Definite Integration Question 71
Question 71
- Integrate $\int_{0}^{\pi / 4} \log (1+\tan x) d x$.
(1997C, 2M)
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Solution:
- Let $I=\int_{0}^{\pi / 4} \log (1+\tan x) d x$
$$ \begin{aligned} & I=\int_{0}^{\pi / 4} \log \left(1+\tan \left(\frac{\pi}{4}-x\right)\right) d x \ \therefore \quad & I=\int_{0}^{\pi / 4} \log 1+\frac{1-\tan x}{1+\tan x} d x \end{aligned} $$
$$ \begin{aligned} & \qquad \int_{0}^{\pi / 4} \log \frac{1+\tan x+1-\tan x}{1+\tan x} d x \ & \qquad I=\int_{0}^{\pi / 4} \log \frac{2}{1+\tan x} d x \Rightarrow I=\int_{0}^{\pi / 4} \log 2 d x-I \ & \Rightarrow \quad 2 I=\frac{\pi}{4} \log 2 \Rightarrow \quad I=\frac{\pi}{8}(\log 2) \ & \text { 73. Let } \quad I=\int_{-\pi}^{\pi} \frac{2 x(1+\sin x)}{1+\cos ^{2} x} d x \ & \quad I=\int_{-\pi}^{\pi} \frac{2 x}{1+\cos ^{2} x} d x+\int_{-\pi}^{\pi} \frac{2 x \sin x}{1+\cos ^{2} x} d x \ & \quad I=I_{1}+I_{2} \end{aligned} $$
Now, $I_{1}=\int_{-\pi}^{\pi} \frac{2 x}{1+\cos ^{2} x} d x$
Let $f(x)=\frac{2 x}{1+\cos ^{2} x}$
$\Rightarrow f(-x)=\frac{-2 x}{1+\cos ^{2}(-x)}=\frac{-2 x}{1+\cos ^{2} x}=-f(x)$
$\Rightarrow \quad f(-x)=-f(x)$ which shows that $f(x)$ is an odd function.
$$ \begin{aligned} \therefore & I_{1} & =0 \ \text { Again, let } & g(x) & =\frac{2 x \sin x}{1+\cos ^{2} x} \end{aligned} $$
$\Rightarrow g(-x)=\frac{2(-x) \sin (-x)}{1+\cos ^{2}(-x)}=\frac{2 x \sin x}{1+\cos ^{2} x}=g(x)$
$\Rightarrow g(-x)=g(x)$ which shows that $g(x)$ is an even function.
$\therefore \quad I_{2}=\int_{-\pi}^{\pi} \frac{2 x \sin x}{1+\cos ^{2} x} d x=2 \cdot 2 \int_{0}^{\pi} \frac{x \sin x}{1+\cos ^{2} x} d x$
$$ \begin{aligned} & =4 \int_{0}^{\pi} \frac{(\pi-x) \sin (\pi-x)}{1+[\cos (\pi-x)]^{2}} d x=4 \int_{0}^{\pi} \frac{(\pi-x) \sin x}{1+\cos ^{2} x} d x \ & =4 \int_{0}^{\pi} \frac{\pi \sin x}{1+\cos ^{2} x} d x-4 \int_{0}^{\pi} \frac{x \sin x}{1+\cos ^{2} x} d x \end{aligned} $$
$\Rightarrow I_{2}=4 \pi \int_{0}^{\pi} \frac{\sin x}{1+\cos ^{2} x} d x-I_{2}$
$\Rightarrow 2 I_{2}=4 \pi \int_{0}^{\pi} \frac{\sin x}{1+\cos ^{2} x} d x$
Put $\quad \cos x=t \quad \Rightarrow \quad-\sin x d x=d t$
$\therefore \quad I_{2}=-2 \pi \int_{1}^{-1} \frac{d t}{1+t^{2}}=2 \pi \int_{-1}^{1} \frac{d t}{1+t^{2}}=4 \pi \int_{0}^{1} \frac{d t}{1+t^{2}}$
$$ =4 \pi\left[\tan ^{-1} t\right]_{0}^{1}=4 \pi\left[\tan ^{-1} 1-\tan ^{-1} 0\right] $$
$$ =4 \pi(\pi / 4-0)=\pi^{2} $$
$\therefore \quad I=I_{1}+I_{2}=0+\pi^{2}=\pi^{2}$
