Definite Integration Question 68
Question 68
- If $f$ is an even function, then prove that $\int_{0}^{\pi / 2} f(\cos 2 x) \cos x d x=\sqrt{2} \int_{0}^{\pi / 4} f(\sin 2 x) \cos x d x$.
(2003, 2M)
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Solution:
- Let $I=\int_{0}^{\pi / 2} f(\cos 2 x) \cos x d x$
$$ \begin{gathered} \Rightarrow \quad I=\int_{0}^{\pi / 2} f \cos 2 \frac{\pi}{2}-x \cdot \cos \frac{\pi}{2}-x \quad d x \ \text { using } \int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x \ \Rightarrow \quad I=\int_{0}^{\pi / 2} f(\cos 2 x) \sin x d x \end{gathered} $$
On adding Eqs. (i) and (ii), we get
$$ \begin{aligned} 2 I & =\int_{0}^{\pi / 2} f(\cos 2 x)(\sin x+\cos x) d x \ & =\sqrt{2} \int_{0}^{\pi / 2} f(\cos 2 x)[\cos (x-\pi / 4)] d x \end{aligned} $$
Put $-x+\frac{\pi}{4}=t \quad \Rightarrow \quad-d x=d t$ $\therefore \quad 2 I=-\sqrt{2} \int_{\pi / 4}^{-\pi / 4} f \cos \frac{\pi}{2}-2 t \quad \cos t d t$
$\Rightarrow \quad 2 I=\sqrt{2} \int_{-\pi / 4}^{\pi / 4} f(\sin 2 t) \cos t d t$
$\therefore \quad I=\sqrt{2} \int_{0}^{\pi / 4} f(\sin 2 t) \cos t d t$
