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Question: Integrate using \(u = 1 - \sin x:\int {2\cos x{e^{1 - \sin x}}dx} \)...

Integrate using u=1−sin⁡x:∫2cos⁡xe1−sin⁡xdxu = 1 - \sin x:\int {2\cos x{e^{1 - \sin x}}dx}

Explanation

Solution

We have been clearly told that the integral must be solved by substitution method. Also the substitute is given to us. You have to start solving the integral by substituting the value of 1−sin⁡x1 - \sin x as u first and then integrate. after integrating, substitute the value of u again to get the final value.

Complete step by step answer:
We have been given to integrate ∫2cos⁡xe1−sin⁡xdx\int {2\cos x{e^{1 - \sin x}}dx} .Also we have been told to solve this question using a substitution method. And the substitute for 1−sin⁡x1 - \sin x is already given to us as u. Hence we will directly start integrating the given integral using the substitute.
Let I = ∫2cos⁡xe1−sin⁡xdx{\text{I = }}\int {2\cos x{e^{1 - \sin x}}dx}
Since u=1−sin⁡xu = 1 - \sin x, differentiating on both sides ,
du=−cos⁡xdx ⇒du−cos⁡x=dx du = - \cos xdx \\\ \Rightarrow \dfrac{{du}}{{ - \cos x}} = dx \\\
Using this values in the integral we get
I = ∫2cos⁡xeudu−cos⁡x{\text{I = }}\int {2\cos x{e^u}\dfrac{{du}}{{ - \cos x}}}
⇒I=∫−2eudu\Rightarrow{\text{I}} = \int { - 2{e^u}du}
Integrating with respect to u, we get
I=−2∫eudu ⇒I=−2eu+c {\text{I}} = - 2\int {{e^u}du} \\\ \Rightarrow{\text{I}}= - 2{e^u} + c \\\
Where cc is constant of integration.
Substituting the value of u=1−sin⁡xu = 1 - \sin x, we get
∴I=−2e1−sin⁡x+c\therefore{\text{I}} = - 2{e^{1 - \sin x}} + c

Hence, ∫2cos⁡xe1−sin⁡xdx\int {2\cos x{e^{1 - \sin x}}dx} is −2e1−sin⁡x+c- 2{e^{1 - \sin x}} + c.

Note: In this particular question they have mentioned that the integral must be solved by using substitution method. Hence we have no option other than to solve by substitution method. Luckily we have been given the substitute which makes it a little easier as substituting the right value is the tricky part.