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Question: If \(I=\int{\dfrac{dx}{\tan x\log \cos ecx}}\) then \(I\) equal to A) \(\log \left| \log \left. \c...

If I=∫dxtan⁡xlog⁡cos⁡ecxI=\int{\dfrac{dx}{\tan x\log \cos ecx}} then II equal to
A) log⁡∣log⁡cos⁡ecx∣+c\log \left| \log \left. \cos ecx \right| \right.+c
B) log⁡∣log⁡cos⁡x∣+c\log \left| \log \left. \cos x \right| \right.+c
C) −log⁡∣log⁡(cos⁡ecx)∣+c-\log \left| \left. \log \left( \cos ecx \right) \right|+c \right.
D) None of the above

Explanation

Solution

Derivative of log⁡(cos⁡ecx)\log \left( \cos ecx \right) is equal to (−cot⁡x)\left( -\cot x \right)
ddx(log⁡cos⁡ecx)=1cos⁡ecx×−cos⁡ecxcot⁡x=−cot⁡x=−1tan⁡x\dfrac{d}{dx}\left( \log \cos ecx \right)=\dfrac{1}{\cos ecx}\times -\cos ecx\cot x=-\cot x=\dfrac{-1}{\tan x}
We can apply the method of substitution of integration as we observed that both the function and its derivative are present in the above given question.
Also, the general form of integration by substitution method is given as follows;
∫f(g(x)).g′(x).dx=f(t).dt\int{f\left( g\left( x \right) \right).{g}'\left( x \right).dx=f\left( t \right).dt} where t=g(x)t = g(x).

Complete step by step solution:
We are given that, I=∫dxtan⁡xlog⁡cos⁡ecxI=\int{\dfrac{dx}{\tan x\log \cos ecx}} ………(1)\ldots \ldots \ldots \left( 1 \right)
Let us take, ………(2)\ldots \ldots \ldots \left( 2 \right)
Now, we will take derivative on both sides with respect to x
ddx(log⁡cos⁡ecx)=dtdx\dfrac{d}{dx}\left( \log \cos ecx \right)=\dfrac{dt}{dx}
Now we will apply chain rule, which states: ddx[f(g(x))]=f′(g(x))∗g′(x)\dfrac{d}{dx}\left[ f\left( g\left( x \right) \right) \right]={f}'\left( g\left( x \right) \right)*{g}'\left( x \right)
And we get; 1cos⁡ecx×−cos⁡ecxcot⁡x=dtdx\dfrac{1}{\cos ecx}\times -\cos ecx\cot x=\dfrac{dt}{dx}
Formulas used in the above step are
ddx(log⁡x)=1x\dfrac{d}{dx}\left( \log x \right)=\dfrac{1}{x}
ddx(cos⁡ecx)=−cos⁡ecxcot⁡x\dfrac{d}{dx}\left( \cos ecx \right)=-\cos ecx\cot x
Hence, ………(3)\ldots \ldots \ldots \left( 3 \right)
Using the substitution method of integration, substituting (2) in equation (1),
Using the substitution method of integration, substituting (3) in equation (1),
We get,
I=∫dxt×−dtdxI=\int{\dfrac{dx}{t}}\times \dfrac{-dt}{dx}
I=∫−dttI=\int{\dfrac{-dt}{t}}
Now, using formula ∫dxx=log⁡∣x∣\int{\dfrac{dx}{x}}=\log \left| \left. x \right| \right.
We get I=−log⁡∣t∣+cI=-\log \left| \left. t \right| \right. +c
Substituting the value of t from equation (2) in the above step, we get
I=−log⁡∣log⁡cos⁡ecx∣+cI=-\log \left| \left. \log \cos ecx \right| \right.+c
This gives us our required answer.
Hence, from the given options, the correct option is (C).

Note:
Questions of such a pattern where there forms a relation between the two functions, we always use the method of substitution of integration. The substitution method is used when an integral contains some function and its derivative. In this case we can set t equal to the function and rewrite the integral in terms of the new variable t. This makes integral easy to solve.
The general form of integration by substitution method is given as follows;
∫f(g(x)).g′(x).dx=f(t).dt\int{f\left( g\left( x \right) \right).{g}'\left( x \right).dx=f\left( t \right).dt} where t = g(x).