If -65 -4d x-4d x-1-2cos x1-2sin x-k - then value of k is

I=∫_π4^65π4dx(1+2^cos x)(1+2^sin x) ⇒ I=8∫_0^2πdx(1+2^cos x)(1+2^sin x) ⇒2I=8∫_0^2πdx(1+2^cos x) nbsp; ⇒2I=16∫_0^πdx(1+2^cos x) nbsp; ⇒ I=8∫_0^πdx(1+2^cos x ...

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

Byju's Answer

Standard XII

Mathematics

Theorems for Continuity

If ∫π65 π/4d ...

Question

If $\frac{65 π}{4} \int \frac{π}{4} \frac{d x}{(1 + 2^{cos x}) (1 + 2^{sin x})} = k π$ , then value of $k$ is

Open in App

Solution

$I = \frac{65 π}{4} \int \frac{π}{4} \frac{d x}{(1 + 2^{cos x}) (1 + 2^{sin x})}$
$\Rightarrow I = 8 2 π \int 0 \frac{d x}{(1 + 2^{cos x}) (1 + 2^{sin x})}$
$\Rightarrow 2 I = 8 2 π \int 0 \frac{d x}{(1 + 2^{cos x})}$
$\Rightarrow 2 I = 16 π \int 0 \frac{d x}{(1 + 2^{cos x})}$
$\Rightarrow I = 8 π \int 0 \frac{d x}{(1 + 2^{cos x})}$
$\Rightarrow 2 I = 8 π \int 0 1 d x = 8 π$
$\Rightarrow I = 4 π$

Suggest Corrections

Similar questions

Q. If

I = \int_{0}^{2 π} e^{x / 2} sin (\frac{x}{2} + \frac{π}{4}) d x,

then

I

equals

\int_{0}^{π / 4} d x = k

. Then the value of k=

\int e^{\frac{x}{2}} s i n (\frac{x}{2} + \frac{π}{4}) d x =

Q. Solve it:

\int sin \frac{x}{2} sin \frac{x}{3} cos \frac{x}{4} d x

\int e^{x / 2} sin (\frac{x}{2} + \frac{π}{4}) d x

is equal to.

Join BYJU'S Learning Program

If 65π4∫π4dx(1+2cosx)(1+2sinx)=kπ, then value of k is

I=65π4∫π4dx(1+2cosx)(1+2sinx) ⇒I=82π∫0dx(1+2cosx)(1+2sinx) ⇒2I=82π∫0dx(1+2cosx) ⇒2I=16π∫0dx(1+2cosx) ⇒I=8π∫0dx(1+2cosx) ⇒2I=8π∫01 dx=8π ⇒I=4π

If $\frac{65 π}{4} \int \frac{π}{4} \frac{d x}{(1 + 2^{cos x}) (1 + 2^{sin x})} = k π$ , then value of $k$ is