Question

If we are given the integral expression as $\int{\dfrac{1-5{{\sin }^{2}}x}{{{\cos }^{5}}x{{\sin }^{2}}x}dx}=\dfrac{f\left( x \right)}{{{\cos }^{5}}x}+C$, then find $f\left( x \right)$.
A. $-\text{cosec}x$
B. $\text{cosec}x$
C. $\cot x$
D. $-\cot x$

Answer 1

We have been given the integral solution and we differentiate both side with respect to x. the differentiation takes the form of $\dfrac{d}{dx}\left[ \dfrac{u\left( x \right)}{v\left( x \right)} \right]=\dfrac{v\left( x \right).{{u}^{'}}\left( x \right)-u\left( x \right)\left[ {{v}^{'}}\left( x \right) \right]}{{{\left( v\left( x \right) \right)}^{2}}}$. Then we equate both sides of the differentiation to find the function and its differential form. We equate with the options to find the solution.

Complete step-by-step solution:

It’s given that $\int{\dfrac{1-5{{\sin }^{2}}x}{{{\cos }^{5}}x{{\sin }^{2}}x}dx}=\dfrac{f\left( x \right)}{{{\cos }^{5}}x}+C$.
We try to take the differentiation on both sides and equate the respective functions to find $f\left( x \right)$.
The differentiation of $\int{\dfrac{1-5{{\sin }^{2}}x}{{{\cos }^{5}}x{{\sin }^{2}}x}dx}$ gives $\dfrac{d}{dx}\left[ \int{\dfrac{1-5{{\sin }^{2}}x}{{{\cos }^{5}}x{{\sin }^{2}}x}dx} \right]=\dfrac{1-5{{\sin }^{2}}x}{{{\cos }^{5}}x{{\sin }^{2}}x}$.
Now we try to find the differentiation of $\dfrac{f\left( x \right)}{{{\cos }^{5}}x}+C$ and equate that with $\dfrac{1-5{{\sin }^{2}}x}{{{\cos }^{5}}x{{\sin }^{2}}x}$.
The function $\dfrac{f\left( x \right)}{{{\cos }^{5}}x}+C$ is of the form $\dfrac{u\left( x \right)}{v\left( x \right)}$ where $\dfrac{d}{dx}\left[ \dfrac{u\left( x \right)}{v\left( x \right)} \right]=\dfrac{v\left( x \right).{{u}^{'}}\left( x \right)-u\left( x \right)\left[ {{v}^{'}}\left( x \right) \right]}{{{\left( v\left( x \right) \right)}^{2}}}$.
Differentiating we get $\dfrac{d}{dx}\left[ \dfrac{f\left( x \right)}{{{\cos }^{5}}x}+C \right]=\dfrac{{{\cos }^{5}}x.{{f}^{'}}\left( x \right)-f\left( x \right)\left[ 5{{\cos }^{4}}x\left( -\sin x \right) \right]}{{{\left( {{\cos }^{5}}x \right)}^{2}}}$.
We divide both the numerator and denominator with ${{\cos }^{5}}x$.
So, $\dfrac{d}{dx}\left[ \dfrac{f\left( x \right)}{{{\cos }^{5}}x}+C \right]=\dfrac{{{f}^{'}}\left( x \right)+f\left( x \right)\left[ 5\tan x \right]}{{{\cos }^{5}}x}$. We equate this with $\dfrac{1-5{{\sin }^{2}}x}{{{\cos }^{5}}x{{\sin }^{2}}x}$.
It gives $\dfrac{{{f}^{'}}\left( x \right)+f\left( x \right)\left[ 5\tan x \right]}{{{\cos }^{5}}x}=\dfrac{1-5{{\sin }^{2}}x}{{{\cos }^{5}}x{{\sin }^{2}}x}$.
Simplifying we get ${{f}^{'}}\left( x \right)+f\left( x \right)\left[ 5\tan x \right]=\text{cosec}^{2}x-5$.
Equating we get ${{f}^{'}}\left( x \right)=\text{cosec}^{2}x,f\left( x \right)\left[ 5\tan x \right]=-5$.
Differentiation of $-\cot x$ is $\text{cosec}^{2}x$. So, we take $f\left( x \right)=-\cot x$.
So, $f\left( x \right)\left[ 5\tan x \right]=\left( -\cot x \right)\left( 5\tan x \right)=-5$
The correct option is D.

Note: We also could have taken ${{f}^{'}}\left( x \right)=-5,f\left( x \right)\left[ 5\tan x \right]=\text{cosec}^{2}x$. But in that case $f\left( x \right)$ would have been $f\left( x \right)=-5x$ and $\left( -5x \right)\left[ 5\tan x \right]=-25x\tan x\ne \text{cosec}^{2}x$. So, the assumption ${{f}^{'}}\left( x \right)=-5,f\left( x \right)\left[ 5\tan x \right]=\text{cosec}^{2}x$ is incorrect.