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Please find integral 1/1+cos^2x dx ............................

Please find integral 1/1+cos^2x dx ............................

Grade:12

1 Answers

Arun
25750 Points
5 years ago
∫1/(1+cos²x)dx 
=∫1/(cos²x(sec²x+1))dx 
=∫sec²x/(2+tan²x)dx ... as sec²x = 1+tan²x 
=∫sec²x/(2(1+tan²x/2)dx 
=1/2∫sec²x/(1+((tanx)/√2)²)dx 
let u = (tanx)/√2 
so du = sec²x/√2 
So the integral becomes: 
1/2∫1/(1+u²)√2du 
=arctanu/√2 + c where c is a constant 
=arctan((tanx)/√2)/√2 + c

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