Question

HEY GENIUS...MATHS ARYABHATT....ENGLISH SHAKESPEARE....CHEMISTRY LAVOISIER...SOLVE MY QUESTIONS....​

Answer 1

Answer:

C=4

Step-by-step explanation:

Hope this will help you....!

Answer 2

Given,

$\displaystyle\longrightarrow\int\limits_0^{\frac{\pi}{2}}\dfrac{a\sin x+b\cos x}{\sin x+\cos x}\ dx=(a+b)\dfrac{\pi}{C}\quad\quad\dots(1)$

Consider the function in the integral.

$\displaystyle\longrightarrow f(x)=\dfrac{a\sin x+b\cos x}{\sin x+\cos x}$

Dividing numerator and denominator by $\cos x,$

$\displaystyle\longrightarrow f(x)=\dfrac{\left(\dfrac{a\sin x+b\cos x}{\cos x}\right)}{\left(\dfrac{\sin x+\cos x}{\cos x}\right)}$

$\displaystyle\longrightarrow f(x)=\dfrac{a\tan x+b}{\tan x+1}$

Adding and subtracting $a$ in the numerator as,

$\displaystyle\longrightarrow f(x)=\dfrac{a\tan x+a+b-a}{\tan x+1}$

$\displaystyle\longrightarrow f(x)=\dfrac{a(\tan x+1)+b-a}{\tan x+1}$

$\displaystyle\longrightarrow f(x)=a+\dfrac{b-a}{\tan x+1}$

Then (1) becomes,

$\displaystyle\longrightarrow\int\limits_0^{\frac{\pi}{2}}\left(a+\dfrac{b-a}{\tan x+1}\right)\ dx=(a+b)\dfrac{\pi}{C}$

$\displaystyle\longrightarrow\int\limits_0^{\frac{\pi}{2}}a\ dx+\int\limits_0^{\frac{\pi}{2}}\dfrac{b-a}{\tan x+1}\ dx=(a+b)\dfrac{\pi}{C}$

$\displaystyle\longrightarrow a\int\limits_0^{\frac{\pi}{2}}dx+(b-a)\int\limits_0^{\frac{\pi}{2}}\dfrac{1}{\tan x+1}\ dx=(a+b)\dfrac{\pi}{C}$

$\displaystyle\longrightarrow a\Big[x\Big]_0^{\frac{\pi}{2}}+(b-a)\int\limits_0^{\frac{\pi}{2}}\dfrac{1}{\tan x+1}\ dx=(a+b)\dfrac{\pi}{C}$

$\displaystyle\longrightarrow a\left(\dfrac{\pi}{2}-0\right)+(b-a)\int\limits_0^{\frac{\pi}{2}}\dfrac{1}{\tan x+1}\ dx=(a+b)\dfrac{\pi}{C}$

$\displaystyle\longrightarrow \dfrac{a\pi}{2}+(b-a)\int\limits_0^{\frac{\pi}{2}}\dfrac{1}{\tan x+1}\ dx=(a+b)\dfrac{\pi}{C}\quad\quad\dots(2)$

Consider the integral in the LHS.

Taking $\tan x=\dfrac{\sin x}{\cos x},$

$\displaystyle\longrightarrow\int\limits_0^{\frac{\pi}{2}}\dfrac{1}{\tan x+1}\ dx=\int\limits_0^{\frac{\pi}{2}}\dfrac{1}{\left(\dfrac{\sin x}{\cos x}+1\right)}\ dx$

$\displaystyle\longrightarrow\int\limits_0^{\frac{\pi}{2}}\dfrac{1}{\tan x+1}\ dx=\int\limits_0^{\frac{\pi}{2}}\dfrac{\cos x}{\sin x+\cos x}\ dx\quad\quad\dots(3)$

We know that,

• $\displaystyle\int\limits_a^bf(x)\ dx=\int\limits_a^bf(a+b-x)\ dx$

Using this formula, we have,

$\displaystyle\longrightarrow\int\limits_0^{\frac{\pi}{2}}\dfrac{\cos x}{\sin x+\cos x}\ dx=\int\limits_0^{\frac{\pi}{2}}\dfrac{\cos\left(0+\dfrac{\pi}{2}-x\right)}{\sin\left(0+\dfrac{\pi}{2}-x\right)+\cos\left(0+\dfrac{\pi}{2}-x\right)}\ dx$

$\displaystyle\longrightarrow\int\limits_0^{\frac{\pi}{2}}\dfrac{\cos x}{\sin x+\cos x}\ dx=\int\limits_0^{\frac{\pi}{2}}\dfrac{\cos\left(\dfrac{\pi}{2}-x\right)}{\sin\left(\dfrac{\pi}{2}-x\right)+\cos\left(\dfrac{\pi}{2}-x\right)}\ dx$

$\displaystyle\longrightarrow\int\limits_0^{\frac{\pi}{2}}\dfrac{\cos x}{\sin x+\cos x}\ dx=\int\limits_0^{\frac{\pi}{2}}\dfrac{\sin x}{\cos x+\sin x}\ dx$

$\displaystyle\longrightarrow\int\limits_0^{\frac{\pi}{2}}\dfrac{\cos x}{\sin x+\cos x}\ dx=\int\limits_0^{\frac{\pi}{2}}\dfrac{\sin x}{\sin x+\cos x}\ dx$

Then from (3),

$\displaystyle\longrightarrow\int\limits_0^{\frac{\pi}{2}}\dfrac{1}{\tan x+1}\ dx=\int\limits_0^{\frac{\pi}{2}}\dfrac{\sin x}{\sin x+\cos x}\ dx\quad\quad\dots(4)$

Adding (3) and (4),

$\displaystyle\longrightarrow2\int\limits_0^{\frac{\pi}{2}}\dfrac{1}{\tan x+1}\ dx=\int\limits_0^{\frac{\pi}{2}}\dfrac{\sin x+\cos x}{\sin x+\cos x}\ dx$

$\displaystyle\longrightarrow2\int\limits_0^{\frac{\pi}{2}}\dfrac{1}{\tan x+1}\ dx=\int\limits_0^{\frac{\pi}{2}}dx$

$\displaystyle\longrightarrow2\int\limits_0^{\frac{\pi}{2}}\dfrac{1}{\tan x+1}\ dx=\Big[x\Big]_0^{\frac{\pi}{2}}$

$\displaystyle\longrightarrow2\int\limits_0^{\frac{\pi}{2}}\dfrac{1}{\tan x+1}\ dx=\dfrac{\pi}{2}$

$\displaystyle\longrightarrow\int\limits_0^{\frac{\pi}{2}}\dfrac{1}{\tan x+1}\ dx=\dfrac{\pi}{4}$

Then (2) becomes,

$\displaystyle\longrightarrow \dfrac{a\pi}{2}+\dfrac{(b-a)\pi}{4}=(a+b)\dfrac{\pi}{C}$

$\displaystyle\longrightarrow \dfrac{2a\pi+(b-a)\pi}{4}=(a+b)\dfrac{\pi}{C}$

$\displaystyle\longrightarrow \dfrac{2a\pi+b\pi-a\pi}{4}=(a+b)\dfrac{\pi}{C}$

$\displaystyle\longrightarrow \dfrac{a\pi+b\pi}{4}=(a+b)\dfrac{\pi}{C}$

$\displaystyle\longrightarrow (a+b)\dfrac{\pi}{4}=(a+b)\dfrac{\pi}{C}$

$\displaystyle\Longrightarrow\underline{\underline{C=4}}$

Hence 4 is the answer.