# Find the integral;

$\int{\sec x\left( \sec x+\tan x \right)dx}$

Answer

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Hint: Simplify the expression within the integral sign and use the following results:

$\int{{{\sec }^{2}}xdx}=\tan x+C$ and $\int{\sec x\tan xdx}=\sec x+C$. Also, use the property that integral of the sum = sum of the integrals.

Complete step-by-step answer:

We, first of all, will simplify the integrand

$\begin{align}

& =\sec x(\left( \sec x+\tan x \right) \\

& =\sec x\sec x+\sec x\tan x \\

& ={{\sec }^{2}}x+\sec x\tan x \\

\end{align}$

Hence $\int{\sec x\left( \sec x+\tan x \right)dx}=\int{\left( {{\sec }^{2}}x+\sec x\tan x \right)dx}$

Since integral of the sum of functions is equal to the sum of integral of the functions, we have

$\int{\sec x\left( \sec x+\tan x \right)dx}=\int{{{\sec }^{2}}xdx}+\int{\sec x\tan xdx}$

We know that $\int{{{\sec }^{2}}xdx}=\tan x+C$ and $\int{\sec x\tan xdx}=\sec x+C$

Using the above formulae, we get

$\int{\sec x\left( \sec x+\tan x \right)dx}=\tan x+\sec x+C$.

Hence $\int{\sec x\left( \sec x+\tan x \right)dx}=\tan x+\sec x+C$.

Note: Alternatively, we can solve the above question by writing the integrand in terms of sine and cosine and then integrating.

Using $\sec x=\dfrac{1}{\cos x}$ and $\tan x=\dfrac{\sin x}{\cos x}$, we have

$\begin{align}

& \sec x\left( \sec x+\tan x \right) \\

& =\dfrac{1}{\cos x}\dfrac{1+\sin x}{\cos x} \\

& =\dfrac{1+\sin x}{{{\cos }^{2}}x} \\

\end{align}$

We know that ${{\cos }^{2}}x=1-{{\sin }^{2}}x$.

Using the above formula, we get

$\sec x\left( \sec x+\tan x \right)=\dfrac{1+\sin x}{1-{{\sin }^{2}}x}$

We know that ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$

Using the above formula, we get

\[\begin{align}

& \sec x\left( \sec x+\tan x \right)=\dfrac{1+\sin x}{\left( 1-\sin x \right)\left( 1+\sin x \right)} \\

& \Rightarrow \sec x\left( \sec x+\tan x \right)=\dfrac{1}{1-\sin x} \\

\end{align}\]

Hence \[\int{\sec x\left( \sec x+\tan x \right)dx}=\int{\dfrac{1}{1-\sin x}dx}\]

In integrals of type $\int{\left( \dfrac{dx}{a\sin x+b\cos x} \right)}$ we substitute $t=\tan \dfrac{x}{2}$

So, let $t=\tan \dfrac{x}{2}$

Differentiating both sides, we get

$dt=\dfrac{1}{2}{{\sec }^{2}}\dfrac{x}{2}dx$

We know that ${{\sec }^{2}}x=1+{{\tan }^{2}}x$

Using the above formula, we get

\[\begin{align}

& dt=\dfrac{1+{{\tan }^{2}}\dfrac{x}{2}}{2}=\dfrac{1+{{t}^{2}}}{2}dx \\

& \Rightarrow dx=\dfrac{2dt}{1+{{t}^{2}}} \\

\end{align}\]

Also, we know $\sin x=\dfrac{2\tan \dfrac{x}{2}}{1+{{\tan }^{2}}\dfrac{x}{2}}=\dfrac{2t}{1+{{t}^{2}}}$

Hence we have

$\begin{align}

& \int{\sec x\left( \sec x+\tan x \right)}=\int{\dfrac{\dfrac{2dt}{1+{{t}^{2}}}}{1-\dfrac{2t}{1+{{t}^{2}}}}} \\

& =\int{\dfrac{2dt}{1+{{t}^{2}}-2t}} \\

\end{align}$

We know that ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$

Using the above formula, we get

$\begin{align}

& \int{\sec x\left( \sec x+\tan x \right)}=\int{\dfrac{2dt}{{{\left( t-1 \right)}^{2}}}=2\int{\dfrac{dt}{{{\left( t-1 \right)}^{2}}}}} \\

& =\dfrac{2}{1-t}+C \\

\end{align}$

Reverting to the original variable, we get

$\int{\sec x\left( \sec x+\tan x \right)}=\dfrac{2}{1-\tan \dfrac{x}{2}}+C$

[2] Although the two results look completely different from each other, by Lagrange mean value theorem, they differ only by a constant.

Hence $\dfrac{2}{1-\tan \dfrac{x}{2}}=\tan x+\sec x+C$ for some constant C.

[3] Proof of [2]: Let F(x) and G(x) be two antiderivatives of the function f(x).

Applying Lagrange mean value theorem to the function F(x) – G(x) in the interval [0,y]{Assuming continuity at 0}, we get

\[\dfrac{F(y)-G(y)-F(0)+G(0)}{y}=f(c)-f(c)=0\]

i.e. $F(y)-G(y)-F(0)+G(0)=0$

Since F(0)-G(0) is a constant let C = F(0) – G(0), we get

$F(y)=G(y)+C$ or in other words F(x) and G(x) differ only by a constant.

$\int{{{\sec }^{2}}xdx}=\tan x+C$ and $\int{\sec x\tan xdx}=\sec x+C$. Also, use the property that integral of the sum = sum of the integrals.

Complete step-by-step answer:

We, first of all, will simplify the integrand

$\begin{align}

& =\sec x(\left( \sec x+\tan x \right) \\

& =\sec x\sec x+\sec x\tan x \\

& ={{\sec }^{2}}x+\sec x\tan x \\

\end{align}$

Hence $\int{\sec x\left( \sec x+\tan x \right)dx}=\int{\left( {{\sec }^{2}}x+\sec x\tan x \right)dx}$

Since integral of the sum of functions is equal to the sum of integral of the functions, we have

$\int{\sec x\left( \sec x+\tan x \right)dx}=\int{{{\sec }^{2}}xdx}+\int{\sec x\tan xdx}$

We know that $\int{{{\sec }^{2}}xdx}=\tan x+C$ and $\int{\sec x\tan xdx}=\sec x+C$

Using the above formulae, we get

$\int{\sec x\left( \sec x+\tan x \right)dx}=\tan x+\sec x+C$.

Hence $\int{\sec x\left( \sec x+\tan x \right)dx}=\tan x+\sec x+C$.

Note: Alternatively, we can solve the above question by writing the integrand in terms of sine and cosine and then integrating.

Using $\sec x=\dfrac{1}{\cos x}$ and $\tan x=\dfrac{\sin x}{\cos x}$, we have

$\begin{align}

& \sec x\left( \sec x+\tan x \right) \\

& =\dfrac{1}{\cos x}\dfrac{1+\sin x}{\cos x} \\

& =\dfrac{1+\sin x}{{{\cos }^{2}}x} \\

\end{align}$

We know that ${{\cos }^{2}}x=1-{{\sin }^{2}}x$.

Using the above formula, we get

$\sec x\left( \sec x+\tan x \right)=\dfrac{1+\sin x}{1-{{\sin }^{2}}x}$

We know that ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$

Using the above formula, we get

\[\begin{align}

& \sec x\left( \sec x+\tan x \right)=\dfrac{1+\sin x}{\left( 1-\sin x \right)\left( 1+\sin x \right)} \\

& \Rightarrow \sec x\left( \sec x+\tan x \right)=\dfrac{1}{1-\sin x} \\

\end{align}\]

Hence \[\int{\sec x\left( \sec x+\tan x \right)dx}=\int{\dfrac{1}{1-\sin x}dx}\]

In integrals of type $\int{\left( \dfrac{dx}{a\sin x+b\cos x} \right)}$ we substitute $t=\tan \dfrac{x}{2}$

So, let $t=\tan \dfrac{x}{2}$

Differentiating both sides, we get

$dt=\dfrac{1}{2}{{\sec }^{2}}\dfrac{x}{2}dx$

We know that ${{\sec }^{2}}x=1+{{\tan }^{2}}x$

Using the above formula, we get

\[\begin{align}

& dt=\dfrac{1+{{\tan }^{2}}\dfrac{x}{2}}{2}=\dfrac{1+{{t}^{2}}}{2}dx \\

& \Rightarrow dx=\dfrac{2dt}{1+{{t}^{2}}} \\

\end{align}\]

Also, we know $\sin x=\dfrac{2\tan \dfrac{x}{2}}{1+{{\tan }^{2}}\dfrac{x}{2}}=\dfrac{2t}{1+{{t}^{2}}}$

Hence we have

$\begin{align}

& \int{\sec x\left( \sec x+\tan x \right)}=\int{\dfrac{\dfrac{2dt}{1+{{t}^{2}}}}{1-\dfrac{2t}{1+{{t}^{2}}}}} \\

& =\int{\dfrac{2dt}{1+{{t}^{2}}-2t}} \\

\end{align}$

We know that ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$

Using the above formula, we get

$\begin{align}

& \int{\sec x\left( \sec x+\tan x \right)}=\int{\dfrac{2dt}{{{\left( t-1 \right)}^{2}}}=2\int{\dfrac{dt}{{{\left( t-1 \right)}^{2}}}}} \\

& =\dfrac{2}{1-t}+C \\

\end{align}$

Reverting to the original variable, we get

$\int{\sec x\left( \sec x+\tan x \right)}=\dfrac{2}{1-\tan \dfrac{x}{2}}+C$

[2] Although the two results look completely different from each other, by Lagrange mean value theorem, they differ only by a constant.

Hence $\dfrac{2}{1-\tan \dfrac{x}{2}}=\tan x+\sec x+C$ for some constant C.

[3] Proof of [2]: Let F(x) and G(x) be two antiderivatives of the function f(x).

Applying Lagrange mean value theorem to the function F(x) – G(x) in the interval [0,y]{Assuming continuity at 0}, we get

\[\dfrac{F(y)-G(y)-F(0)+G(0)}{y}=f(c)-f(c)=0\]

i.e. $F(y)-G(y)-F(0)+G(0)=0$

Since F(0)-G(0) is a constant let C = F(0) – G(0), we get

$F(y)=G(y)+C$ or in other words F(x) and G(x) differ only by a constant.

Last updated date: 28th May 2023

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