Question:

Match List I with List II
List IList II
A.\(\int\limits^{\frac{\pi}{2}}_0\frac{\sin^{\frac{7}{2}}x}{\sin^{\frac{7}{2}}+\cos^{\frac{7}{2}}}dx\)I.\(\frac{\pi}{4}-\frac{1}{2}\)
B.\(\int\limits_0^{\pi}\frac{x\sin x}{1+\cos^2x}dx\)II.0
C.\(\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}x\cos x\ dx\)III.\(\frac{\pi}{4}\)
D.\(\int\limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\sin^2x\ dx\)IV.\(\frac{\pi^2}{4}\)
Choose the correct answer from the options given below :

Updated On: May 11, 2025
  • A-III, B-IV, C-II, D-I
  • A-I, B-IV, C-II, D-III
  • A-I, B-IV, C-II, D-III
  • A-IV, B-III, C-II, D-I
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The Correct Option is A

Solution and Explanation

To solve the problem of matching List I with List II, we evaluate each integral and match it with the corresponding value.
List IList II
A.\(\int\limits^{\frac{\pi}{2}}_0\frac{\sin^{\frac{7}{2}}x}{\sin^{\frac{7}{2}}x+\cos^{\frac{7}{2}}x}dx\)I.\(\frac{\pi}{4}-\frac{1}{2}\)
B.\(\int\limits_0^{\pi}\frac{x\sin x}{1+\cos^2x}dx\)II.0
C.\(\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}x\cos x\ dx\)III.\(\frac{\pi}{4}\)
D.\(\int\limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\sin^2x\ dx\)IV.\(\frac{\pi^2}{4}\)
  1. Integral A: \(\int\limits^{\frac{\pi}{2}}_0\frac{\sin^{\frac{7}{2}}x}{\sin^{\frac{7}{2}}x+\cos^{\frac{7}{2}}x}dx\) evaluates to \(\frac{\pi}{4}\) by symmetry and properties of definite integrals. Thus, A matches with III.
  2. Integral B: \(\int\limits_0^{\pi}\frac{x\sin x}{1+\cos^2x}dx\) is evaluated using symmetry and properties, resulting in \(\frac{\pi^2}{4}\). Thus, B matches with IV.
  3. Integral C: \(\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}x\cos x\ dx\) calculates to 0 as it is an odd function over a symmetric interval. Thus, C matches with II.
  4. Integral D: \(\int\limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\sin^2x\ dx\) results in \(\frac{\pi}{4}-\frac{1}{2}\) by trigonometric identities and integration. Thus, D matches with I.
The correct matching is A-III, B-IV, C-II, D-I.
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