Question

Evaluate the Integration, $\int{{{\sec }^{2}}\theta {{\left( \sec \theta +\tan \theta \right)}^{2}}d\theta }$ .

Answer 1

Assume that $t=\left( \sec \theta +\tan \theta \right)$ . Differentiate $t$ with respect to $d\theta$ and get the relation between $dt$ and $d\theta$ . We know the identity, $\left( {{\sec }^{2}}\theta -{{\tan }^{2}}\theta \right)=1$ . Now, get the value of $\left( \sec \theta -\tan \theta \right)$ in terms of $t$ by expanding the identity $\left( {{\sec }^{2}}\theta -{{\tan }^{2}}\theta \right)=1$ using the formula, ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ . Now, arrange the given expression as $\int{\sec \theta \left( \sec \theta +\tan \theta \right)\sec \theta \left( \sec \theta +\tan \theta \right)d\theta }$ and then modify it in terms of $t$ . Solve it further by using the formula, $\int{{{t}^{a}}dt=\dfrac{{{t}^{a+1}}}{a+1}}$ .

Complete step by step solution:
According to the question, we have to integrate, $\int{{{\sec }^{2}}\theta {{\left( \sec \theta +\tan \theta \right)}^{2}}d\theta }$ ……………………………………..(1)
We have to simplify the above equation into a simpler form.
First of all, let us assume that $t=\left( \sec \theta +\tan \theta \right)$ ………………………………..(2)
Now, on differentiating the LHS and RHS with respect to $d\theta$ of equation (2), we get
$\dfrac{dt}{d\theta }=\dfrac{d}{d\theta }\left( \sec \theta +\tan \theta \right)$ ……………………………….(3)
We know the formula, $\dfrac{d\left( \sec \theta \right)}{d\theta }=\sec \theta \tan \theta$ and $\dfrac{d\left( \tan \theta \right)}{d\theta }={{\sec }^{2}}\theta$ …………………………………..(4)
Now, using the formula shown in equation (4) and on simplifying equation (4), we get

& \Rightarrow \dfrac{dt}{d\theta }=\left( \sec \theta \tan \theta +{{\sec }^{2}}\theta \right) \\\ & \Rightarrow \dfrac{dt}{d\theta }=\sec \theta \left( \sec \theta +\tan \theta \right) \\\ \end{aligned}$$ $$\Rightarrow dt=\sec \theta \left( \sec \theta +\tan \theta \right)d\theta $$ ………………………………………….(5) We know the identity, $$\left( {{\sec }^{2}}\theta -{{\tan }^{2}}\theta \right)=1$$ We also know the formula, $${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$$ . Now, from the above identity and formula, we get $$\Rightarrow \left( \sec \theta +\tan \theta \right)\left( \sec \theta -\tan \theta \right)=1$$ ………………………………………..(6) Using equation (2) and on substituting $$\left( \sec \theta +\tan \theta \right)$$ by t in equation (6), we get $$\Rightarrow t\left( \sec \theta -\tan \theta \right)=1$$ $$\Rightarrow \left( \sec \theta -\tan \theta \right)=\dfrac{1}{t}$$ ………………………………………….(7) Now, on adding equation (2) and equation (7), we get $$\begin{aligned} & \Rightarrow \left( \sec \theta +\tan \theta \right)+\left( \sec \theta -\tan \theta \right)=t+\dfrac{1}{t} \\\ & \Rightarrow 2\sec \theta =\left( t+\dfrac{1}{t} \right) \\\ \end{aligned}$$ $$\Rightarrow \sec \theta =\dfrac{1}{2}\left( t+\dfrac{1}{t} \right)$$…………………………………………(8) Now, arranging equation (1), we get $$=\int{\sec \theta \left( \sec \theta +\tan \theta \right)\sec \theta \left( \sec \theta +\tan \theta \right)d\theta }$$ ……………………………………………………(9) Using equation (2), equation (5), equation (8), and on substituting $$\sec \theta $$ by $$\dfrac{1}{2}\left( t+\dfrac{1}{t} \right)$$, $$\left( \sec \theta +\tan \theta \right)$$ by $$t$$ , and $$\sec \theta \left( \sec \theta +\tan \theta \right)d\theta $$ by $$dt$$ , we get $$\begin{aligned} & =\int{\dfrac{1}{2}\left( t+\dfrac{1}{t} \right)tdt} \\\ & =\dfrac{1}{2}\int{\left( {{t}^{2}}+\dfrac{1}{t}\times t \right)dt} \\\ & =\dfrac{1}{2}\int{\left( {{t}^{2}}+1 \right)dt} \\\ \end{aligned}$$ $$=\dfrac{1}{2}\left( \int{{{t}^{2}}dt+\int{dt}} \right)$$ …………………………………..(10) We also know the formula, $$\int{{{t}^{a}}dt=\dfrac{{{t}^{a+1}}}{a+1}}$$ ………………………………………(11) Using equation (11) and simplifying equation (10), we get $$=\dfrac{1}{2}\left( \dfrac{{{t}^{3}}}{3}+t \right)+c$$ ……………………………………………………..(12) From equation (2), we have $$t=\left( \sec \theta +\tan \theta \right)$$ . Now, on substituting $$t$$ by $$\left( \sec \theta +\tan \theta \right)$$ in equation (12), we get $$=\dfrac{1}{2}\left\\{ \dfrac{{{\left( \sec \theta +\tan \theta \right)}^{3}}}{3}+\left( \sec \theta +\tan \theta \right) \right\\}+c$$ ………………………………………………….(13) Therefore, $$\int{{{\sec }^{2}}\theta {{\left( \sec \theta +\tan \theta \right)}^{2}}d\theta }=\dfrac{1}{2}\left\\{ \dfrac{{{\left( \sec \theta +\tan \theta \right)}^{3}}}{3}+\left( \sec \theta +\tan \theta \right) \right\\}+c$$ . **Note:** For this question, one might think to expand the expression and then simplify it. This approach will not work here because on expanding we get more terms which will make the solution more complex to be simplified further.