Question

How do you integrate $\int {\dfrac{x}{{\sqrt {{x^2} + 1} }}} $ by trigonometric substitution?

Answer 1

Hint:The above question is based on the concept of integration. Since it is an indefinite integral which has no upper and lower limits, we can apply integration properties by integrating it where the power increases by one and we can find the antiderivative of the above expression.

Complete step by step answer:
Integration is a way of finding the antiderivative of any function. It is the inverse of differentiation. It denotes the summation of discrete data. Calculation of small problems is an easy task but for adding big problems which include higher limits, integration method is used. The above given expression is an indefinite integral which means there are no upper or lower limits given.The above expression after integrating should be in the below form.
\[\int {f(x) = F(x) + C} \]
where C is constant.
So, the above expression is given as
$\int {\dfrac{x}{{\sqrt {{x^2} + 1} }}} $
Now let \[x = \tan \theta \].Then \[dx = {\sec ^2}\theta d\theta \]
Then by substituting the value of x and dx we get,
$\int {\dfrac{{\tan \theta }}{{\sqrt {{{\tan }^2}\theta + 1} }}} $
By applying the trigonometric identity of tangent and secant in the denominator we get,
$\Rightarrow \int {\dfrac{{\tan \theta }}{{\sqrt {{{\tan }^2}\theta + 1} }}} {\sec ^2}\theta d\theta \\
\Rightarrow \int {\dfrac{{\tan \theta }}{{\sqrt {{{\sec }^2}\theta } }}{{\sec }^2}\theta d\theta } \\ $
Now by taking square root of secant function in denominator we get,
\[\Rightarrow \int {\dfrac{{\tan \theta }}{{\sec \theta }} \times {{\sec }^2}} \theta \\
\Rightarrow \int {\tan \theta } \sec \theta \\ \]
The common integral is \[\int {\tan \theta \sec \theta = \sec \theta + C} \]
Now since $\sec \theta = \dfrac{{Hypotenuse}}{{adjacent}}$.But according trigonometric identity \[\sec x = \dfrac{1}{{\cos x}}\].

So accordingly we get \[\sec \theta = \sqrt {{x^2} + 1} + C\] according to Pythagoras theorem.

Note: An important thing to note is that the trigonometric ratio of right angle triangle for cosec function is \[\cos x = \dfrac{{adjacent}}{{hypotenuse}}\] which is the inverse of secant function and the Pythagoras theorem will add the opposite and adjacent side so we get the answer as \[\sec \theta = \sqrt {{x^2} + 1} + C\].