1
$\begingroup$

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: $$u=2\tan\ \theta$$

$\endgroup$
  • 3
    $\begingroup$ The substitution performed is $x=\color{red}{2}\tan\theta$. $\endgroup$ – projectilemotion Jan 9 '18 at 15:49
2
$\begingroup$

Here,

$ 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.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.