
Find the integral of \[\int {\dfrac{{\cos ecx}}{{\log \tan (\dfrac{x}{2})}}dx} \].
Answer
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Hint: The ratios of sides of a right-angled triangle with respect to any of its acute angles are known as the trigonometric ratios of that particular angle. Here we will assume the denominator with some variable and then we will find the derivative of it. We will also use trigonometric ratio \[\tan x = \dfrac{{\sin x}}{{\cos x}}\] . Since, cot is inverse of tan i.e. \[\tan x = \dfrac{1}{{\cot x}}\] and so will use \[\cot x = \dfrac{{\cos x}}{{\sin x}}\]. Substituting the value we will get derivative. Also we will use the trigonometric ratio \[\cos ecx = \dfrac{1}{{\sin x}}\] . After that, we will find the integral of the given function to get the final output.
Complete step-by-step answer:
Given that,
\[\int {\dfrac{{\cos ecx}}{{\log \tan (\dfrac{x}{2})}}dx} \]
Let \[\log \tan (\dfrac{x}{2}) = t\]
\[ \Rightarrow \dfrac{1}{{\tan (\dfrac{x}{2})}} \times {\sec ^2}(\dfrac{x}{2}) \times \dfrac{1}{2}dx = dt\]
\[ \Rightarrow \dfrac{1}{{\tan (\dfrac{x}{2})}} \times \dfrac{1}{{{{\cos }^2}(\dfrac{x}{2})}} \times \dfrac{1}{2}dx = dt\]
\[ \Rightarrow \cot (\dfrac{x}{2}) \times \dfrac{1}{{{{\cos }^2}(\dfrac{x}{2})}} \times \dfrac{1}{2}dx = dt\] \[(\because \tan x = \dfrac{1}{{\cot x}})\]
\[ \Rightarrow \dfrac{{\cos (\dfrac{x}{2})}}{{\sin (\dfrac{x}{2})}} \times \dfrac{1}{{{{\cos }^2}(\dfrac{x}{2})}} \times \dfrac{1}{2}dx = dt\]\[(\because \cot x = \dfrac{{\cos x}}{{\sin x}})\]
\[ \Rightarrow \dfrac{1}{{\sin (\dfrac{x}{2})}} \times \dfrac{1}{{\cos (\dfrac{x}{2})}} \times \dfrac{1}{2}dx = dt\]
\[ \Rightarrow \dfrac{1}{{2\sin (\dfrac{x}{2})\cos (\dfrac{x}{2})}}dx = dt\]
\[ \Rightarrow \dfrac{1}{{\sin x}}dx = dt\]
\[ \Rightarrow \cos ecxdx = dt\]
Here, we will find the final output as below:
\[\therefore \int {\dfrac{{\cos ecx}}{{\log \tan (\dfrac{x}{2})}}dx} \]
\[ = \int {\dfrac{{dt}}{t}} \]
\[ = \log t + C\]
\[ = \log \{ \log \tan (\dfrac{x}{2})\} + C\]
Hence, the integral value of \[\int {\dfrac{{\cos ecx}}{{\log \tan (\dfrac{x}{2})}}dx} = \log \{ \log \tan (\dfrac{x}{2})\} + C\] .
So, the correct answer is “Option B”.
Note: The differentiation is one of the two important concepts apart from integration. Differentiation is a method of finding the derivative of a function. Derivatives are fundamental to the solution of problems in calculus and differential equations. The logarithm is the inverse function to exponentiation. A logarithm is the power to which a number must be raised in order to get some other number. Trigonometry is a branch of mathematics that deals with the sides and angles of a right-angled triangle. Therefore, trig ratios are evaluated with respect to sides and angles. The six trigonometric ratios are sine (sin), cosine (cos), tangent (tan), cotangent (cot), cosecant (cosec), and secant (sec). Opposite side is the perpendicular side and the adjacent side is the base of the right-triangle.
Complete step-by-step answer:
Given that,
\[\int {\dfrac{{\cos ecx}}{{\log \tan (\dfrac{x}{2})}}dx} \]
Let \[\log \tan (\dfrac{x}{2}) = t\]
\[ \Rightarrow \dfrac{1}{{\tan (\dfrac{x}{2})}} \times {\sec ^2}(\dfrac{x}{2}) \times \dfrac{1}{2}dx = dt\]
\[ \Rightarrow \dfrac{1}{{\tan (\dfrac{x}{2})}} \times \dfrac{1}{{{{\cos }^2}(\dfrac{x}{2})}} \times \dfrac{1}{2}dx = dt\]
\[ \Rightarrow \cot (\dfrac{x}{2}) \times \dfrac{1}{{{{\cos }^2}(\dfrac{x}{2})}} \times \dfrac{1}{2}dx = dt\] \[(\because \tan x = \dfrac{1}{{\cot x}})\]
\[ \Rightarrow \dfrac{{\cos (\dfrac{x}{2})}}{{\sin (\dfrac{x}{2})}} \times \dfrac{1}{{{{\cos }^2}(\dfrac{x}{2})}} \times \dfrac{1}{2}dx = dt\]\[(\because \cot x = \dfrac{{\cos x}}{{\sin x}})\]
\[ \Rightarrow \dfrac{1}{{\sin (\dfrac{x}{2})}} \times \dfrac{1}{{\cos (\dfrac{x}{2})}} \times \dfrac{1}{2}dx = dt\]
\[ \Rightarrow \dfrac{1}{{2\sin (\dfrac{x}{2})\cos (\dfrac{x}{2})}}dx = dt\]
\[ \Rightarrow \dfrac{1}{{\sin x}}dx = dt\]
\[ \Rightarrow \cos ecxdx = dt\]
Here, we will find the final output as below:
\[\therefore \int {\dfrac{{\cos ecx}}{{\log \tan (\dfrac{x}{2})}}dx} \]
\[ = \int {\dfrac{{dt}}{t}} \]
\[ = \log t + C\]
\[ = \log \{ \log \tan (\dfrac{x}{2})\} + C\]
Hence, the integral value of \[\int {\dfrac{{\cos ecx}}{{\log \tan (\dfrac{x}{2})}}dx} = \log \{ \log \tan (\dfrac{x}{2})\} + C\] .
So, the correct answer is “Option B”.
Note: The differentiation is one of the two important concepts apart from integration. Differentiation is a method of finding the derivative of a function. Derivatives are fundamental to the solution of problems in calculus and differential equations. The logarithm is the inverse function to exponentiation. A logarithm is the power to which a number must be raised in order to get some other number. Trigonometry is a branch of mathematics that deals with the sides and angles of a right-angled triangle. Therefore, trig ratios are evaluated with respect to sides and angles. The six trigonometric ratios are sine (sin), cosine (cos), tangent (tan), cotangent (cot), cosecant (cosec), and secant (sec). Opposite side is the perpendicular side and the adjacent side is the base of the right-triangle.
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