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#### renyikouniao

##### Member

- Jun 1, 2013

- 41

f(x)=integral(upper bound: x lower bound: 0) (cos^2(x)+4cos(x)+1)^0.5 dx

Here's my solution:

I use the equation L=integral ( upper bound: a lower bound: b ) sqrt[1+(f'(x))^2]

f'(x)=(cos^2(x)+4cos(x)+1)^0.5

1+[f'(x)]^2=2+cos^(x)+4cos(x)=[cos(x)+2]^2-2

L=integral( upper bound: x lower bound: 0 ) [2+cos^2(x)+4cos(x)]^0.5

This is where I am stuck...Am I right so far?