Question

Integration

1/a+bcosx

urgent........​

Answer 1

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integrate dx/(a+bcosx)

put cosx=(1-tan^2(x/2))/(1+tan^(x/2))

integrate sec^2(x/2) dx/a(1+tan^2(x/2)+b(1-tan^2(x/2))

=integrate sec^2(x/2)dx/(a+b)+(a-b)tan^2(x/2)

put tan(x/2)=t

diff w. r. to x we get

sec^2(x/2).1/2 dx=dt

=integrate 2dt/(a+b)+(a-b)t^2

=2/sqrt(a+b) arctan( sqrt(a-b)t/sqrt(a+b))

=2/sqrt(a+b) arctan(sqrt(a-b)tan(x/2)/sqrt(a+b))

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Answer 2

Answer:

hiii

your answer is here. !

Step-by-step explanation:

=> There are three cases here.

1. |a|>|b|

=> let c=(a^2-b^2).

=> The result is then (2/c)*arctan(c*tan(x/2))/(a+b)

+C

2. |a|<|b|

=> let c=sqrt(b^2-c^2)

=> The result is then

(1/c)*ln((c*tan(x/2)+a+b)/(c*tan(x/2) -a -b)) +C

3. a=b

=> If you replace 1 +cos(x) with (sin(x/2))^2 the integral reduces to integral of

=> dx/(sin(x/2))^2.

=> This is a standard integral.

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