Please find integral 1/1+cos^2x dx ............................
Ashok Eswar , 6 Years ago
Grade 12
Integral Calculus> Please find integral 1/1+cos^2x dx .........
1 Answers
Arun
Last Activity: 6 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
=∫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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