A) 3/2
B) ?8/3
C) 3/8
D) 8/3
Correct Answer: B
Solution :
\[\int_{-\pi /4}^{\pi /4}{{{\sin }^{-4}}x\,dx=2\int_{0}^{\pi /4}{\frac{{{\cos }^{4}}x}{{{\sin }^{4}}x}{{\sec }^{4}}x\,dx}}=2\int_{0}^{\pi /4}{\frac{{{\sec }^{4}}xdx}{{{\tan }^{4}}x}}\] Put \[\tan x=t\], we get \[2\int_{0}^{1}{\frac{1+{{t}^{2}}}{{{t}^{4}}}dt}\] \[=2\left[ \int_{0}^{1}{{{t}^{-4}}dt+\int_{0}^{1}{{{t}^{-2}}dt}} \right]\]\[=2\left[ \left| -\frac{1}{3{{t}^{3}}} \right|_{0}^{1}+\left| -\frac{1}{t} \right|_{0}^{1} \right]=-\frac{8}{3}\].
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