Question

The value of the integral $\int {x\sin x{{\sec }^3}xdx} $ is equal to
$\left( a \right)\dfrac{1}{2}\left[ {{{\sec }^2}x - \tan x} \right] + c$
$\left( b \right)\dfrac{1}{2}\left[ {x{{\sec }^2}x - \tan x} \right] + c$
$\left( c \right)\dfrac{1}{2}\left[ {x{{\sec }^2}x + \tan x} \right] + c$
$\left( d \right)\dfrac{1}{2}\left[ {{{\sec }^2}x + \tan x} \right] + c$

Answer 1

Hint: In this particular type of question use the concept that, sec x = (1/cos x) and (sin x/cos x) = tan x so first simplify the integral according to this property than apply integration by parts formula i.e. $\int {{1^{st}}{{.2}^{nd}}dx} = {2^{nd}}\int {{1^{st}}dx} - \int {\left( {\left( {\dfrac{d}{{dx}}{2^{nd}}} \right)\int {{1^{st}}dx} } \right)dx} + C$ so use these concepts to reach the solution of the question.

Complete step by step answer:
Given integral is
$\int {x\sin x{{\sec }^3}xdx} $
Let,
I = $\int {x\sin x{{\sec }^3}xdx} $
This integral is also written as
$ \Rightarrow I = \int {x\sin x\sec x{{\sec }^2}xdx} $
Now as we know that sec x = (1/cos x), so use this property in the above equation we have,
$ \Rightarrow I = \int {x\dfrac{{\sin x}}{{\cos x}}{{\sec }^2}xdx} $
Now as we know that (sin x/cos x) = tan x, so use this property in the above equation we have,
$ \Rightarrow I = \int {x\tan x{{\sec }^2}xdx} $..................... (1)
Now let, tan x = t............... (2)
Now differentiate equation (2) w.r.t x we have,
$ \Rightarrow \dfrac{d}{{dx}}\tan x = \dfrac{{dt}}{{dx}}$
Now as we know that the differentiation of tan x is ${\sec ^2}x$ so we have,
$ \Rightarrow {\sec ^2}xdx = dt$................ (3)
Now from equation (1), $x = {\tan ^{ - 1}}t$................. (4)
Now substitute the value from equation (2), (3) and (4) in equation (1) we have,
\[ \Rightarrow I = \int {t{{\tan }^{ - 1}}tdt} \]
Now integrate the above equation by integration by parts we have,
As we know that the formula for integration by parts is given as,
$ \Rightarrow \int {{1^{st}}{{.2}^{nd}}dx} = {2^{nd}}\int {{1^{st}}dx} - \int {\left( {\left( {\dfrac{d}{{dx}}{2^{nd}}} \right)\int {{1^{st}}dx} } \right)dx} + C$, where C is some integration constant.
Noe in the above equation ${\tan ^{ - 1}}t$ is the ${1^{st}}$ function and t is the ${2^{nd}}$ function so we have,
\[ \Rightarrow I = \int {t{{\tan }^{ - 1}}tdt} = {\tan ^{ - 1}}t\int {tdt} - \int {\left( {\left( {\dfrac{d}{{dx}}{{\tan }^{ - 1}}t} \right)\int {tdt} } \right)dt} + C\]
Now apply integration using, $\int {tdt = \dfrac{{{t^2}}}{2} + c{\text{ and }}\dfrac{d}{{dt}}{{\tan }^{ - 1}}t} = \dfrac{1}{{1 + {t^2}}}$ so we have,
\[ \Rightarrow I = \int {t{{\tan }^{ - 1}}tdt} = \dfrac{{{t^2}}}{2}{\tan ^{ - 1}}t - \int {\left( {\left( {\dfrac{1}{{1 + {t^2}}}} \right)\dfrac{{{t^2}}}{2}} \right)dt} + C\]
Now add and subtract by 1 in the numerator of the above integral we have,
\[ \Rightarrow I = \int {t{{\tan }^{ - 1}}tdt} = \dfrac{{{t^2}}}{2}{\tan ^{ - 1}}t - \dfrac{1}{2}\int {\left( {\left( {\dfrac{{{t^2} + 1 - 1}}{{1 + {t^2}}}} \right)} \right)dt} + C\]
\[ \Rightarrow I = \int {t{{\tan }^{ - 1}}tdt} = \dfrac{{{t^2}}}{2}{\tan ^{ - 1}}t - \dfrac{1}{2}\int {\left( {\left( {1 - \dfrac{1}{{1 + {t^2}}}} \right)} \right)dt} + C\]
Now separate the second integral we have,
\[ \Rightarrow I = \int {t{{\tan }^{ - 1}}tdt} = \dfrac{{{t^2}}}{2}{\tan ^{ - 1}}t - \dfrac{1}{2}\int {dt + \dfrac{1}{2}} \int {\dfrac{1}{{1 + {t^2}}}dt} + C\]
Now integrate using the property that, \[\int {dt = t + c{\text{ and }}} \int {\dfrac{1}{{1 + {t^2}}}dt = {{\tan }^{ - 1}}t + c} \] so we have,
\[ \Rightarrow I = \int {t{{\tan }^{ - 1}}tdt} = \dfrac{{{t^2}}}{2}{\tan ^{ - 1}}t - \dfrac{t}{2} + \dfrac{1}{2}{\tan ^{ - 1}}t + C\]
Now from equation (2) and (4) we have,
\[ \Rightarrow I = \int {t{{\tan }^{ - 1}}tdt} = \dfrac{1}{2}x{\tan ^2}x - \dfrac{1}{2}\tan x + \dfrac{1}{2}x + C\]
\[ \Rightarrow I = \int {t{{\tan }^{ - 1}}tdt} = \dfrac{1}{2}x\left[ {1 + {{\tan }^2}x} \right] - \dfrac{1}{2}\tan x + C\]
\[ \Rightarrow I = \int {x\sin x{{\sec }^3}xdx} = \dfrac{1}{2}x\left[ {{{\sec }^2}x} \right] - \dfrac{1}{2}\tan x + C\], $\left[ {\because \left( {1 + {{\tan }^2}x} \right) = {{\sec }^2}x} \right]$
\[ \Rightarrow I = \int {x\sin x{{\sec }^3}xdx} = \dfrac{1}{2}\left( {x{{\sec }^2}x - \tan x} \right) + C\]
So this is the required value of the integral.
Hence option (b) is the correct answer.

Note: Whenever we face such types of questions the key concept we have to remember is that always remember basic integration and basic differentiation formula which is used in that question and which is all stated above so first simplify the integration and then integrate it by parts as above then apply these formulas and simplify as above we will get the required answer.