Integrate using $u = 1 - \sin x:\int {2\cos x{e^{1 - \sin x}}dx} $
Answer
579.6k+ views
Hint: 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 $1 - \sin x$ 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 $\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 $1 - \sin x$ is already given to us as u. Hence we will directly start integrating the given integral using the substitute.
Let ${\text{I = }}\int {2\cos x{e^{1 - \sin x}}dx} $
Since $u = 1 - \sin x$, differentiating on both sides ,
$du = - \cos xdx \\
\Rightarrow \dfrac{{du}}{{ - \cos x}} = dx \\ $
Using this values in the integral we get
${\text{I = }}\int {2\cos x{e^u}\dfrac{{du}}{{ - \cos x}}} $
$\Rightarrow{\text{I}} = \int { - 2{e^u}du} $
Integrating with respect to u, we get
${\text{I}} = - 2\int {{e^u}du} \\
\Rightarrow{\text{I}}= - 2{e^u} + c \\ $
Where $c$ is constant of integration.
Substituting the value of $u = 1 - \sin x$, we get
$\therefore{\text{I}} = - 2{e^{1 - \sin x}} + c$
Hence, $\int {2\cos x{e^{1 - \sin x}}dx}$ is $- 2{e^{1 - \sin x}} + c$.
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.
Complete step by step answer:
We have been given to integrate $\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 $1 - \sin x$ is already given to us as u. Hence we will directly start integrating the given integral using the substitute.
Let ${\text{I = }}\int {2\cos x{e^{1 - \sin x}}dx} $
Since $u = 1 - \sin x$, differentiating on both sides ,
$du = - \cos xdx \\
\Rightarrow \dfrac{{du}}{{ - \cos x}} = dx \\ $
Using this values in the integral we get
${\text{I = }}\int {2\cos x{e^u}\dfrac{{du}}{{ - \cos x}}} $
$\Rightarrow{\text{I}} = \int { - 2{e^u}du} $
Integrating with respect to u, we get
${\text{I}} = - 2\int {{e^u}du} \\
\Rightarrow{\text{I}}= - 2{e^u} + c \\ $
Where $c$ is constant of integration.
Substituting the value of $u = 1 - \sin x$, we get
$\therefore{\text{I}} = - 2{e^{1 - \sin x}} + c$
Hence, $\int {2\cos x{e^{1 - \sin x}}dx}$ is $- 2{e^{1 - \sin x}} + c$.
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.
Recently Updated Pages
Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 12 Business Studies: Engaging Questions & Answers for Success

Trending doubts
Which are the Top 10 Largest Countries of the World?

Draw a labelled sketch of the human eye class 12 physics CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Why is the cell called the structural and functional class 12 biology CBSE

Draw ray diagrams each showing i myopic eye and ii class 12 physics CBSE

Which is the correct genotypic ratio of mendel dihybrid class 12 biology CBSE

