How do you integrate \[{e^{\sin x}}\cos xdx\] ?
Answer
605.4k+ views
Hint: Here the given product of functions is to be integrated. We can observe that the expression given is having function and its derivative both present. This is a hint that we can use substitution for the solution. So we will put \[\sin x = u\] .Then the derivative of the substitution will be found and will be replaced in the given original problem.
Complete step by step solution:
Given that
\[\int {{e^{\sin x}}\cos xdx} \]
Substitute \[\sin x = u\]
Taking derivative on both sides with respect to x we get,
\[\cos xdx = du\]
Now changing the substitutions in the original integral we get,
\[\int {{e^u}du} \]
We know that \[\int {{e^x}dx} = {e^x} + C\]
Using the formula above,
\[ = {e^u} + C\]
Substituting the value of u w get,
\[ = {e^{\sin x}} + C\]
This is our integral answer.
So, the correct answer is “ \[ = {e^{\sin x}} + C\] ”.
Note: Note that if students are facing the question in how to identify the method of solution it only comes with practice. For example in the above question the method used is substitution. We saw that integral has function and derivative both so we used substitution. Don’t forget to write the constant C after finding the integral.
Complete step by step solution:
Given that
\[\int {{e^{\sin x}}\cos xdx} \]
Substitute \[\sin x = u\]
Taking derivative on both sides with respect to x we get,
\[\cos xdx = du\]
Now changing the substitutions in the original integral we get,
\[\int {{e^u}du} \]
We know that \[\int {{e^x}dx} = {e^x} + C\]
Using the formula above,
\[ = {e^u} + C\]
Substituting the value of u w get,
\[ = {e^{\sin x}} + C\]
This is our integral answer.
So, the correct answer is “ \[ = {e^{\sin x}} + C\] ”.
Note: Note that if students are facing the question in how to identify the method of solution it only comes with practice. For example in the above question the method used is substitution. We saw that integral has function and derivative both so we used substitution. Don’t forget to write the constant C after finding the integral.
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