Question

The value of $\int{\left( x{{e}^{\ln \sin x}}-\cos x \right)dx}$ is equal to:

(a) $-x\cos x+c$
(b) $\sin x-x\cos x+c$
(c) $-{{e}^{\ln x}}\cos x+c$
(d) $\sin x+x\cos x+c$

Answer 1

Hint: Remove the bracket and write the two terms separately under integral sign. Now, leave the integral of cosine as it is and simplify the exponential term. Then use integration by parts to solve the first term.

We already know about integration but here we are going to use a method called integration by parts. Let us know about this method. In calculus, and more generally in mathematical analysis, integration by parts or partial integration is a process that finds the integral of a product of functions in terms of the integral of the product of their derivative and antiderivative. It is frequently used to transform the antiderivative of a product of functions into an antiderivative for which a solution can be found more easily.

Complete step-by-step answer:
If $u=u(x)$ and $du=u'(x)dx$ while $v=v(x)$ and $dv=v'(x)dx$, then the integration by parts formula states that: $\int{uvdx}=u\int{vdx}-\int{\left[ \dfrac{du}{dx}\times \left( \int{vdx} \right) \right]dx}$. What we have done is, in the first part $u$ is multiplied to the integral of $v$ and in the second part we have taken the integration of product of differentiation of $u$and integration of v. now, the problem is which function is to be considered as $u$ and which function as $v$. So, there is a rule called ILATE rule for this.

I- Inverse function

L- Logarithmic function

A- Algebraic function

T- Trigonometric function

E- Exponential function

If there is product of any of these two functions then consider then consider ‘$u$’ as the function which comes first according to ILATE. For example, in a product of inverse and logarithmic function, consider inverse function as ‘$u$’.

Now, let us come to the question. We have,

$I=$$\int{\left( x{{e}^{\ln \sin x}}-\cos x \right)dx}$

There is an important property of logarithm which is: ${{a}^{{{\log }_{a}}x}}=x$. When the base of log is the same as the base of the whole exponent, then the result is the argument of log. Hence, ${{e}^{\ln \sin x}}=\sin x$.

$\begin{align}
  & \therefore I=\int{\left( x\sin x-\cos x \right)dx} \\
 & =\int{x\sin xdx-\int{\cos xdx}}..................(i) \\
\end{align}$

Now, considering $x$ as ‘$u$’ using ILATE, we have,

$\begin{align}
  & I=x\int{\sin xdx}-\int{\dfrac{dx}{dx}\times \left( \int{\sin xdx} \right)dx-\int{\cos xdx}} \\
 & =x(-\cos x)-\int{(-\cos x)dx-}\int{\cos xdx} \\
 & =-x\cos x+\int{\cos xdx}-\int{\cos xdx}+c \\
 & =-x\cos x+c \\
\end{align}$

Here, $c$ is any arbitrary constant.

Hence, option (a) is the correct answer.


Note: You can see that we have left the integration of $\cos x$ initially and later it gets cancelled. You can also integrate it initially, that will not bother the answer. We have simplified the exponential term because it would be difficult for us to integrate ${{e}^{\ln \sin x}}$.