The correct option is B 1I=∫_ π/2^π/2 cos x/1+e^xdx=∫_ π/2^π/2cos( x)/1+e^ xdx=∫_π/2^π/2 e^x.cos x/e^x 1dx 2I=∫_ π/2^π/2(1+e^x)cos x/1+e^xdx=2 ∫_ π/2^0 cos x d ...

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

Integration of Piecewise Continuous Functions

∫-π/2π/105× x...

Question

$\int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{c o s x}{1 + e^{x}} d x =$

0.0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

-2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D 1
$I = \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{c o s x}{1 + e^{x}} d x = \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{c o s (- x)}{1 + e^{- x}} d x = \int_{\frac{π}{2}}^{\frac{π}{2}} \frac{e^{x} . c o s x}{e^{x} - 1} d x$
$2 I = \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{(1 + e^{x}) c o s x}{1 + e^{x}} d x = 2 \int_{- \frac{π}{2}}^{0} c o s x d x = 2 [s i n x]_{0}^{\frac{π}{2}} = 2 \Rightarrow I = 1$

Suggest Corrections

Similar questions

I = \int_{- \frac{π}{2}}^{\frac{π}{2}} {sin}^{7} x d x

Then

I = ?

Q. Integrate

\int_{- π / 2}^{π / 2} e^{{sin}^{- 2} x} . {sin}^{2 n + 1} x d x =

\int_{\frac{π}{4}}^{\frac{π}{2}} e^{x} (l o g s i n x + c o t x) d x =

Q. lf

f : R \to R

and

g : R \to R

are continuous then

\int_{- π / 2}^{π / 2} (f (x) + f (- x)) (g (x) - g (- x)) d x =

Q. Let

f

and

g

be two continuous functions. Then

\int_{- π / 2}^{π / 2} {f (x) + f (- x)} {g (x) - g (- x)} d x

is equal to

Join BYJU'S Learning Program

∫π2−π2 cos x1+exdx=

The correct option is D 1I=∫π2−π2 cos x1+exdx=∫π2−π2cos(−x)1+e−xdx=∫π2π2 ex.cos xex−1dx 2I=∫π2−π2(1+ex)cos x1+exdx=2 ∫0−π2 cos x dx=2[sin x]π20=2⇒I=1

$\int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{c o s x}{1 + e^{x}} d x =$