What's the integral of sec^2(x/2)tan(x/2)dx?

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I want to know if I solved this integral correctly,

$$\int\sec^2\Big(\frac{x}{2}\Big) \tan \Big(\frac{x}{2}\Big)dx$$

I set $u = \tan\left(\frac x2\right)$, so $du = \frac12\sec^2\left(\frac x2\right)dx$.

My answer is $\tan\left(\frac x2\right)+c$, but the answers you got me a little confused.

Sorry, my English is very poor

Thanks.

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2 Answers

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Try substituting $u=tan(x/2)$.

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$\int \sec^2\frac x2 \tan \frac x2 dx= \int \frac{\sin \frac x2 dx}{ sin^2\frac x2 \cos \frac x2}=\int \frac{dx}{ sin\frac x2 \cos \frac x2}=2\int \frac{dx}{ sin x}=2\int \frac{\sin xdx}{ 1-\cos^2 x}=-2\int \frac{d(\cos x)}{ 1-\cos^2 x}$

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