First slide

Evaluation of definite integrals

Question

$\int_{0}^{2 π} \frac{1}{1 + \tan^{4} x} d x is equal to$

Easy

A

$π$
B

$\frac{π}{2}$
C

$2 π$
D

$3 π$

Solution

$\begin{array}{l} I = \int_{0}^{2 π} \frac{1}{1 + \tan^{4} x} d x \\ \Rightarrow I = 2 \int_{0}^{π} \frac{d x}{1 + \tan^{4} x} a p p l y \int_{0}^{2 a} f (x) d x = 2 \int_{0}^{a} f (x) d x i f f (2 a - x) = f (x) \\ \Rightarrow I = 4 \int_{0}^{π / 2} \frac{d x}{1 + \tan^{4} x} \dots \dots \dots (1) s a m e f o r m u l a \\ Or I = 4 \int_{0}^{π / 2} \frac{d x}{1 + \cot^{4} x} \dots (2) a p p l y \int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x \end{array}$
Adding (1) and (2), we get
$\begin{array}{l} 2 I = 4 \int_{0}^{π / 2} d x = 4 \times \frac{π}{2} \\ \Rightarrow I = π \end{array}$

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