Question

Tell me fast.... integrity maths​

Answer 1

Given to evaluate,

$\displaystyle\longrightarrow I=\int\limits_0^{\frac{\pi}{2}}\int\limits_{\frac{\pi}{2}}^{\pi}\cos(x+y)\ dy\ dx$

or,

$\displaystyle\longrightarrow I=\int\limits_0^{\frac{\pi}{2}}\left[\int\limits_{\frac{\pi}{2}}^{\pi}\cos(x+y)\ dy\right]\ dx\quad\quad\dots (1)$

Note that $x$ and $y$ are treated as variables independent to each other.

Substitute,

$\longrightarrow u=x+y$

Differentiating wrt $y,$ we get,

$\longrightarrow dy=du$

At $y=\dfrac{\pi}{2} ,$

$\longrightarrow u=x+\dfrac{\pi}{2}$

At $y=\pi,$

$\longrightarrow u=x+\pi$

Then (1) becomes,

$\displaystyle\longrightarrow I=\int\limits_0^{\frac{\pi}{2}}\left[\int\limits_{x+\frac{\pi}{2}}^{x+\pi}\cos u\ du\right]\ dx$

$\displaystyle\longrightarrow I=\int\limits_0^{\frac{\pi}{2}}\Big[\sin u\ du\Big]_{x+\frac{\pi}{2}}^{x+\pi}\ dx$

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

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

$\displaystyle\longrightarrow I=-\Big[\cos(x+\pi)\Big]_0^{\frac{\pi}{2}}+\left[\cos\left(x+\dfrac{\pi}{2}\right)\right]_0^{\frac{\pi}{2}}$

$\displaystyle\longrightarrow I=-\cos\left(\dfrac{3\pi}{2}\right)+\cos\pi+\cos\pi-\cos\left(\dfrac{\pi}{2}\right)$

$\displaystyle\longrightarrow\underline{\underline{I=-2}}$

Hence -2 is the answer.

Answer 2

$\large\underline\purple{\bold{Solution :- }}$

─━─━─━─━─━─━─━─━─━─━─━─━─

$\tt \: I=\int\limits_0^{\frac{\pi}{2}}\int\limits_{\frac{\pi}{2}}^{\pi}\tt \: cos(x+y)\tt dy\tt dx$

$\tt \: I=\int\limits_0^{\frac{\pi}{2}} \bigg(\int\limits_{\frac{\pi}{2}}^{\pi}\tt \: cos(x+y)\tt dy \bigg)\tt dx$

$\tt \: I=\int\limits_0^{\frac{\pi}{2}}\Big[\sin (x \: + \:y )\ \Big]_{\frac{\pi}{2}}^{\pi}\ dx$

$\tt \: I=\int\limits_0^{\frac{\pi}{2}}\Big(\sin\left(x+\pi\right)-\sin\left(x+\dfrac{\pi}{2 }\right)\Big)\ dx$

$\tt \: I=\int\limits_0^{\frac{\pi}{2}}\Big( - sinx- \: cosx \Big)\ dx$

$\tt \: I= \: [cosx \: - \: sinx]_0^ \frac{\pi}{2}$

$\tt \: I= \: (cos \frac{\pi}{2} - sin \frac{\pi}{2} ) - (cos0 \: - sin0)$

$\tt \: I= \: (0 - 1) - (1 - 0)$

$\tt \: I= \: - \: 2$

$\tt \: \therefore\int\limits_0^{\frac{\pi}{2}}\int\limits_{\frac{\pi}{2}}^{\pi}\tt \: cos(x+y)\tt dy\tt dx \: = \: - \: 2$