Question

How do you integrate $ \int {x\sqrt {{x^2} + 4} dx} $ by trigonometric substitution?

Answer 1

Hint: To evaluate the given integral with help of trigonometric substitution, find any trigonometric parameter which is suitable for the given integrand and then substitute that trigonometric function in the integrand and simplify it, after simplifying the integrand you have to take substitution one more time to solve the integral. And after solving the integral, put back the considered variables to get the required integral.
Substitute $ x = 2\tan a $ to evaluate the given integral first.
Formula used: Trigonometric identity: $ {\sec ^2}x - {\tan ^2}x = 1 $

Complete step-by-step answer:
In order to evaluate the given integral $ \int {x\sqrt {{x^2} + 4}.dx} $ by trigonometric substitution, we have to consider the trigonometric substitution in the integrand such that it will simplify the integral to be integrated simply.
On observing the integrand, we can see that we can assume our substitution as follows
Substituting $ x = 2\tan a \Rightarrow dx = 2{\sec ^2}a.da $ so we can write the integral as
 \[
  \int {x\sqrt {{x^2} + 4}.dx} = \int {2\tan a\sqrt {{{(2\tan a)}^2} + 4} {{\sec }^2}a.da} \\
   = \int {2\tan a.\sqrt {4{{\tan }^2}a + 4} .{{\sec }^2}a.da} \\
   = \int {4\tan a.\sqrt {{{\tan }^2}a + 1} .{{\sec }^2}a.da} \;
 \]
Now, using the trigonometric identity giving the relation $ {\sec ^2}x = {\tan ^2}x + 1 $ we will get
 \[
  \int {4\tan a.\sqrt {{{\tan }^2}a + 1}. {{\sec }^2}ada} = \int {4\tan a.\sec a.{{\sec }^2}a.da} \\
   = \int {4\tan a.{{\sec }^3}ada} \;
 \]
Here we will again take the substitution,
Taking $ t = \sec a \Rightarrow dt = \tan a.\sec a.da $ we will get
 \[
  \int {4\tan a.{{\sec }^3}a.da} = \int {4{{\sec }^2}a.\tan a.\sec a.da} \\
   = \int {4{t^2}dt} \\
   = \dfrac{{4{t^3}}}{3} + c \;
 \]
Now putting the substitution back, we will get
 \[
  \dfrac{{4{t^3}}}{3} + c = \dfrac{{4{{\sec }^3}a}}{3} + c \\
   = \dfrac{{4{{\sec }^2}a\sec a}}{3} + c \\
   = \dfrac{{4\left( {1 + {{\tan }^2}a} \right)\sqrt {1 + {{\tan }^2}a} }}{3} + c \\
   = \dfrac{{4\left( {1 + {x^2}} \right)\sqrt {1 + {x^2}} }}{3} + c \\
   = \dfrac{{4{{\left( {1 + {x^2}} \right)}^{\dfrac{3}{2}}}}}{3} + c \;
 \]
Therefore \[\dfrac{{4{{\left( {1 + {x^2}} \right)}^{\dfrac{3}{2}}}}}{3} + c\] is the required integral of the given integral by trigonometric substitution.
So, the correct answer is “\[\dfrac{{4{{\left( {1 + {x^2}} \right)}^{\dfrac{3}{2}}}}}{3} + c\]”.

Note: When using the substitution method in order to simplify the integrand and eventually integrate the given integral, at the last step or after evaluating the integration, must replace the considered variables to original ones to get the required integration. Also after doing indefinite integral, always add the constant which is also known as arbitrary constant because indefinite integration has arbitrary constant and can have any constant value.