Can you help me solve this trigonometric/integration Question?

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1 Answer
Apr 22, 2017

#int_0^(1/4pi) cos^2theta# #d theta= 1/4+1/4pi#

Explanation:

We're trying to find #int_0^2 8/(4+x^2)^2 dx#.

First let #x=2tantheta#

#dx/(d theta) = 2sec^2theta#

#dx= 2sec^2theta d theta#

#x = 2, theta = 1/4 pi#

#x = 0, theta=0#

#int_0^2 8/(4+x^2) dx = int_0^2 8/(4+4tan^2theta)^2 dx = int_0^(1/4pi) 8/(4(1+tan^2theta))^2 2sec^2theta d theta=#
#int_0^(1/4pi) 8/(16sec^4theta) 2sec^2theta d theta= int_0^(1/4pi) 1/sec^2theta d theta=#
#int_0^(1/4pi) cos^2theta d theta#

Recall that #int cos^2theta d theta= 1/4(2theta + sin2theta) + "c"#

#int_0^(1/4pi) cos^2theta d theta = [1/4(2theta + sin2theta)]_0^(1/4pi)=#
#1/4 + 1/4pi#