First slide

Evaluation of definite integrals

Question

$If I = \int_{0}^{\frac{π}{2}} \cos (\sin x) d x, J = \int_{0}^{\frac{π}{2}} \sin (\cos x) d x and K = \int_{0}^{\frac{π}{2}} \cos x d x, then$

Difficult

A

$K > I > J$
B

$J > I > K$
C

$I > J > K$
D

$I > K > J$

Solution

$\begin{array}{l} Given that I = \int_{0}^{\frac{π}{2}} \cos (\sin x) d x, J = \int_{0}^{\frac{π}{2}} \sin (\cos x) d x, k = \int_{0}^{\frac{π}{2}} \cos x d x \\ We know that \cos (\sin x) > \cos x > \sin (\cos x) for x \in (0, \frac{π}{2}) \\ (\sin c e \sin x < x f o r x \in (0, \frac{π}{2}) \Rightarrow \sin (\cos x) < \cos x (r e p l a c e x b y \cos x)) \\ (a n d a g a i n \sin x < x \Rightarrow \cos (\sin x) > \cos x (\sin c e \cos f u n c t i o n i s d e c r e a \sin g i n (0, \frac{π}{2}))) \\ \Rightarrow \int_{0}^{\frac{π}{2}} \cos (\sin x) d x > \int_{0}^{\frac{π}{2}} \cos x d x > \int_{0}^{\frac{π}{2}} \sin (\cos x) d x \Rightarrow I > K > J \end{array}$

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