Match List-I with List-II
| List-I | List-II | ||
| (A) | $\int\limits^{\frac{\pi }{2}}_{-\frac{\pi }{2}}sin^5xdx$ | (I) | $\pi $ |
| (B) | $\int\limits^{\frac{\pi }{2}}_{-\frac{\pi }{2}}(x^3+tan^3x+1)dx$ | (II) | $\frac{\pi }{12}$ |
| (C) | $\int\limits^{\frac{\pi }{2}}_{0}\frac{cos^5x}{cos^5x+sin^5x}dx$ | (III) | $\frac{\pi }{4}$ |
| (D) | $\int\limits^{\sqrt{3}}_{1}\frac{dx}{1+x^2}$ | (IV) | 0 |
Choose the correct answer from the options given below :
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A)-(IV), (B)-(I), (C)-(III), (D)-(II)
(A) $\int\limits^{\frac{\pi }{2}}_{-\frac{\pi }{2}}\sin^5xdx=0$ (IV) odd function
(B) $\int\limits^{\frac{\pi }{2}}_{-\frac{\pi }{2}}(x^3+\tan^3x+1)dx=0+0+\left(\frac{\pi }{2}+\frac{\pi }{2}\right)=\pi$ (I)
(C) $I=\int\limits^{\frac{\pi }{2}}_{0}\frac{\cos^5x}{\cos^5x+\sin^5x}dx$
so $2I=\int\limits^{\frac{\pi }{2}}_{0}dx⇒I=\frac{\pi}{4}$ (III)
(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)