# Difficulty in integrating

I tried to integrate this by parts but it didn't work out. Any simple means of doing it.$$\int\sin^{-1}\biggl(\frac{2x+2}{\sqrt{4x^{2}+8x+13}}\biggr) \ dx$$

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Actually, integration by parts does seem to help. – Srivatsan Aug 13 '11 at 13:11

Put $2x+2 = 3 \tan(\theta)$ and see what happens.

$\textbf{Added.}$ First observe that $$4x^{2}+8x+13= (2x+2)^{2} + 3^{2}.$$ So I hope you are aware of the fact that $\text{if you have an integral of the form}$, $1+x^{2}$, then one generally substitutes, $x= \tan(\theta)$. That's the case here as well. By doing that we get,

\begin{align*} \int \sin^{-1}\biggl(\frac{2x+2}{\sqrt{4x^{2}+8x+13}}\biggr) \ dx &= \int\sin^{-1}\biggl(\frac{3\cdot \tan\theta}{3 \cdot \sec\theta}\biggr) \cdot 3 \sec^{2}\theta \ d\theta \\ &= 3\cdot\int \sin^{-1}(\sin\theta) \cdot \sec^{2}\theta \ d\theta \\ &= 3 \cdot \int \theta \cdot\sec^{2}\theta \ d \theta \end{align*}

Use integration by parts to evaluate the last integral. Put $u = \theta$ and $dv = \sec^{2}(\theta) \ d\theta$. So the answer for the last part should be $$\int \theta \cdot \sec^{2}\theta \ d \theta = \theta\cdot\tan\theta + \ln(\cos\theta) + C$$

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Let's not start to integrate in too much of a hurry.

Completing the square is a natural move: $4x^2+8x+13=(2x+2)^2+9$. In a right triangle, a certain angle has sine equal to $$\frac{2x+2}{\sqrt{(2x+2)^2 +9}}.$$ If the "opposite" side is $2x+2$ and the hypotenuse is $\sqrt{(2x+2)^2+9}$, the remaining side of the triangle must be $3$. So we are trying to integrate $$\arctan\left(\frac{2x+2}{3}\right).$$ After the natural substitution, we arrive at $$\int\frac{3}{2}\arctan t \,dt,$$ a mild variant of a standard integral.

Added for completeness: We find $\int \arctan t\;dt$, using integration by parts. Let $u=\arctan t$, $dv=dt$. Then $du=\frac{dt}{1+t^2}$ and $v$ can be taken to be $t$. It follows that $$\int \arctan t\;dt=t\arctan t -\int \frac{t}{1+t^2} dt.$$ The remaining integral yields easily to the substitution $w=1+t^2$. Finally, we remember about our constant factor $3/2$ and obtain $$\frac{3}{2}t \arctan t -\frac{3}{4}\ln(1+t^2) + C.$$

We omit the back substitution $t=(2x+2)/3$.

The issue of signs: What about if $2x+3$ is negative? For indefinite integrals, there is almost a calculus tradition not to worry about such things. But let's. The usual meaning of $\arcsin u$ is the number between $-\pi/2$ and $\pi/2$ whose sine is $u$. For $\arctan u$, there are two common choices of interval, $(-\pi/2,\pi/2)$ and $[0,\pi)$. If we use $(-\pi/2,\pi/2)$, the answer obtained above is correct for negative $2x+3$. If we use $[0,\pi)$, a small adjustment needs to be made.

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This may not be the accepted answer, but I like this more; considering the geometric interpretation is always a good thing. – J. M. Aug 13 '11 at 17:43