integral (1/cosx +2)

Question

for solve this question

 

Answer 1

 
 integral (1/cosx +2)
for solve this question

 

Solution: let me rewrite this in the form

 

∫1/[cos(x)+2]dx

 

What you need to do is rewrite this in trigonometric identities

 

lets start with Cos(x) from trigonometric identities

 

cos(x/2)=√[1+cos(x)/2] square both sides

 

cos²(x/2)=1+cos(x)/2, and cos(x)=2cos²(x/2)-1 substitute in integral

 

∫1/[cos(x)+2] = ∫1/[2cos²(x/2)-1+2]=∫1/[2cos²(x/2)+1]

 

Factor cos²(x/2) , ∫1/cos²(x/2)(2+1/cos²(x/2)

 

but 1/cos²(x/2)=sec²(x/2) substitute and the integral becomes

 

∫[sec²(x/2)/2+1/cos²(x/2)]= ∫[sec²(x/2)/2+sec²(x/2)]

 

Again from trigonometric identities sec²(x/2)=tan²(x/2)+1

 

substitute the integral becomes ∫[sec²(x/2)/2+tan²(x/2)+1)

 

or ∫[sec²(x/2)/tan²(x/2)+3] dx

 

but tan²(x/2)+3 is of the form u²+ a² factoring

 

let u = tan(x/2)/√3 then dx= [2√3/sec²(x/2)] du, du=

 

Substitute = ∫[2√3sec²(x/2)du/3sec²(x/2)u²+3]

 

Simplify ∫[2√3)du/3u²+3] = ∫[2√3)du/3(u²+1)]

 

taking constants outside integral = (2√3/3)∫du/(u²+1)]

 

=2/√3)∫du/(u²+1) = (2/√3) arc tan (u) but u= tan(x/2)/√3

 

= (2/√3) arc tan[ tan(x/2)/√3)+ C

 

∫1/[cos(x)+2]dx = (2/√3) arc tan[tan(x/2)/√3)]+ C