Match List - I with List - II. Evaluate the integrals. List - I List…

A. $I=\int\limits_0^{\pi / 2} \frac{(\sin x-\cos x)}{1+\sin x \cos x} d x-1$          ......(1)

as $\int\limits_0^a f(x) dx$

= $\int\limits_0^a f(a-x) dx$

$\Rightarrow \int\limits_0^{\pi / 2} \frac{(\sin (\pi / 2-x)-\cos (\pi / 2-x))}{1+\sin (\pi / 2-x) \cos \left(\pi / 2-x\right)}$

$I=\int\limits_0^{\pi / 2} \frac{(\cos x-\sin x) d x-\text { (2) }}{1+\sin x \cos x}$          ........(2)

adding eq (1) and (2)

$2 I=\int\limits_0^{\pi / 2} \frac{\sin x-\cos x+\cos x-\sin x d x}{1+\sin x \cos x}$

$2 I=\int\limits_0^{\pi / 2} \frac{0}{1+\sin x \cos x} d x=\int\limits_0^{\pi / 2} 0 d x$

$2 I=0 \Rightarrow I=0$

(A) → (III) (0)

B. $I=\int\limits_0^{1 / 2} \frac{d x}{\sqrt{x-x^2}}=\int\limits_0^{1 / 2} \frac{d x}{\sqrt{\frac{1}{4}-\frac{1}{4}+\frac{1}{2} \times 2 x-x^2}} =\int\limits_0^{1 / 2} \frac{d x}{\sqrt{\frac{1}{4}-\left(x-\frac{1}{2}\right)^2}}$

Completing square process

let $x - \frac{1}{2} = y \Rightarrow d x=d y$

as x → 0, y = $-\frac{1}{2}$

x → $\frac{1}{2}$,  y → 0

substituting y in I

$I=\int\limits_{-\frac{1}{2}}^0 \frac{d y}{\sqrt{\frac{1}{2}^2-y^2}}=\left[\sin ^{-1} \frac{y}{\frac{1}{2}}\right]_{-\frac{1}{2}}^0$

$=\sin ^{-1}(0)-\sin ^{-1}\left(\frac{-1 / 2}{1 / 2}\right)$

$=-\sin ^{-1}(-1)$

$I=-\left(-\frac{\pi}{2}\right)$

$I=\frac{\pi}{2}$

(B) → (IV)

C. $I=\int\limits_{-\pi}^\pi x \sin x d x$

$I=\int\limits_{-\pi}^\pi \underbrace{x}_I \underbrace{\sin x}_{II} d x$

x → first function

sin x → second function

using $\int\limits u v d x=u \int\limits v d x-\int\limits u' \int\limits v d x d x$

$=\left[x \int\limits \sin x d x\right]_{-\pi}^\pi-\int\limits_{-\pi}^\pi \frac{d(x)}{d x} \int\limits \sin x d x d x$

$=[-x \cos x]_{-\pi}^\pi+\int\limits_{-\pi}^\pi \cos x d x$

$=[-x \cos x]_{-\pi}^\pi+[\sin x]_{-\pi}^\pi$

$=\pi+\pi+0$

$=2 \pi$

(C) → (I)

D. $\int\limits_0^1 \frac{1}{1+x^2} d x=\left[\tan ^{-1}(x)\right]_0^1=\frac{\pi}{4}$

D → (II)

A → III, B → IV, C → I, D → II