Question

If we have an integral expression as $I=\int{{{\sec }^{2}}x{{\operatorname{cosec}}^{4}}xdx}=K{{\cot }^{3}}x+L\tan x+M\cot x+C$ then
(a) $K=-\dfrac{1}{3}$
(b) $L=2$
(c) $M=-2$
(d) none of these

Answer 1

We have to integrate the expression present on the LHS of the given equation. For this, we need to use the integration by parts. For this we need to take ${{\operatorname{cosec}}^{4}}x$ as the first function and ${{\sec }^{2}}x$ as the second function. Then simplifying the obtained result using trigonometric identities, and again applying the by parts integration rule, we will obtain the LHS in the form of the RHS. By comparing the coefficients of the trigonometric terms on the LHS and the RHS, we will obtain the values of K, L and M.

Complete step by step solution:
The equation given in the above question is
$\Rightarrow I=\int{{{\sec }^{2}}x{{\operatorname{cosec}}^{4}}xdx}=K{{\cot }^{3}}x+M\tan x+C$
Let us consider the LHS of the above question.
$\Rightarrow I=\int{{{\sec }^{2}}xcose{{c}^{4}}xdx}$
Let us choose $cose{{c}^{4}}x$ as the first function and ${{\sec }^{2}}x$ as the second function to integrate by parts the above integral as

& \Rightarrow I=cose{{c}^{4}}x\int{se{{c}^{2}}xdx}-\int{\dfrac{d\left( cose{{c}^{4}}x \right)}{dx}\left( \int{{{\sec }^{2}}xdx} \right)dx} \\\ & \Rightarrow I=cose{{c}^{4}}x\int{se{{c}^{2}}xdx}-\int{4cose{{c}^{3}}x\left( -\operatorname{cosec}x\cot x \right)\left( \int{{{\sec }^{2}}xdx} \right)dx} \\\ & \Rightarrow I=cose{{c}^{4}}x\int{se{{c}^{2}}xdx}+4\int{cose{{c}^{4}}x\cot x\left( \int{{{\sec }^{2}}xdx} \right)dx} \\\ \end{aligned}$$ We know that $\int{se{{c}^{2}}xdx}=\tan x$. Putting this above, we get $$\Rightarrow I=cose{{c}^{4}}x\tan x+4\int{cose{{c}^{4}}x\cot x\left( \tan x \right)dx}$$ Now, putting $\cot x=\dfrac{1}{\tan x}$ we get $$\begin{aligned} & \Rightarrow I=cose{{c}^{4}}x\tan x+4\int{cose{{c}^{4}}x\cot x\left( \dfrac{1}{\cot x} \right)dx} \\\ & \Rightarrow I=cose{{c}^{4}}x\tan x+4\int{cose{{c}^{4}}xdx} \\\ \end{aligned}$$ Now, we can write $cose{{c}^{4}}x=\left( cose{{c}^{2}}x \right)\left( cose{{c}^{2}}x \right)$ to get $$\Rightarrow I=cose{{c}^{4}}x\tan x+4\int{\left( cose{{c}^{2}}x \right)\left( cose{{c}^{2}}x \right)dx}$$ Choosing $$cose{{c}^{2}}x$$ as the first function, we again integrate by parts as $$\begin{aligned} & \Rightarrow I=cose{{c}^{4}}x\tan x+4\left( cose{{c}^{2}}x\int{cose{{c}^{2}}xdx}-\int{\dfrac{d\left( cose{{c}^{2}}x \right)}{dx}\left( \int{cose{{c}^{2}}xdx} \right)dx} \right) \\\ & \Rightarrow I=cose{{c}^{4}}x\tan x+4\left( cose{{c}^{2}}x\int{cose{{c}^{2}}xdx}-\int{2cosecx\left( -\operatorname{cosec}x\cot x \right)\left( \int{cose{{c}^{2}}xdx} \right)dx} \right) \\\ & \Rightarrow I=cose{{c}^{4}}x\tan x+4\left( cose{{c}^{2}}x\int{cose{{c}^{2}}xdx}+2\int{cose{{c}^{2}}x\cot x\left( \int{cose{{c}^{2}}xdx} \right)dx} \right) \\\ \end{aligned}$$ Now, we know that $$\int{cose{{c}^{2}}xdx}=-\cot x$$. Putting this above, we get $$\begin{aligned} & \Rightarrow I=cose{{c}^{4}}x\tan x+4\left( cose{{c}^{2}}x\left( -\cot x \right)+2\int{cose{{c}^{2}}x\cot x\left( -\cot x \right)dx} \right) \\\ & \Rightarrow I=cose{{c}^{4}}x\tan x-4cose{{c}^{2}}x\cot x-8\int{cose{{c}^{2}}x{{\cot }^{2}}xdx}......\left( i \right) \\\ \end{aligned}$$ Putting $t=\cot x$ so that $$\begin{aligned} & \Rightarrow \dfrac{dt}{dx}=-cose{{c}^{2}}x \\\ & \Rightarrow dt=-cose{{c}^{2}}xdx \\\ & \Rightarrow cose{{c}^{2}}xdx=-dt \\\ \end{aligned}$$ Substituting these inside the integral in (i) we get $$\begin{aligned} & \Rightarrow I=cose{{c}^{4}}x\tan x-4cose{{c}^{2}}x\cot x-8\int{{{t}^{2}}\left( -dt \right)} \\\ & \Rightarrow I=cose{{c}^{4}}x\tan x-4cose{{c}^{2}}x\cot x+8\int{{{t}^{2}}dt} \\\ & \Rightarrow I=cose{{c}^{4}}x\tan x-4cose{{c}^{2}}x\cot x+8\left( \dfrac{{{t}^{3}}}{3} \right)+C \\\ \end{aligned}$$ Substituting $t=\cot x$ back into the above equation, we get $$\begin{aligned} & \Rightarrow I=cose{{c}^{4}}x\tan x-4cose{{c}^{2}}x\cot x+8\left( \dfrac{{{\cot }^{3}}x}{3} \right)+C \\\ & \Rightarrow I=cose{{c}^{4}}x\tan x-4cose{{c}^{2}}x\cot x+\dfrac{8}{3}{{\cot }^{3}}x+C.......\left( ii \right) \\\ \end{aligned}$$ Now, we know that $$\Rightarrow {{\operatorname{cosec}}^{2}}x=1+{{\cot }^{2}}x........\left( iii \right)$$ On squaring both the sides, we get $$\begin{aligned} & \Rightarrow {{\left( {{\operatorname{cosec}}^{2}}x \right)}^{2}}={{\left( 1+{{\cot }^{2}}x \right)}^{2}} \\\ & \Rightarrow {{\operatorname{cosec}}^{4}}x={{\left( 1+{{\cot }^{2}}x \right)}^{2}} \\\ \end{aligned}$$ Using $${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$$ we can simplify the RHS as $$\begin{aligned} & \Rightarrow {{\operatorname{cosec}}^{4}}x={{1}^{2}}+2\left( 1 \right)\left( {{\cot }^{2}}x \right)+{{\left( {{\cot }^{2}}x \right)}^{2}} \\\ & \Rightarrow {{\operatorname{cosec}}^{4}}x=1+2{{\cot }^{2}}x+{{\cot }^{4}}x.......\left( iv \right) \\\ \end{aligned}$$ Putting (iii) and (iv) into the equation (ii) we get $$\begin{aligned} & \Rightarrow I=\left( 1+2{{\cot }^{2}}x+{{\cot }^{4}}x \right)\tan x-4\left( 1+{{\cot }^{2}}x \right)\cot x+\dfrac{8}{3}{{\cot }^{3}}x+C \\\ & \Rightarrow I=\tan x+2{{\cot }^{2}}x\tan x+{{\cot }^{4}}x\tan x-4\left( \cot x+{{\cot }^{3}}x \right)+\dfrac{8}{3}{{\cot }^{3}}x+C \\\ & \Rightarrow I=\tan x+2\cot x+{{\cot }^{3}}x-4\cot x-4{{\cot }^{3}}x+\dfrac{8}{3}{{\cot }^{3}}x+C \\\ & \Rightarrow I=\tan x-2\cot x+{{\cot }^{3}}x\left( 1-4+\dfrac{8}{3} \right)+C \\\ & \Rightarrow I=\tan x-2\cot x-\dfrac{1}{3}{{\cot }^{3}}x+C......\left( v \right) \\\ \end{aligned}$$ Now, according to the question $$\Rightarrow I=K{{\cot }^{3}}x+L\tan x+M\cot x+C......\left( vi \right)$$ From (v) and (vi) we can write $$\Rightarrow \tan x-2\cot x-\dfrac{1}{3}{{\cot }^{3}}x+C=K{{\cot }^{3}}x+L\tan x+M\cot x+C$$ On comparing the coefficients, we get $\begin{aligned} & \Rightarrow L=1 \\\ & \Rightarrow M=-2 \\\ & \Rightarrow K=-\dfrac{1}{3} \\\ \end{aligned}$ **Hence, the correct options are (a) and (c).** **Note:** In the question, we were given the integral whose value was given in the form of unknown coefficients. If we find the integration to be tough, then we can also differentiate both the sides to get the values of the coefficients by comparison.