Question

Which of the following integrals are evaluated correctly ?
 (i) $\int {\dfrac{{x{e^x}}}{{{{(1 + x)}^2}}}} dx = \dfrac{{{e^x}}}{{{{(1 + x)}^2}}} + C$
(ii) \[\int {{e^x}(} \dfrac{{1 + \sin x}}{{1 + \cos x}})dx = {e^x}{\sec ^2}\dfrac{x}{2} + C\]
(iii) \[\int {\dfrac{{{e^x}(x - 3)}}{{{{(x - 1)}^3}}}} dx = \dfrac{{{e^x}}}{{{{(x - 1)}^2}}} + C\]
(iv) $\int {{e^{2x}}} \sin xdx = \dfrac{{{e^{2x}}}}{5}(cosx - 2sinx) + C$
(a) Only(iii) b) (ii) and(iii)
  (c) (i), (ii) & (iv) d) All of these

Answer 1

Hint: Whenever integral function consists of two functions or more then reduce it into standard from . Use integration by part method of integration to solve it easily.

Complete step-by-step answer:
 We must check every integral one by one to get an answer, and one thing we can observe from every option is that every integral consists of more than one type of function.
Every integral in the option consists of two kinds of function i.e. one exponential function and other algebraic function so, it’s clear that we must reduce it into more standard form, so we use integration by part method.
In integration by part method, the integral of the two functions are taken, by considering one function as first and other as a second function using ILATE method which is a method of choosing functions of integration by parts.
I stand for inverse function.
L stands for logarithmic function.
A stands for algebraic function.
T stands for trigonometric function.
E stands for exponential function.
$\int {uvdx = u\int {vdx - \int {(\dfrac{{du}}{{dx}}} } } \int {vdx} )dx$ …(1.1)
Here u is taken as the first function and v as a second function.
Now,
(i) $\int {\dfrac{{x{e^x}}}{{{{(1 + x)}^2}}}} dx = \dfrac{{{e^x}}}{{{{(1 + x)}^2}}} + C$
Taking LEFT HAND SIDE,
\[\]
We will rearrange the expression,
$I = \int {x{e^x}(\dfrac{1}{{{{(1 + x)}^2}}})} dx$
Now,
$u = x{e^x}$ and $v = \dfrac{1}{{{{(1 + x)}^2}}}$
Use equation (1.1) method of integration by part.
$I = x{e^x}\int {\dfrac{1}{{{{(1 + x)}^2}}}} - \int {\dfrac{{d(x{e^x})}}{{dx}}} (\int {\dfrac{1}{{{{(1 + x)}^2}}}dx)dx} $
We know, $\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} $ and product rule of derivation $\dfrac{{d(uv)}}{{dx}} = u\dfrac{{dv}}{{dx}} + v\dfrac{{du}}{{dx}}$
So,
$I = x{e^x}(\dfrac{{ - 1}}{{1 + x}}) - \int {({e^x} + x{e^x})} (\dfrac{{ - 1}}{{1 + x}})dx$ , using these formula of derivation$\dfrac{{d{e^x}}}{{dx}} = {e^x}$ &$\dfrac{{d{x^n}}}{{dx}} = n{x^{n - 1}}$
Now, taking common usage from the term.
$I = x{e^x}(\dfrac{{ - 1}}{{1 + x}}) - \int {{e^x}(1 + x)} (\dfrac{{ - 1}}{{1 + x}})dx + C$
\[I = x{e^x}(\dfrac{{ - 1}}{{1 + x}}) - \int {( - {e^x}} )dx = C\]
We know ,
So,
\[I = x{e^x}(\dfrac{{ - 1}}{{1 + x}}) + {e^x} + C\]
It’s clear that,
LEFT HAND SIDE$ \ne $ RIGHT HAND SIDE.
(ii) \[\int {{e^x}(} \dfrac{{1 + \sin x}}{{1 + \cos x}})dx = {e^x}{\sec ^2}\dfrac{x}{2} + C\]
LEFT HAND SIDE$ = \int {{e^x}(} \dfrac{{1 + \sin x}}{{1 + \cos x}})dx$
$I = \int {{e^x}(} \dfrac{{1 + \sin x}}{{1 + \cos x}})dx$
Here, we need to use half angle formula i.e.
$\begin{gathered}
  \cos 2x = 2{\cos ^2}x - 1 \\
  \cos 2x + 1 = 2{\cos ^2}x \\
\end{gathered} $
Now, replace $x$ with
So, $\cos x + 1 = 2{\cos ^2}\dfrac{x}{2}$
And
$\sin 2x = 2\sin x\cos x$
$\sin x = 2\sin \dfrac{x}{2}\cos \dfrac{x}{2}$
And we know $1 = {\cos ^2}x + {\sin ^2}x$ and we can write it as $1 = {\sin ^2}\dfrac{x}{2} + {\cos ^2}\dfrac{x}{2}$
Now,
I\[I = \int {{e^x}\dfrac{{(co{s^2}\dfrac{x}{2} + {{\sin }^2}\dfrac{x}{2} + 2\sin \dfrac{x}{2}\cos \dfrac{x}{2})}}{{2{{\cos }^2}\dfrac{x}{2}}}} dx\]
We know, ${(a + b)^2} = {a^2} + {b^2} + 2ab$
\[\begin{gathered}
  I = \int {{e^x}\dfrac{{{{(\cos \dfrac{x}{2} + \sin \dfrac{x}{2})}^2}}}{{2{{(cos\dfrac{x}{2})}^2}}}} dx \\
  I = \dfrac{1}{2}\int {{e^x}{{(\dfrac{{\cos \dfrac{x}{2} + \sin \dfrac{x}{2}}}{{cos\dfrac{x}{2}}})}^2}dx} \\
\end{gathered} \]
Separating the denominator,
\[I = \dfrac{1}{2}\int {{e^x}{{(1 + tan\dfrac{x}{2})}^2}} dx\]
We know $1 + {\tan ^2}x = {\sec ^2}x$ ,
\[\begin{gathered}
  I = \dfrac{1}{2}\int {{e^x}(1 + ta{n^2}\dfrac{x}{2} + 2\tan \dfrac{x}{2}} )dx \\
  I = \dfrac{1}{2}\int {{e^x}({{\sec }^2}\dfrac{x}{2} + 2\tan \dfrac{x}{2}} )dx \\
  I = \dfrac{1}{2}\int {{e^x}{{\sec }^2}\dfrac{x}{2}dx + \dfrac{1}{2}\int {{e^x}} \tan \dfrac{x}{2}} dx \\
\end{gathered} \]
Using by part method in the second term using equation (1.1).
\[I = \dfrac{1}{2}\int {{e^x}{{\sec }^2}\dfrac{x}{2}dx + \dfrac{1}{2}(\tan \dfrac{x}{2}} \int {{e^x}} - \int {\dfrac{{d\tan \dfrac{x}{2}}}{{dx}}(\int {{e^x}dx)dx)} } \]
We know, $\dfrac{{d\tan \dfrac{x}{2}}}{{dx}} = {\sec ^2}\dfrac{x}{2}$
So,
\[\begin{gathered}
  I = \dfrac{1}{2}\int {{e^x}{{\sec }^2}\dfrac{x}{2}dx + \dfrac{1}{2}(\tan \dfrac{x}{2}} {e^x} - \int {se{c^2}\dfrac{x}{2}{e^x}} dx) \\
  I = \dfrac{1}{2}\int {{e^x}{{\sec }^2}\dfrac{x}{2}dx + \dfrac{1}{2}\tan \dfrac{x}{2}} {e^x} - \dfrac{1}{2}\int {{e^x}se{c^2}\dfrac{x}{2}dx} \\
\end{gathered} \]
In this two terms are same with opposite sign, we get
$I = \dfrac{1}{2}\tan \dfrac{x}{2}{e^x} + C$
So, It’s clear that
LEFT HAND SIDE RIGHT HAND SIDE.
(iii) \[\]
LEFT HAND SIDE$ = \int {\dfrac{{{e^x}(x - 3)}}{{{{(x - 1)}^3}}}} dx$
\[I = \int {\dfrac{{{e^x}(x - 3)}}{{{{(x - 1)}^3}}}} dx\]
Here, we can’t use the integration by part directly, so we will alter the equation.
$I = \int {\dfrac{{{e^x}(x - 1 - 2)}}{{{{(x - 1)}^3}}}} dx$
Separating the denominator,
$I = \int {{e^x}} (\dfrac{{x - 1}}{{{{(x - 1)}^3}}} - \dfrac{2}{{{{(x -1)}^3}}})dx$
$I = \int {{e^x}} \dfrac{1}{{{{(x - 1)}^2}}}dx - 2\int {\dfrac{{{e^x}}}{{{{(x - 1)}^{^3}}}}} dx$
We use integration by part in the first term using equation (1.1).
 $\begin{gathered}
  I = \dfrac{1}{{{{(x - 1)}^2}}}\int {{e^x}} dx - \int {\dfrac{{d(\dfrac{1}{{{{(x - 1)}^2}}})}}{{dx}}(\int {{e^x}dx)dx} } - 2\int {\dfrac{{{e^x}}}{{{{(x - 1)}^{^3}}}}} dx \\
  I = \dfrac{1}{{{{(x - 1)}^2}}}{e^x} - \int {\dfrac{{ - 2{e^x}}}{{{{(x - 1)}^3}}}} dx - 2\int {\dfrac{{{e^x}}}{{{{(x - 1)}^{^3}}}}} dx + C \\
\end{gathered} $
Since, We know $\dfrac{{d{x^n}}}{{dx}} = n{x^{n - 1}}$
Here two terms are same in magnitude with opposite sign so they must be cancelled out so we get,
$I = \dfrac{1}{{{{(x + 1)}^2}}}{e^x} + C$
It’s clear that LEFT HAND SIDE$ = $ RIGHT HAND SIDE.
(iv) $\int {{e^{2x}}} \sin xdx = \dfrac{{{e^{2x}}}}{5}(cosx - 2sinx) + C$
LEFT HAND SIDE$ = \int {{e^{2x}}} \sin xdx$
$I = \int {{e^{2x}}} \sin xdx$ …(1.2)
Using by part using equation (1.1),
$I = \sin x\int {{e^{2x}}} dx - \int {\dfrac{{d(sinx)}}{{dx}}} (\int {{e^{2x}}dx)dx} $
We know $\dfrac{{d\sin x}}{{dx}} = \cos x$
So,
\[\begin{gathered}
  I = \dfrac{{\sin x{e^{2x}}}}{2} - \int {\cos x\dfrac{{{e^{2x}}}}{2}} dx \\
  I = \dfrac{{\sin x{e^{2x}}}}{2} - \cos x\int {\dfrac{{{e^{2x}}}}{2}dx + \int {\dfrac{{d\cos x}}{{dx}}} } (\int {\dfrac{{{e^{2x}}}}{2}dx)dx.} \\
\end{gathered} \]
Here, we used the part formula again in the second term.
And we know $\dfrac{{d\cos x}}{{dx}} = - \sin x$
\[\begin{gathered}
  I = \dfrac{{\sin x{e^{2x}}}}{2} - \dfrac{{\cos x{e^{2x}}}}{4} + \int {( - sinx)\dfrac{{{e^{2x}}}}{4}dx} \\
  I = \dfrac{{\sin x{e^{2x}}}}{2} - \dfrac{{\cos x{e^{2x}}}}{4} - \dfrac{1}{4}\int {sinx{e^{2x}}dx} \\
\end{gathered} \]
From equation (1.2)
\[I = \dfrac{{\sin x{e^{2x}}}}{2} - \dfrac{{\cos x{e^{2x}}}}{4} - \dfrac{1}{4}I\]
\[I + \dfrac{I}{4} = \dfrac{{\sin x{e^{2x}}}}{2} - \dfrac{{\cos x{e^{2x}}}}{4}\]
\[\dfrac{{5I}}{4} = \dfrac{{\sin x{e^{2x}}}}{2} - \dfrac{{\cos x{e^{2x}}}}{4}\]
\[I = \dfrac{4}{5}(\dfrac{{\sin x{e^{2x}}}}{2} - \dfrac{{\cos x{e^{2x}}}}{4})\]
Taking common and $\dfrac{1}{4}$ common from each term and solve it further,
\[I = \dfrac{{{e^{2x}}}}{5}(2sinx - \cos x) + C\]
It’s clear that LEFT HAND SIDE$ \ne $ RIGHT HAND SIDE.
So, only option (iii) is right.
Which means option a) is right.
Note: We can use a shortcut formula rather than using integration by part always to save our time.
If integral is in this form:
$\int {{e^x}[f(x) + \dfrac{{df\left( x \right)}}{{dx}}]dx} = {e^x}f\left( x \right) + C$
Carefully use the integration by part, considering the right function as first and second.