First slide

Evaluation of definite integrals

Question

$The value of \int_{0}^{2 π} \frac{\cos^{2 n} x}{\cos^{2 n} x + \sin^{2 n} x} d x is$

Moderate

A

$\frac{π}{2}$
B

$π$
C

$2 π$
D

$4 π$

Solution

$\begin{array}{l} Here f (x) = \frac{\cos^{2 n} x}{\cos^{2 n} x + \sin^{2 n} x} for which f (x) = f (2 π - x) \\ \Rightarrow I = \int_{0}^{2 π} \frac{\cos^{2 n} x}{\cos^{2 n} x + \sin^{2 n} x} d x = 2 \int_{0}^{π} \frac{\cos^{2 n} x}{\cos^{2 n} x + \sin^{2 n} x} d x \\ Again, f (x) = f (π - x) \end{array}$
$\Rightarrow I = 4 \int_{0}^{π / 2} \frac{\cos^{2 n} x}{\cos^{2 n} x + \sin^{2 n} x} d x - - - (1)$
$\begin{array}{l} = 4 \int_{0}^{π / 2} \frac{\cos^{2 n} (\frac{π}{2} - x)}{\cos^{2 n} (\frac{π}{2} - x) + \sin^{2 n} (\frac{π}{2} - x)} d x \\ (by property IV) \end{array}$
$= 4 \int_{0}^{π / 2} \frac{\sin^{2 n} x}{\sin^{2 n} x + \cos^{2 n} x} d x - - - - - - - (2)$
$Adding (1) and (2), we have 2 I = 4 \int_{0}^{π / 2} 1 d x \Rightarrow I = π$

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