Match List I with List II

List - I	List - II
A. $\int_{-\pi/2}^{\pi/2} \sin^7 x \, dx$	I. $\frac{\pi}{2}$
B. $\int_{-\pi/2}^{\pi/2} \sin^2 x \, dx$	II. $\frac{\pi}{4}$
C. $\int_0^{\pi/2} \frac{1}{1 + \tan x} \, dx$	III. 0
D. $\int_0^{\pi} \lvert \cos x \rvert \, dx$	IV. 2

$\textit{A.}$ $\displaystyle\int_{-\pi/2}^{\pi/2} \sin^7 x \, dx$

Property: If $f(x)$ is an odd function, i.e., $f(-x) = -f(x)$, then $\displaystyle\int_{-a}^{a} f(x)\, dx = 0$

$\sin^7(-x) = (-\sin x)^7 = -\sin^7 x$ → odd function

Since we are integrating over symmetric limits $[-\pi/2,\, \pi/2]$:

$\displaystyle\int_{-\pi/2}^{\pi/2} \sin^7 x \, dx = 0$

$\therefore$ A $\rightarrow$ III

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$\textit{B.}$ $\displaystyle\int_{-\pi/2}^{\pi/2} \sin^2 x \, dx$

Property: If $f(x)$ is an even function, i.e., $f(-x) = f(x)$, then $\displaystyle\int_{-a}^{a} f(x)\, dx = 2\int_{0}^{a} f(x)\, dx$

$\sin^2(-x) = (-\sin x)^2 = \sin^2 x$ → even function

$\displaystyle\int_{-\pi/2}^{\pi/2} \sin^2 x \, dx = 2\int_0^{\pi/2} \sin^2 x \, dx$

Using the identity $\sin^2 x = \dfrac{1 - \cos 2x}{2}$:

$= 2 \displaystyle\int_0^{\pi/2} \dfrac{1 - \cos 2x}{2}\, dx$

$= \left[x - \dfrac{\sin 2x}{2}\right]_0^{\pi/2}$

$= \left(\dfrac{\pi}{2} - 0\right) - (0 - 0)$

$= \dfrac{\pi}{2}$

$\therefore$ B $\rightarrow$ I

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$\textit{C.}$ $\displaystyle\int_0^{\pi/2} \dfrac{1}{1 + \tan x} \, dx$

Let $I = \displaystyle\int_0^{\pi/2} \dfrac{1}{1 + \tan x} \, dx$

Using King's Rule: $\displaystyle\int_0^a f(x)\, dx = \int_0^a f(a - x)\, dx$

Replace $x$ with $\dfrac{\pi}{2} - x$:

$I = \displaystyle\int_0^{\pi/2} \dfrac{1}{1 + \tan\left(\dfrac{\pi}{2} - x\right)} dx = \int_0^{\pi/2} \dfrac{1}{1 + \cot x} dx$

Since $\tan\left(\dfrac{\pi}{2} - x\right) = \cot x$

Simplifying:

$\dfrac{1}{1 + \cot x} = \dfrac{1}{1 + \dfrac{\cos x}{\sin x}} = \dfrac{\sin x}{\sin x + \cos x}$

So $I = \displaystyle\int_0^{\pi/2} \dfrac{\sin x}{\sin x + \cos x}\, dx$

Also, the original integral can be rewritten as:

$I = \displaystyle\int_0^{\pi/2} \dfrac{\cos x}{\sin x + \cos x}\, dx$

Adding both forms:

$2I = \displaystyle\int_0^{\pi/2} \dfrac{\cos x + \sin x}{\sin x + \cos x}\, dx = \int_0^{\pi/2} 1\, dx = \dfrac{\pi}{2}$

$I = \dfrac{\pi}{4}$

$\therefore$ C $\rightarrow$ II

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$\textit{D.}$ $\displaystyle\int_0^{\pi} |\cos x| \, dx$

$\cos x$ changes sign at $x = \dfrac{\pi}{2}$:

On $\left[0,\, \dfrac{\pi}{2}\right]$: $\cos x \geq 0$, so $|\cos x| = \cos x$

On $\left[\dfrac{\pi}{2},\, \pi\right]$: $\cos x \leq 0$, so $|\cos x| = -\cos x$

$\displaystyle\int_0^{\pi} |\cos x|\, dx = \int_0^{\pi/2} \cos x\, dx + \int_{\pi/2}^{\pi} (-\cos x)\, dx$

$= [\sin x]_0^{\pi/2} + [-\sin x]_{\pi/2}^{\pi}$

$= (1 - 0) + (-\sin\pi + \sin\dfrac{\pi}{2})$

$= 1 + (0 + 1) = 2$

$\therefore$ D $\rightarrow$ IV

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Final Matching:

$A \rightarrow III \;(0)$

$B \rightarrow I \;(\pi/2)$

$C \rightarrow II \;(\pi/4)$

$D \rightarrow IV \;(2)$