Practicing Success
Match List - I with List - II. Evaluate the integrals.
Choose the correct answer from the options given below : |
(A) - (I), (B) - (IV), (C) - (II), (D) - (III) (A) - (III), (B) - (IV), (C) - (I), (D) - (II) (A) - (III), (B) - (II), (C) - (I), (D) - (IV) (A) - (I), (B) - (III), (C) - (IV), (D) - (II) |
(A) - (III), (B) - (IV), (C) - (I), (D) - (II) |
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 |