# Derivates with trigonometric substitution

In this integral how do they get $$dx=2\sec^2\theta \ d\theta$$ Here is the integral: $$\int\frac{dx}{x^2\sqrt{x^2+4}}=\int\frac{2\sec^2d\theta}{\mathrm{4\tan^2\ \theta}\!\cdot\!\mathrm{2\sec\ \theta}}=\frac{1}{4}\int\frac{sec\ \theta}{tan^2\ \theta} \ d\theta$$

If $$u=tan\ \theta$$ then it follows that $$d\theta=sec^2\ d\theta$$

With the substitution therein lays my confusion.

The correct substitution: $$u=2\tan\ \theta$$

• The substitution performed is $x=\color{red}{2}\tan\theta$. – projectilemotion Jan 9 '18 at 15:49

$x = 2 \tan θ$ is the substitution, which implies that, $dx = 2\sec ^2θ .dθ$
The first target here is to get rid of the square root in the denominator, which is the primary reason why $2 \tan θ$ is substituted instead of $\tan θ$.
Now, $$\int \frac {dx}{x^2 \sqrt {x^2 + 4}}$$ $$= \int \frac {2\sec^2 θ \,dθ } {4 \tan^2 θ \sqrt{4(\tan^2 θ + 1)}}$$ $$= \int \frac {2\sec^2θ \, dθ}{4 \tan^2 θ \,2\secθ}$$ $$= \frac {1}{4} \int \frac {\sec θ}{\tan^2 θ} \,dθ$$ (as you said. Here I am assuming that till now, you have done the integral correctly. If yes, then this is how it is continued:) $$= \frac {1}{4} \int \operatorname{cosec} \,θ \cot θ \,dθ$$ $$= \frac {1}{4} (-\sin θ) + c$$ $$= \frac {-1}{4} \sin (\tan ^{-1}{(x/2))} + c$$ $$= \frac {-x}{4(\sqrt{x^2 + 4})} + c$$ (In last step, I changed tan inverse to sin inverse to simplify it.)