Match List-I woth List-II
| List-I | List-II | ||
| A | $\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}sin^7x\, dx$ | I | $\pi $ |
| B | $\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}(x\, cos \, x+1)dx$ | II | $\frac{\pi}{12}$ |
| C | $\int\limits^{\frac{\pi}{2}}_{0}\frac{\sqrt{sinx}}{\sqrt{sin\, x}+\sqrt{cos\, x}}dx$ | III | 0 |
| D | $\int\limits^{\sqrt{3}}_{1}\frac{dx}{1+x^2}$ | IV | $\frac{\pi}{4}$ |
Choose the correct answer from the options given below :
Answer & explanation
Correct answer: option 2
The correct answer is option (2) → A-III, B-I, C-IV, D-II
(A) $\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}sin^7x\, dx=0$ (III)
(B) $\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}(x\, \cos x+1)dx$
(C) $\int\limits^{\frac{\pi}{2}}_{0}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx$ ...(1)
$\int\limits^{\frac{\pi}{2}}_{0}\frac{\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx$ ...(2)
$⇒2I=\int\limits^{\frac{\pi}{2}}_{0}1dx⇒I=\frac{\pi}{4}$ (IV)
(D) $\int\limits^{\sqrt{3}}_{1}\frac{dx}{1+x^2}=\left[\tan^{-1}x\right]^{\sqrt{3}}_{1}=\frac{\pi}{3}-\frac{\pi}{4}=\frac{\pi}{12}$ (II)