds = u + 4\int \frac{s}{(s+1)^3 } \, ds = $ $ u + 4\int \frac{1}{(s+1)^2 } - \frac{1}{(s+1)^3 } \, ds = u + 4\left(\frac{-1}{s+1} +\frac{1}{2(s+1)^2} \right) + C =$ $ u + 4\left(\frac{-1}{u^2+1} +\frac{1}{2(u^2+1)^2} \right) + C = $ $ \tan(x/2) + 4\left(\frac{-1}{(\tan(x/2))^2+1} +\frac{1}{2((\tan(x/2))^2+1)^2} \right) + C := F(x)$

$$F(\pi/2) - F(-\pi/2) = 2 $$ Wolfram tells that result ($=2$) is okay, but my integral is different from wolfram result: $$ \frac{1}{8} \sec \left(\frac{x}{2}\right) \left(8 \sin \left(\frac{x}{2}\right)-4 \cos \left(\frac{x}{2}\right)-3 \cos \left(\frac{3 x}{2}\right)+\cos \left(\frac{5 x}{2}\right)\right) $$ Where I failed?

Answer 1

There's no error in your solution. Both results (yours and Wolfram's) are equal up to a constant (which for an indefinite integral is arbitrary). You can check that using the equalities $$ \cos (5\alpha) = \cos^5 \alpha - 10\cos^3\alpha \sin\alpha + 5\cos\alpha \sin^4\alpha = 16\cos^5\alpha -20\cos^3\alpha + 5\cos\alpha$$ $$ \cos (3\alpha) = \cos^3\alpha - 3\cos\alpha\sin^2\alpha = 4\cos^3\alpha -3\cos\alpha$$ $$ \frac{1}{\tan^2\alpha +1} = \cos^2\alpha$$ $$ \sec\alpha = \frac{1}{\cos\alpha}$$ Just write both results using only $\cos(\frac{x}{2})$ and compare.

Answer 2

Use $$\frac{\sin^3x+1}{\cos{x}+1}=\frac{8\sin^3\frac{x}{2}\cos^3\frac{x}{2}+1}{2\cos^2\frac{x}{2}}=4\sin^3\frac{x}{2}\cos\frac{x}{2}+\frac{1}{2\cos^2\frac{x}{2}}.$$