SATHEE: Definite Integration Question 68

Definite Integration Question 68

Question 68

If $f$ is an even function, then prove that $\int_{0}^{π / 2} f (\cos 2 x) \cos x d x = \sqrt{2} \int_{0}^{π / 4} f (\sin 2 x) \cos x d x$ .

(2003, 2M)

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Solution:

Let $I = \int_{0}^{π / 2} f (\cos 2 x) \cos x d x$

$\begin{matrix} \Rightarrow I = \int_{0}^{π / 2} f \cos 2 \frac{π}{2} - x \cdot \cos \frac{π}{2} - x d x using \int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x \Rightarrow I = \int_{0}^{π / 2} f (\cos 2 x) \sin x d x \end{matrix}$

On adding Eqs. (i) and (ii), we get

$\begin{aligned} 2 I & = \int_{0}^{π / 2} f (\cos 2 x) (\sin x + \cos x) d x & = \sqrt{2} \int_{0}^{π / 2} f (\cos 2 x) [\cos (x - π / 4)] d x \end{aligned}$

Put $- x + \frac{π}{4} = t \Rightarrow - d x = d t$ $∴ 2 I = - \sqrt{2} \int_{π / 4}^{- π / 4} f \cos \frac{π}{2} - 2 t \cos t d t$

$\Rightarrow 2 I = \sqrt{2} \int_{- π / 4}^{π / 4} f (\sin 2 t) \cos t d t$

$∴ I = \sqrt{2} \int_{0}^{π / 4} f (\sin 2 t) \cos t d t$

Definite Integration Question 68

Question 68

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Definite Integration Question 68

Question 68

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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