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Integral Calculus> integrate 1/(a+bcosx)^2 dx...

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integrate 1/(a+bcosx)^2 dx

Gunda , 11 Years ago

Grade 12

2 Answers

Jitender Singh

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

$I = \frac{2a.tanh^{-1}(\frac{(a-b)tan\frac{x}{2}}{\sqrt{b^{2}-a^{2}}})}{(b^{2}-a^{2})^{3/2}}+ \frac{bsin(x)}{(b^{2}-a^{2})(a+bcos(x))} + cons.$

Last Activity: 11 Years ago

Nishant Dahma

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.

Last Activity: 6 Years ago

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