Solveeit Logo
Login

Question

Question: Value of the integral \(\int_{{}}^{{}}{\tan x.\tan 2x.\tan 3xdx}\) is equal to (a) \(\dfrac{1}{3}...

Value of the integral ∫tan⁡x.tan⁡2x.tan⁡3xdx\int_{{}}^{{}}{\tan x.\tan 2x.\tan 3xdx} is equal to
(a) 13log⁡∣sec⁡3x∣−12log⁡∣sec⁡2x∣+log⁡∣sec⁡x∣+C\dfrac{1}{3}\log |\sec 3x|-\dfrac{1}{2}\log |\sec 2x|+\log |\sec x|+C
(b) 13log⁡∣sec⁡3x∣−12log⁡∣sec⁡2x∣−log⁡∣sec⁡x∣+C\dfrac{1}{3}\log |\sec 3x|-\dfrac{1}{2}\log |\sec 2x|-\log |\sec x|+C
(c) 13log⁡∣sec⁡3x∣+12log⁡∣sec⁡2x∣+log⁡∣sec⁡x∣+C\dfrac{1}{3}\log |\sec 3x|+\dfrac{1}{2}\log |\sec 2x|+\log |\sec x|+C
(d) None of these

Explanation

Solution

Hint: Here, we will use the trigonometric formula for tan3x and with the help of that we will reduce the given expression into simpler form. After that we will use the formula for integration to arrive at the answer.

Step-by-step answer:
Given a function ff of a real variable x, then the integral ∫f(x).dx\int_{{}}^{{}}{f\left( x \right).dx} can be interpreted as the area of the region in the x-y plane that is bounded by the graph of ff and the x-axis.
Since, we are not given any boundaries in the given integration, it means that it is an indefinite integration.
Since, the function given to us in the question to integrate is tan⁡x.tan⁡2x.tan⁡3x\tan x.\tan 2x.\tan 3x.
Let this function be denoted as f(x)f\left( x \right). So, we have:
f(x)=tan⁡x.tan⁡2x.tan⁡3xf\left( x \right)=\tan x.\tan 2x.\tan 3x
Now, we know that tan3x can be written as:
tan⁡3x=tan⁡(2x+x)\tan 3x=\tan \left( 2x+x \right)
We, also that if A and B are two angles then tan(A+B) can be written as:
tan⁡(A+B)=tan⁡A+tan⁡B1−tan⁡Atan⁡B\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}
Here, we have A as 2x and B as x. Therefore, tan3x can be written as:
tan⁡3x=tan⁡2x+tan⁡x1−tan⁡2x.tan⁡x\tan 3x=\dfrac{\tan 2x+\tan x}{1-\tan 2x.\tan x}
On simplifying the above equation, we can write:
tan⁡3x(1−tan⁡2x.tan⁡x)=tan⁡2x+tan⁡x\tan 3x\left( 1-\tan 2x.\tan x \right)=\tan 2x+\tan x
On simplifying the left hand side of this expression, we get:
tan⁡3x−tan⁡3x.tan⁡2x.tan⁡x=tan⁡2x+tan⁡x ⇒tan⁡3x−tan⁡2x−tan⁡x=tan⁡x.tan⁡2x.tan⁡3x \begin{aligned} & \tan 3x-\tan 3x.\tan 2x.\tan x=\tan 2x+\tan x \\\ & \Rightarrow \tan 3x-\tan 2x-\tan x=\tan x.\tan 2x.\tan 3x \\\ \end{aligned}
So, the given function f(x)=tan⁡x.tan⁡2x.tan⁡3xf\left( x \right)=\tan x.\tan 2x.\tan 3x can be written as f(x)=tan⁡3x−tan⁡2x−tan⁡xf\left( x \right)=\tan 3x-\tan 2x-\tan x.
Therefore, we have:
∫f(x).dx=∫(tan⁡3x−tan⁡2x−tan⁡x)dx ⇒∫f(x)=∫tan⁡3x.dx−∫tan⁡2x.dx−∫tan⁡x.dx..........(1) \begin{aligned} & \int_{{}}^{{}}{f\left( x \right).dx=\int_{{}}^{{}}{\left( \tan 3x-\tan 2x-\tan x \right)dx}} \\\ & \Rightarrow \int_{{}}^{{}}{f\left( x \right)=\int_{{}}^{{}}{\tan 3x.dx}-\int_{{}}^{{}}{\tan 2x.dx}-\int_{{}}^{{}}{\tan x.dx}}..........\left( 1 \right) \\\ \end{aligned}
We know that:
∫tan⁡x.dx=log⁡∣sec⁡x∣+c1\int_{{}}^{{}}{\tan x.dx}=\log |\sec x|+{{c}_{1}} , ∫tan⁡2x.dx=12log⁡∣sec⁡x∣+c2\int_{{}}^{{}}{\tan 2x.dx}=\dfrac{1}{2}\log |\sec x|+{{c}_{2}} and ∫tan⁡3x.dx=13log⁡∣sec⁡x∣+c2\int_{{}}^{{}}{\tan 3x.dx}=\dfrac{1}{3}\log |\sec x|+{{c}_{2}}
Here, c1,c2 {{c}_{1}},{{c}_{2}}\text{ }and c3{{c}_{3}} are the constants of integration.
On substituting these values in equation (1), we get:
∫f(x).dx=13log⁡∣sec⁡3x∣−12log⁡∣sec⁡2x∣−log !!∣!! secx !!∣!! +C\int_{{}}^{{}}{f\left( x \right).dx}=\dfrac{1}{3}\log |\sec 3x|-\dfrac{1}{2}\log |\sec 2x|-\text{log }\\!\\!|\\!\\!\text{ secx }\\!\\!|\\!\\!\text{ +C}
Here C is a new integration constant.
Hence, option (b) is the correct answer.

Note: Students should be careful while choosing which function has to be decomposed. We can arrive at the correct result only if we decompose tan3x into simpler form. Students should also remember the formulas of integration correctly to avoid mistakes.