integrate 1/(a+bcosx)^2 dx

integrate 1/(a+bcosx)^2 dx


2 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
8 years ago
Ans:Hello student, please find answer to your question
I = \int \frac{1}{(a+bcosx)^{2}}dx
I = \int \frac{1}{(a+b(\frac{1-tan^{2}(\frac{x}{2})}{1+tan^{2}(\frac{x}{2})}))^{2}}dx
I = \int \frac{(1+tan^{2}(\frac{x}{2}))^{2}}{((a+b)+(a-b)tan^{2}(\frac{x}{2}))^{2}}dx
I = \int \frac{(1+tan^{2}(\frac{x}{2})).sec^{2}(\frac{x}{2})}{((a+b)+(a-b)tan^{2}(\frac{x}{2}))^{2}}dx
tan(\frac{x}{2}) = t
sec^{2}(\frac{x}{2}).\frac{1}{2}dx = dt
I = \int \frac{2(1+t^{2})}{((a+b)+(a-b)t^{2})^{2}}dt
Now, simply use the standard form & partial farction

I = \frac{2a.tanh^{-1}(\frac{(a-b)tan\frac{x}{2}}{\sqrt{b^{2}-a^{2}}})}{(b^{2}-a^{2})^{3/2}}+ \frac{bsin(x)}{(b^{2}-a^{2})(a+bcos(x))} + cons.
Nishant Dahma
13 Points
3 years ago
Put 1/(a+bcosx)=t
Then, bsinx/(a+bcosx)^2=dt/dx
Therefore, {1/(a+bcosx)^2}dx=(cosecx/b)dt
Now find the value of bsinx from the substitution. And then differentiate using partial fractions to get the answer.

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