The value of - cos2x-1-ax dx- a-0 is

Let I=∫_ π^π cos^2x/1+a^x dx=∫_ π^πcos^2 x/1+a^ π+π xdx= ∫_ π^πa^x.cos^2x/1+a^xdx ∴ 2I=∫_ π^πcos^2x(1+a^x)/1+a^xdx=∫_ π^π cos^2x dx=2 ∫_0^π cos^2x dx ⇒ I=2 ∫_0^ ...

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Byju's Answer

Standard XII

Mathematics

Integration of Piecewise Continuous Functions

The value of ...

Question

The value of $\int_{- π}^{π} \frac{c o s^{2} x}{1 + a^{x}} d x, a > 0 i s$

$a π$

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$\frac{π}{2}$

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$\frac{π}{a}$

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$2 π$

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Solution

The correct option is B
$\frac{π}{2}$

$L e t I = \int_{- π}^{π} \frac{c o s^{2} x}{1 + a^{x}} d x = \int_{- π}^{π} \frac{c o s^{2} x}{1 + a^{- π + π - x}} d x = \int_{- π}^{π} \frac{a^{x} . c o s^{2} x}{1 + a^{x}} d x$
$∴ 2 I = \int_{- π}^{π} \frac{c o s^{2} x (1 + a^{x})}{1 + a^{x}} d x = \int_{- π}^{π} c o s^{2} x d x = 2 \int_{0}^{π} c o s^{2} x d x \Rightarrow I = 2 \int_{0}^{\frac{π}{2}} c o s^{2} x d x$
$= 2. \frac{1}{2} . \frac{π}{2} = \frac{π}{2}$

Suggest Corrections

The value of ∫π−π cos2x1+ax dx, a>0 is

The correct option is B π2 Let I=∫π−π cos2x1+ax dx=∫π−πcos2 x1+a−π+π−xdx=∫π−πax.cos2x1+axdx ∴2I=∫π−πcos2x(1+ax)1+axdx=∫π−πcos2x dx=2∫π0 cos2x dx⇒I=2∫π20cos2x dx =2.12.π2=π2

The value of $\int_{- π}^{π} \frac{c o s^{2} x}{1 + a^{x}} d x, a > 0 i s$