Definite Integration Question 83

Question 83

  1. Evaluate 01/2xsin1x1x2dx.

(1984,2M)

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Answer:

Correct Answer: 85. 312π+12

Solution:

  1. Let I=01/2xsin1x1x2dx Put 1x=θx=sinθ

dx=cosθdθ

I=0π/6θsinθ1sin2θcosθdθ=0π/6θsinθdθ

$=[-\theta \cos \theta]{0}^{\pi / 6}+\int{0}^{\pi / 6} \cos \theta d \theta$

=π6cosπ6+0+sinπ6sin0=3π12+12



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