The value of int_{-frac{pi}{4}}^{frac{pi}{4}} | vedclass

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Question

(C) Let $I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} (\sin x)^{-4} \,dx$.
Since $f(x) = (\sin x)^{-4} = \frac{1}{\sin^4 x}$ is an even function,we have

Vedclass · Accepted Answer

$\infty$

Vedclass · Answer

(C) Let $I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} (\sin x)^{-4} \,dx$.
Since $f(x) = (\sin x)^{-4} = \frac{1}{\sin^4 x}$ is an even function,we have:
$I = 2 \int_{0}^{\frac{\pi}{4}} \csc^4 x \,dx$.
Using the identity $\csc^2 x = 1 + \cot^2 x$,we get:
$I = 2 \int_{0}^{\frac{\pi}{4}} (1 + \cot^2 x) \csc^2 x \,dx$.
Let $u = \cot x$,then $du = -\csc^2 x \,dx$.
As $x 	o 0^+$,$u 	o \infty$. When $x = \frac{\pi}{4}$,$u = 1$.
$I = 2 \int_{\infty}^{1} (1 + u^2) (-du) = 2 \int_{1}^{\infty} (1 + u^2) \,du$.
Evaluating this integral: $2 [u + \frac{u^3}{3}]_{1}^{\infty} = \infty$.
The integral is divergent.

The value of $\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} (\sin x)^{-4} \,dx$ is

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