Question

Solve the following: $\int{{{e}^{\log x}}.\cos xdx}$.
(A). $x\sin x-\cos x+C$
(B). $\dfrac{x}{2}\sin x+\cos x+C$
(C). $x\sin x+\cos x+C$
(D). $x\sin x+\cos 2x+C$

Answer 1

Hint: Simplify the given expression and apply integration by parts, which corresponds to product rule for differentiation. Substitute the values in the formula and simplify it.

Complete step-by-step solution -
We have been given the expression, $\int{{{e}^{\log x}}.\cos xdx}$.
Let us put, $I=\int{{{e}^{\log x}}.\cos xdx}$.
We know, ${{e}^{\log x}}=x$.
Thus, $I=\int{x.\cos x.dx}-(1)$
We can solve it by doing integration by parts. It corresponds to the product rule for differentiation.
By using integration by parts, the formula is,

& \int{uv.dx}=u\int{vdx}-\int{u'\left( \int{v.dx} \right)dx} \\\ & \int{udv}=uv-\int{vdu}-(2) \\\ \end{aligned}$$ By using ILATE Rule, the integral of two functions is taken, by considering the left term as the first function and the second term as the second function. In the given integral, $$\int{x\cos x.dx}$$. Let us take, $$u=x$$, thus by differentiation, $$u'=\dfrac{du}{dx}=1$$ Let us take, $$v=\cos x$$, We know $$\int{vdx}=\sin x$$ and $$dv=\cos x.dx$$. Now put all the values in (2). $$\therefore \int{x\cos xdx}=x\sin x-\int{\sin xdx}$$ We know, $$\int{\sin xdx}=-\cos x+C$$ $$\begin{aligned} & \therefore \int{x\cos xdx}=x\sin x-\left( -\cos x \right)+C \\\ & \int{x\cos xdx}=x\sin x+\cos x+C \\\ \end{aligned}$$ Thus we got the $$\int{{{e}^{\log x}}.\cos x.dx}=x\sin x+\cos x+C$$. $$\therefore $$ We got the required solution i.e. Option (c) is the correct answer. Note: If you choose, $$u=\cos x$$ and $$v=x$$, then you may complicate the solution, thus always use ILATE rule by doing Integration by parts. We have to choose the first function in such a way that the derivative of the function could be easily integrated.