Indefinite Integration 1 Question 1
2. If $\int _0^{\pi / 2} \frac{\cot x}{\cot x+\operatorname{cosec} x} d x=m(\pi+n)$, then $m \cdot n$ is equal to
(a) $-\frac{1}{2}$
(b) 1
(c) $\frac{1}{2}$
(d) -1
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Solution:
- Let $I=\int _0^{\pi / 2} \frac{\cot x}{\cot x+\operatorname{cosec} x} d x$
$$ \begin{aligned} & =\int _0^{\pi / 2} \frac{\frac{\cos x}{\sin x}}{\frac{\cos x}{\sin x}+\frac{1}{\sin x}} d x=\int _0^{\pi / 2} \frac{\cos x}{1+\cos x} d x \\ & =\int _0^{\pi / 2} \frac{\cos x(1-\cos x)}{1-\cos ^{2} x} d x \\ & =\int _0^{\pi / 2} \frac{\cos x-\cos ^{2} x}{\sin ^{2} x} d x \\ & =\int _0^{\pi / 2}\left(\operatorname{cosec} x \cot x-\cot ^{2} x\right) d x \\ & =\int _0^{\pi / 2}\left(\operatorname{cosec} x \cot x-\operatorname{cosec}^{2} x+1\right) d x \\ & =[-\operatorname{cosec} x+\cot x+x] _0^{\pi / 2} \\ & =x+\frac{\cos x-1}{\sin x} \quad 0 \quad x+\frac{-2 \sin ^{2} \frac{x}{2}}{2 \sin \frac{x}{2} \cos \frac{x}{2}} 0 \\ & =x-\tan \frac{x}{2}=\frac{\pi}{2}-1=\frac{1}{2}[\pi-2] \\ & =m[\pi+n] \end{aligned} $$
On comparing, we get $m=\frac{1}{2}$ and $n=-2$
[given]
$$ \therefore \quad m \cdot n=-1 $$
(given)
Alternate Solution
Let $I=\int _0^{\pi / 2} \frac{\cot x}{\cot x+\operatorname{cosec} x} d x$
$=\int _0^{\pi / 2} \frac{\frac{\cos x}{\sin x}}{\frac{\cos x}{\sin x}+\frac{1}{\sin x}} d x$
$=\int _0^{\pi / 2} \frac{\cos x}{\cos x+1} d x$
$=\int _0^{\pi / 2} \frac{2 \cos ^{2} \frac{x}{2}-1}{2 \cos ^{2} \frac{x}{2}} d x$
$\because \cos \theta=2 \cos ^{2} \frac{\theta}{2}-1$ and $\cos \theta+1=2 \cos ^{2} \frac{\theta}{2}$
$=\int _0^{\pi / 2} 1-\frac{1}{2} \sec ^{2} \frac{x}{2} d x$
$=x-\tan \frac{x}{2} _0^{\pi / 2}=\frac{\pi}{2}-1=\frac{1}{2}(\pi-2)$
Since, $\quad I=m(\pi-n)$
$\therefore m(\pi-n)=\frac{1}{2}(\pi-2)$
On comparing both sides, we get
$$ m=\frac{1}{2} \text { and } n=-2 $$
Now, $m n=\frac{1}{2} \times-2=-1$
