- - 2 - - 2 log 2-sin x 2-sin x d x

Let #160;I= #8747; #960;2 #960;2log2 sinx2+sinxdxHere, #160;fx=log2 sinx2+sinxf x=log2 sin x2+sin x=log2+sinx2 sinx= log2 sinx2+sinx= fxHence #160; ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Differentiability

∫-π / 2 π / 2...

Question

$\int_{- π / 2}^{π / 2} \log (\frac{2 - \sin x}{2 + \sin x}) d x$

Open in App

Solution

$L e t I = \int_{- \frac{π}{2}}^{\frac{π}{2}} \log (\frac{2 - \sin x}{2 + \sin x}) d x Here, f (x) = l og (\frac{2 - \sin x}{2 + \sin x}) f (- x) = l og (\frac{2 - \sin (- x)}{2 + \sin (- x)}) = l og (\frac{2 + \sin x}{2 - \sin x}) = - l og (\frac{2 - \sin x}{2 + \sin x}) = - f (x) Hence f (x) is an odd function ∴ I = 0$

Suggest Corrections

Similar questions

Q. Solve:

\int_{- π / 2}^{π / 2} log (\frac{2 - sin x}{2 + sin x}) d x

\int_{- π / 2}^{π / 2} log (\frac{2 - sin θ}{2 + sin θ}) d θ = ?

\int_{- π / 2}^{π / 2} log (\frac{2 - sin x}{2 + sin x}) d x =

Q. Evaluate :

\int_{- π / 2}^{π / 2} log (\frac{2 - sin x}{2 + sin x}) d x

Q. The value of

\int_{- π / 2}^{π / 2} sin x d x

is equal to

Join BYJU'S Learning Program

∫-π/2π/2log2-sin x2+sin x dx

Let I=∫-π2π2log2-sinx2+sinxdxHere, fx=log2-sinx2+sinxf-x=log2-sin-x2+sin-x=log2+sinx2-sinx=-log2-sinx2+sinx=-fxHence fx is an odd function∴ I =0

$\int_{- π / 2}^{π / 2} \log (\frac{2 - \sin x}{2 + \sin x}) d x$