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Question

Question: Evaluate the following integral: \(\int{{{\tan }^{\dfrac{3}{2}}}x{{\sec }^{2}}xdx}\)...

Evaluate the following integral: tan32xsec2xdx\int{{{\tan }^{\dfrac{3}{2}}}x{{\sec }^{2}}xdx}

Explanation

Solution

Hint: To find the value of a given integral, use the substitution method to simplify the given integral by assuming t=tanxt=\tan x. Rewrite the given integral in terms of variable ‘t’. Evaluate the value of integral using the fact that xndx=xn+1n+1\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}. Rewrite the value of integral in terms of ‘x’.

Complete step-by-step answer:
We have to evaluate the value of the integral tan32xsec2xdx\int{{{\tan }^{\dfrac{3}{2}}}x{{\sec }^{2}}xdx}. We observe that this is an indefinite integral. An indefinite integral is a function that takes the antiderivative of another function. It represents a family of functions whose derivatives are the function given in the integral.
To find the value of the integral, we will simplify the given integral by substitution method.
Let’s assume that t=tanx.....(1)t=\tan x.....\left( 1 \right). We will now differentiate the equation. Thus, we have dtdx=sec2x\dfrac{dt}{dx}={{\sec }^{2}}x.
Cross multiplying the terms on both sides of the equality, we have dt=sec2xdx.....(2)dt={{\sec }^{2}}xdx.....\left( 2 \right).
Substituting equation (1) and (2) in the given integral, we have tan32xsec2xdx=t32dt\int{{{\tan }^{\dfrac{3}{2}}}x{{\sec }^{2}}xdx}=\int{{{t}^{\dfrac{3}{2}}}dt}.
We know that integral of a function of the form y=xny={{x}^{n}} is xndx=xn+1n+1\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}.
Substituting n=32n=\dfrac{3}{2} in the above formula, we have x32dx=x5252\int{{{x}^{\dfrac{3}{2}}}dx}=\dfrac{{{x}^{\dfrac{5}{2}}}}{\dfrac{5}{2}}.
Thus, we have tan32xsec2xdx=t32dt=t5252\int{{{\tan }^{\dfrac{3}{2}}}x{{\sec }^{2}}xdx}=\int{{{t}^{\dfrac{3}{2}}}dt}=\dfrac{{{t}^{\dfrac{5}{2}}}}{\dfrac{5}{2}}.
Simplifying the above expression, we have tan32xsec2xdx=t32dt=25t52\int{{{\tan }^{\dfrac{3}{2}}}x{{\sec }^{2}}xdx}=\int{{{t}^{\dfrac{3}{2}}}dt}=\dfrac{2}{5}{{t}^{\dfrac{5}{2}}}.
We will again substitute t=tanxt=\tan x in the above equation. Thus, we have tan32xsec2xdx=t32dt=25t52=25tan52x\int{{{\tan }^{\dfrac{3}{2}}}x{{\sec }^{2}}xdx}=\int{{{t}^{\dfrac{3}{2}}}dt}=\dfrac{2}{5}{{t}^{\dfrac{5}{2}}}=\dfrac{2}{5}{{\tan }^{\dfrac{5}{2}}}x.
Hence, the value of the integral tan32xsec2xdx\int{{{\tan }^{\dfrac{3}{2}}}x{{\sec }^{2}}xdx} is 25tan52x\dfrac{2}{5}{{\tan }^{\dfrac{5}{2}}}x.

Note: The substitution method is used when an integral contains some function and its first derivative. It’s important to keep in mind that the first derivative of y=tanxy=\tan x is dydx=sec2x\dfrac{dy}{dx}={{\sec }^{2}}x. Otherwise, we won’t be able to solve this question.