Integral Calculus Question 368
Question: $ \int_{{}}^{{}}{\frac{1}{1+{{\sin }^{2}}x}\ dx=} $
Options:
A) $ \frac{1}{\sqrt{2}}{{\tan }^{-1}}(\sqrt{2}\tan x)+k $
B) $ \sqrt{2}{{\tan }^{-1}}(\sqrt{2}\tan x)+k $
C) $ -\frac{1}{\sqrt{2}}{{\tan }^{-1}}(\sqrt{2}\tan x)+k $
D) $ -\sqrt{2}{{\tan }^{-1}}(\sqrt{2}\tan x)+k $
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Answer:
Correct Answer: A
Solution:
$ I=\int_{{}}^{{}}{\frac{1}{1+{{\sin }^{2}}x},dx}=\int_{{}}^{{}}{\frac{dx}{2{{\sin }^{2}}x+{{\cos }^{2}}x}} $ $ =\int_{{}}^{{}}{\frac{{{\sec }^{2}}x,dx}{2{{\tan }^{2}}x+1}} $ $ =\frac{1}{2}\int_{{}}^{{}}{\frac{{{\sec }^{2}}x,dx}{{{\tan }^{2}}x+\frac{1}{2}}} $ Put $ \tan x=t\Rightarrow {{\sec }^{2}}x,dx=dt, $ then $ I=\frac{1}{2}\int_{{}}^{{}}{\frac{dt}{t^{2}+\frac{1}{2}}=\frac{1}{2}}.\frac{1}{1/\sqrt{2}}{{\tan }^{-1}}\frac{t}{1/\sqrt{2}} $ $ =\frac{1}{\sqrt{2}}{{\tan }^{-1}}(\sqrt{2}\tan x)+k $ .
BETA
