# Where did I go wrong in my working? - integration via substitution $\int_{0}^{\frac{1}{2}\ln3}\frac{1}{e^x+e^{-x}}dx$.

Here is the question:

Using the substitution $u=e^x$, find the exact value of $\int_{0}^{\frac{1}{2}\ln3}\dfrac{1}{e^x+e^{-x}}dx$

The actual to this answer is $\dfrac{\pi}{12}$, and in the official working out it seems they took the arctan of two numbers: (official working)

This surprised me because I've never seen arctan used in these questions before so I didn't even think about doing that: could anyone also explain when to take the arctan in these sort of questions please?

My working:

$u=e^x, x=\ln(u)$

$\int_{0}^{\frac{1}{2}\ln3}\dfrac{1}{e^x+e^{-x}}dx = \int_{0}^{\frac{1}{2}\ln3}\dfrac{1}{u+u^{-1}}dx$

$\frac{du}{dx} = e^x \therefore \frac{du}{e^x}=dx$

$\int_{1}^{\sqrt{3}}\frac{1}{u^2+1} du = \int_{1}^{\sqrt{3}}({u^2+1})^{-1} du$

$\int_{1}^{\sqrt{3}}(u^{-2}+1) du = [-u^{-1}+u]_1^\sqrt{3}$

$(-\sqrt{3}^{-1}+\sqrt{3}) - (-1 + 1) = \frac{2}{\sqrt{3}}$

Hint: See that $(u^2+1)^{-1}\neq u^{-2}+1$, also $$\int\dfrac{1}{u^2+1}du=\arctan u+C$$

• Oh right, okay - so does this mean that any expression of the form $\int_{}^{} \frac{1}{u} du$ can be rewritten as $arctan(u)$?
– Chx
Commented Jul 27, 2018 at 13:52
• Also, why do we use $arctan$ and not $arccos$ or $arcsin$?
– Chx
Commented Jul 27, 2018 at 14:23

HINT:

$$\int\frac{1}{u^2+1}\,du=\arctan(u)+C\ne -\frac1u + u +C$$

$$I=\int_{0}^{\frac{1}{2}\ln(3)}\frac{1}{e^x+e^{-x}}dx=\int_{0}^{\frac{1}{2}\ln(3)}\frac{1}{e^x+\frac{1}{e^x}}dx=\int_{0}^{\frac{1}{2}\ln(3)}\frac{e^x}{(e^x)^2+1}dx$$ $u=e^x\,,\frac{du}{dx}=e^x$ $$\therefore I=\int_{1}^{\sqrt{3}}\frac{1}{u^2+1}du=\arctan(\sqrt{3})-\arctan(1)=\frac{\pi}{3}-\frac{\pi}{4}=\frac{\pi}{12}$$

Hint: Try multiplying the numerator and denominator by $e^x$ and see what happens. Then make the appropiate substitution after doing that.

Hope this really helps! :)