Solveeit Logo
Login

Question

Question: Evaluate \(\int\limits_{0}^{\dfrac{\pi }{2}}{{{e}^{x}}}(\sin x-\cos x)dx.\)...

Evaluate ∫0π2ex(sin⁡x−cos⁡x)dx.\int\limits_{0}^{\dfrac{\pi }{2}}{{{e}^{x}}}(\sin x-\cos x)dx.

Explanation

Solution

Hint: The given integration is of type∫ex[f(x)+f′(x)]dx\int{{{e}^{x}}\left[ f(x)+{{f}^{'}}(x) \right]}dx.
we can use ∫ex[f(x)+f′(x)]dx\int{{{e}^{x}}\left[ f(x)+{{f}^{'}}(x) \right]}dx =ex.f(x)+c={{e}^{x}}.f(x)+c, where c is constant. So first we find integration by using the property and after that we put limits.

Complete step-by-step answer:
Now we have to evaluate ∫0π2ex(sin⁡x−cos⁡x)dx\int\limits_{0}^{\dfrac{\pi }{2}}{{{e}^{x}}}(\sin x-\cos x)dx
⇒−\Rightarrow - ∫0π2ex(cos⁡x−sin⁡x)dx\int\limits_{0}^{\dfrac{\pi }{2}}{{{e}^{x}}}(\cos x-\sin x)dx
Here we take – sign outside of the integral sign. Now suppose f(x)=cos⁡xf(x)=\cos x so we can write f′(x)=−sin⁡x{{f}^{'}}(x)=-\sin x
So we have∫ex[f(x)+f′(x)]dx\int{{{e}^{x}}\left[ f(x)+{{f}^{'}}(x) \right]}dx =ex.f(x)+c={{e}^{x}}.f(x)+c
⇒−\Rightarrow - ∫0π2ex(cos⁡x+(−sin⁡x))dx\int\limits_{0}^{\dfrac{\pi }{2}}{{{e}^{x}}}(\cos x+(-\sin x))dx=∫0π2ex(f(x)+f′(x))dx\int\limits_{0}^{\dfrac{\pi }{2}}{{{e}^{x}}}(f(x)+f'(x))dx
⇒−\Rightarrow - ∫0π2ex(cos⁡x−sin⁡x)dx\int\limits_{0}^{\dfrac{\pi }{2}}{{{e}^{x}}}(\cos x-\sin x)dx=[exf(x)]0π2\left[ {{e}^{x}}f(x) \right]_{0}^{\dfrac{\pi }{2}}
Putting value of f(x) =cosx we can write
⇒−\Rightarrow - ∫0π2ex(cos⁡x−sin⁡x)dx\int\limits_{0}^{\dfrac{\pi }{2}}{{{e}^{x}}}(\cos x-\sin x)dx=[excos⁡x]0π2\left[ {{e}^{x}}\cos x \right]_{0}^{\dfrac{\pi }{2}}

Now substituting lower and upper limits we have
⇒−\Rightarrow - ∫0π2ex(cos⁡x−sin⁡x)dx\int\limits_{0}^{\dfrac{\pi }{2}}{{{e}^{x}}}(\cos x-\sin x)dx=[eπ2cos⁡π2−e0cos⁡0]\left[ {{e}^{\dfrac{\pi }{2}}}\cos \dfrac{\pi }{2}-{{e}^{0}}\cos 0 \right]
As we know cos⁡π2=0;e0=1;and cos0=1\cos \dfrac{\pi }{2}=0;{{e}^{0}}=1;and\text{ cos0=1}
So we can write
⇒−\Rightarrow - ∫0π2ex(cos⁡x−sin⁡x)dx\int\limits_{0}^{\dfrac{\pi }{2}}{{{e}^{x}}}(\cos x-\sin x)dx=[eπ2×0−e0×1]\left[ {{e}^{\dfrac{\pi }{2}}}\times 0-{{e}^{0}}\times 1 \right]
⇒−\Rightarrow - ∫0π2ex(cos⁡x−sin⁡x)dx\int\limits_{0}^{\dfrac{\pi }{2}}{{{e}^{x}}}(\cos x-\sin x)dx=−1-1
⇒∫0π2ex(cos⁡x−sin⁡x)dx=1\Rightarrow \int\limits_{0}^{\dfrac{\pi }{2}}{{{e}^{x}}}(\cos x-\sin x)dx=1
Hence
∫0π2ex(cos⁡x−sin⁡x)dx=1\int\limits_{0}^{\dfrac{\pi }{2}}{{{e}^{x}}}(\cos x-\sin x)dx=1

Note: This integration can be solved by integrating ∫0π2ex(cos⁡x)dx−∫0π2sin⁡xdx\int\limits_{0}^{\dfrac{\pi }{2}}{{{e}^{x}}}(\cos x)dx-\int\limits_{0}^{\dfrac{\pi }{2}}{\sin x}dx separately by using integration by parts for∫0π2ex(cos⁡x)dx\int\limits_{0}^{\dfrac{\pi }{2}}{{{e}^{x}}}(\cos x)dx