
Solve the trigonometric equation :$\int {\dfrac{{\sec x}}{{\sec x + \tan x}}} dx$
Answer
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Hint: Try to convert the expression into the form of trigonometric ratios of which integral is known. Then use the antiderivative to find the integral value of simplified expression.
Complete step-by-step answer:
Given the problem, we need to evaluate the integral
$I = \int {\dfrac{{\sec x}}{{\sec x + \tan x}}} dx{\text{ (1)}}$
We need to evaluate the above indefinite integral.
Since the integral is not in a form of which the antiderivative is known, we need to transform this into a more simplified form.
Multiplying both numerator and denominator of the expression under integral by $\left( {\sec x - \tan x} \right)$,we get,
$I = \int {\dfrac{{\sec x\left( {\sec x - \tan x} \right)}}{{\sec x + \tan x\left( {\sec x - \tan x} \right)}}} dx$
Using the identity $\left( {a - b} \right)\left( {a + b} \right) = {a^2} = {b^2}$ in the denominator above, we get,
$ \Rightarrow I = \int {\dfrac{{{{\sec }^2}x - \sec x\tan x}}{{\left( {{{\sec }^2}x - {{\tan }^2}x} \right)}}} dx$
We know that trigonometric identity,
${\sec ^2}x - {\tan ^2}x = 1$
Using this identity in integral above, we get
$ \Rightarrow I = \int {\left( {{{\sec }^2}x - \sec x\tan x} \right)} dx{\text{ (2)}}$
We know that the derivative of $\sec x$ is $ - \sec x\tan x$and that of $\tan x$is ${\sec ^2}x$, hence by antiderivative approach the integral of $\sec x\tan x$would be $ - \sec x$and that of ${\sec ^2}x$would be $\tan x$.
Using this result in integral $(2)$
\[
\Rightarrow I = \int {\left( {{{\sec }^2}x - \sec x\tan x} \right)} dx = \int {\left( {{{\sec }^2}x} \right)} dx - \int {\left( {\sec x\tan x} \right)} dx \\
\Rightarrow I = \tan x - \left( { - \sec x} \right) + c \\
\Rightarrow I = \tan x + \sec x + c \\
\]
Where $c$is a constant.
Comparing this with equation $(1)$, we get
$I = \int {\dfrac{{\sec x}}{{\sec x + \tan x}}} dx = \tan x + \sec x + c$
Hence the value of integral $I = \int {\dfrac{{\sec x}}{{\sec x + \tan x}}} dx$ is $\tan x + \sec x + c$.
Note: Indefinite integral means integrating a function without any limit but in definite integral there are upper and lower limits. An antiderivative of a function is another function whose derivative is equal to the original function. The integrals of trigonometric should be kept in mind while solving problems like above. There could be many approaches to an integration problem like above each giving the same solution. One should try to solve the problem with different approaches in order to verify the result.
Complete step-by-step answer:
Given the problem, we need to evaluate the integral
$I = \int {\dfrac{{\sec x}}{{\sec x + \tan x}}} dx{\text{ (1)}}$
We need to evaluate the above indefinite integral.
Since the integral is not in a form of which the antiderivative is known, we need to transform this into a more simplified form.
Multiplying both numerator and denominator of the expression under integral by $\left( {\sec x - \tan x} \right)$,we get,
$I = \int {\dfrac{{\sec x\left( {\sec x - \tan x} \right)}}{{\sec x + \tan x\left( {\sec x - \tan x} \right)}}} dx$
Using the identity $\left( {a - b} \right)\left( {a + b} \right) = {a^2} = {b^2}$ in the denominator above, we get,
$ \Rightarrow I = \int {\dfrac{{{{\sec }^2}x - \sec x\tan x}}{{\left( {{{\sec }^2}x - {{\tan }^2}x} \right)}}} dx$
We know that trigonometric identity,
${\sec ^2}x - {\tan ^2}x = 1$
Using this identity in integral above, we get
$ \Rightarrow I = \int {\left( {{{\sec }^2}x - \sec x\tan x} \right)} dx{\text{ (2)}}$
We know that the derivative of $\sec x$ is $ - \sec x\tan x$and that of $\tan x$is ${\sec ^2}x$, hence by antiderivative approach the integral of $\sec x\tan x$would be $ - \sec x$and that of ${\sec ^2}x$would be $\tan x$.
Using this result in integral $(2)$
\[
\Rightarrow I = \int {\left( {{{\sec }^2}x - \sec x\tan x} \right)} dx = \int {\left( {{{\sec }^2}x} \right)} dx - \int {\left( {\sec x\tan x} \right)} dx \\
\Rightarrow I = \tan x - \left( { - \sec x} \right) + c \\
\Rightarrow I = \tan x + \sec x + c \\
\]
Where $c$is a constant.
Comparing this with equation $(1)$, we get
$I = \int {\dfrac{{\sec x}}{{\sec x + \tan x}}} dx = \tan x + \sec x + c$
Hence the value of integral $I = \int {\dfrac{{\sec x}}{{\sec x + \tan x}}} dx$ is $\tan x + \sec x + c$.
Note: Indefinite integral means integrating a function without any limit but in definite integral there are upper and lower limits. An antiderivative of a function is another function whose derivative is equal to the original function. The integrals of trigonometric should be kept in mind while solving problems like above. There could be many approaches to an integration problem like above each giving the same solution. One should try to solve the problem with different approaches in order to verify the result.
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