Question

The value of $\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{(\pi-4\theta) \tan \theta}{1-\tan \theta} d\theta$ is equal to:

• A. $\frac{\pi}{2} \ln 2 - \frac{\pi^2}{4}$
• B. $\frac{\pi}{2} \ln 2 + \frac{\pi^2}{4}$
• C. $\pi \ln 2 - \frac{\pi^2}{4}$
• D. $-\pi \ln 2 - \frac{\pi^2}{4}$

Answer 1

Answer: (C)

$\pi\ln2-\frac{\pi^2}{4}$

Answer 2

We start with

$$I=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{(\pi-4\theta)\tan\theta}{1-\tan\theta}\,d\theta.$$

A very clever change of variable is to use

$$u=\frac{\pi}{4}-\theta.$$

Then when $\theta=-\frac{\pi}{4}$ we have

$$u=\frac{\pi}{4}-\left(-\frac{\pi}{4}\right)=\frac{\pi}{2},$$

and when $\theta=\frac{\pi}{4}$ we get $u=0$. Also, $d\theta=-du$.

Now, express the integrand in terms of $u$. Note that

$$\pi-4\theta=\pi-4\Bigl(\frac{\pi}{4}-u\Bigr)=\pi-(\pi-4u)=4u.$$

Next, using the identity

$$\tan\left(\frac{\pi}{4}-u\right)=\frac{1-\tan u}{1+\tan u},$$

we have

$$1-\tan\left(\frac{\pi}{4}-u\right)=1-\frac{1-\tan u}{1+\tan u}=\frac{(1+\tan u)-(1-\tan u)}{1+\tan u}=\frac{2\tan u}{1+\tan u}.$$

Thus the integrand becomes

$$\frac{(\pi-4\theta)\tan\theta}{1-\tan\theta}\quad \rightarrow\quad \frac{4u\;\tan\left(\frac{\pi}{4}-u\right)}{1-\tan\left(\frac{\pi}{4}-u\right)} =\frac{4u\,\dfrac{1-\tan u}{1+\tan u}}{\dfrac{2\tan u}{1+\tan u}} =\frac{4u\,(1-\tan u)}{2\tan u} =\frac{2u\,(1-\tan u)}{\tan u}.$$

Changing the limits (and noting that the minus sign from $d\theta=-du$ reverses the limits) we write

$$I=\int_{u=\pi/2}^{0}\frac{2u\,(1-\tan u)}{\tan u}(-du) =\int_{0}^{\pi/2}\frac{2u\,(1-\tan u)}{\tan u}\,du.$$

It is convenient to rewrite this expression as

$$I=2\int_{0}^{\pi/2}u\Bigl(\frac{1}{\tan u}-1\Bigr)du =2\int_{0}^{\pi/2}u\Bigl(\cot u-1\Bigr)du.$$

We now separate the integral:

$$I=2\Biggl[\underbrace{\int_{0}^{\pi/2}u\,\cot u\,du}_{L}-\underbrace{\int_{0}^{\pi/2}u\,du}_{M}\Biggr].$$

Step 1. Evaluation of $M$:

$$M=\int_{0}^{\pi/2}u\,du=\left.\frac{u^2}{2}\right|_0^{\pi/2}=\frac{\pi^2}{8}.$$

Step 2. Evaluation of $L$:

Use integration by parts. Let

$$v=u,\quad dw=\cot u\,du\quad \Longrightarrow\quad dv=du,\quad w=\int \cot u\,du=\ln|\sin u|.$$

Then

$$L=u\ln|\sin u|\Big|_0^{\pi/2}-\int_{0}^{\pi/2}\ln|\sin u|\,du.$$

At $u=\pi/2$, $\ln(\sin(\pi/2))=\ln1=0$; at $u=0$ the term $u\ln|\sin u|$ tends to 0 (since $u\ln u\to0$ as $u\to0$). Hence,

$$L=-\int_{0}^{\pi/2}\ln(\sin u)\,du.$$

A standard result is:

$$\int_{0}^{\pi/2}\ln(\sin u)\,du=-\frac{\pi}{2}\ln2.$$

Thus,

$$L=-\Bigl(-\frac{\pi}{2}\ln2\Bigr)=\frac{\pi}{2}\ln2.$$

Step 3. Final answer:

Plug $L$ and $M$ back into the expression for $I$:

$$I=2\left[\frac{\pi}{2}\ln2-\frac{\pi^2}{8}\right] =\pi\ln2-\frac{\pi^2}{4}.$$

Summary:

• Explanation: Substitute $u=\tfrac{\pi}{4}-\theta$, simplify using $\tan(\frac{\pi}{4}-u)=\frac{1-\tan u}{1+\tan u}$ to recast the integral as $2\int_0^{\pi/2} u(\cot u-1)du$. Separate the integral and use integration by parts with the standard formula $\int_0^{\pi/2}\ln(\sin u)du=-\frac{\pi}{2}\ln2$.
• Answer: $\pi\ln2-\frac{\pi^2}{4}$ (Option 3)