
Find the integral \[\int\limits_{0}^{{}^{\pi }/{}_{2}}{\dfrac{{{\sin }^{{}^{3}/{}_{2}}}xdx}{{{\cos }^{{}^{3}/{}_{2}}}x+{{\sin }^{{}^{3}/{}_{2}}}x}}=\]
A. \[0\]
B. \[\pi \]
C. \[\dfrac{\pi }{2}\]
D. \[\dfrac{\pi }{4}\]
Answer
164.4k+ views
Hint: In this question, we are to find the given integral. The given integral is in the interval $\left[ 0,\dfrac{\pi }{2} \right]$. And here the given function in the interval consists trigonometric function. So, by substituting $x=a-x$in the function, we can evaluate the given integral. Here the upper limit $a=\dfrac{\pi }{2}$.
Formula Used:Definite integral:
Consider a function $f(x)$ is defined on $[a,b]$. If the integral of this function, $\int{f(x)dx=F(x)}$, then $F(b)-F(a)$ is called as the definite integral of the function $f(x)$ over $[a,b]$.
I.e.,$\int\limits_{a}^{b}{f(x)dx}=\left. F(x) \right|_{a}^{b}=F(b)-F(a)$
Here $a$ (lower limit) and $b$(upper limit).
\[\int\limits_{a}^{b}{f(x)dx}=\int\limits_{a}^{b}{f(t)dt}\]
Some of the properties of the definite integrals are:
1) Interchanging the limits: \[\int\limits_{a}^{b}{f(x)dx=-}\int\limits_{b}^{a}{f(x)dx}\]
2) If $a3) $\int\limits_{0}^{a}{f(x)dx}=\int\limits_{0}^{a}{f(a-x)dx}$
4) $\int\limits_{-a}^{a}{f(x)dx}=2\int\limits_{0}^{a}{f(x)dx}$ if $f(x)$ is an even function
$\int\limits_{-a}^{a}{f(x)dx}=0$ if $f(x)$ is an odd function
5) $\begin{align}
& \int\limits_{0}^{2a}{f(x)dx}=2\int\limits_{0}^{a}{f(x)dx};\text{if }f(2a-x)=f(x) \\
& \text{ }=0\text{ if }f(2a-x)=-f(x) \\
\end{align}$
6) $\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{f(\sin x)}{f(\sin x)+f(\cos x)}dx=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{f(\cos x)}{f(\sin x)+f(\cos x)}dx=\dfrac{\pi }{4}}}$
7) $\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{f(\tan x)}{f(\tan x)+f(\cot x)}dx=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{f(\cot x)}{f(\tan x)+f(\cot x)}dx=\dfrac{\pi }{4}}}$
Complete step by step solution:Given integral is
\[I=\int\limits_{0}^{{}^{\pi }/{}_{2}}{\dfrac{{{\sin }^{{}^{3}/{}_{2}}}xdx}{{{\cos }^{{}^{3}/{}_{2}}}x+{{\sin }^{{}^{3}/{}_{2}}}x}}\]
Since the limits are $\left[ 0,\dfrac{\pi }{2} \right]$, we can use the formula,
\[\int\limits_{0}^{a}{f(x)dx}=\int\limits_{0}^{a}{f(a-x)dx}\]
So, the given integral become
\[\begin{align}
& I=\int\limits_{0}^{{}^{\pi }/{}_{2}}{\dfrac{{{\sin }^{{}^{3}/{}_{2}}}xdx}{{{\cos }^{{}^{3}/{}_{2}}}x+{{\sin }^{{}^{3}/{}_{2}}}x}} \\
& \text{ }=\int\limits_{0}^{{}^{\pi }/{}_{2}}{\dfrac{{{\sin }^{{}^{3}/{}_{2}}}(\dfrac{\pi }{2}-x)dx}{{{\cos }^{{}^{3}/{}_{2}}}(\dfrac{\pi }{2}-x)+{{\sin }^{{}^{3}/{}_{2}}}(\dfrac{\pi }{2}-x)}} \\
& \text{ }=\int\limits_{0}^{{}^{\pi }/{}_{2}}{\dfrac{{{\cos }^{{}^{3}/{}_{2}}}xdx}{{{\sin }^{{}^{3}/{}_{2}}}x+{{\cos }^{{}^{3}/{}_{2}}}x}} \\
\end{align}\]
So, we can write
\[\begin{align}
& I+I=\int\limits_{0}^{{}^{\pi }/{}_{2}}{\dfrac{{{\sin }^{{}^{3}/{}_{2}}}xdx}{{{\cos }^{{}^{3}/{}_{2}}}x+{{\sin }^{{}^{3}/{}_{2}}}x}}+\int\limits_{0}^{{}^{\pi }/{}_{2}}{\dfrac{{{\cos }^{{}^{3}/{}_{2}}}xdx}{{{\sin }^{{}^{3}/{}_{2}}}x+{{\cos }^{{}^{3}/{}_{2}}}x}} \\
& \Rightarrow 2I=\int\limits_{0}^{{}^{\pi }/{}_{2}}{\dfrac{{{\sin }^{{}^{3}/{}_{2}}}x+{{\cos }^{{}^{3}/{}_{2}}}x}{{{\cos }^{{}^{3}/{}_{2}}}x+{{\sin }^{{}^{3}/{}_{2}}}x}}dx \\
& \Rightarrow 2I=\int\limits_{0}^{{}^{\pi }/{}_{2}}{[1]dx} \\
& \therefore I=\dfrac{1}{2}\left[ x \right]_{0}^{{}^{\pi }/{}_{2}}=\dfrac{\pi }{4} \\
\end{align}\]
Option ‘D’ is correct
Note: In this question, the interval for the given integral is in the form of $[0,a]$. So, we can apply $x=a-x$ in the function of the given integral. By evaluating this, we get the required integral. Here we can also use a direct formula since the function in the given integral has trigonometric functions. The direct formula used for solving this type of integral is $\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{f(\sin x)}{f(\sin x)+f(\cos x)}dx=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{f(\cos x)}{f(\sin x)+f(\cos x)}dx=\dfrac{\pi }{4}}}$. Here we need to remember that $\sin \left( \dfrac{\pi }{2}-\theta \right)=\cos \theta $ and $\cos \left( \dfrac{\pi }{2}-\theta \right)=\sin \theta $. By using this, we can easily evaluate the given integral.
Formula Used:Definite integral:
Consider a function $f(x)$ is defined on $[a,b]$. If the integral of this function, $\int{f(x)dx=F(x)}$, then $F(b)-F(a)$ is called as the definite integral of the function $f(x)$ over $[a,b]$.
I.e.,$\int\limits_{a}^{b}{f(x)dx}=\left. F(x) \right|_{a}^{b}=F(b)-F(a)$
Here $a$ (lower limit) and $b$(upper limit).
\[\int\limits_{a}^{b}{f(x)dx}=\int\limits_{a}^{b}{f(t)dt}\]
Some of the properties of the definite integrals are:
1) Interchanging the limits: \[\int\limits_{a}^{b}{f(x)dx=-}\int\limits_{b}^{a}{f(x)dx}\]
2) If $a
4) $\int\limits_{-a}^{a}{f(x)dx}=2\int\limits_{0}^{a}{f(x)dx}$ if $f(x)$ is an even function
$\int\limits_{-a}^{a}{f(x)dx}=0$ if $f(x)$ is an odd function
5) $\begin{align}
& \int\limits_{0}^{2a}{f(x)dx}=2\int\limits_{0}^{a}{f(x)dx};\text{if }f(2a-x)=f(x) \\
& \text{ }=0\text{ if }f(2a-x)=-f(x) \\
\end{align}$
6) $\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{f(\sin x)}{f(\sin x)+f(\cos x)}dx=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{f(\cos x)}{f(\sin x)+f(\cos x)}dx=\dfrac{\pi }{4}}}$
7) $\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{f(\tan x)}{f(\tan x)+f(\cot x)}dx=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{f(\cot x)}{f(\tan x)+f(\cot x)}dx=\dfrac{\pi }{4}}}$
Complete step by step solution:Given integral is
\[I=\int\limits_{0}^{{}^{\pi }/{}_{2}}{\dfrac{{{\sin }^{{}^{3}/{}_{2}}}xdx}{{{\cos }^{{}^{3}/{}_{2}}}x+{{\sin }^{{}^{3}/{}_{2}}}x}}\]
Since the limits are $\left[ 0,\dfrac{\pi }{2} \right]$, we can use the formula,
\[\int\limits_{0}^{a}{f(x)dx}=\int\limits_{0}^{a}{f(a-x)dx}\]
So, the given integral become
\[\begin{align}
& I=\int\limits_{0}^{{}^{\pi }/{}_{2}}{\dfrac{{{\sin }^{{}^{3}/{}_{2}}}xdx}{{{\cos }^{{}^{3}/{}_{2}}}x+{{\sin }^{{}^{3}/{}_{2}}}x}} \\
& \text{ }=\int\limits_{0}^{{}^{\pi }/{}_{2}}{\dfrac{{{\sin }^{{}^{3}/{}_{2}}}(\dfrac{\pi }{2}-x)dx}{{{\cos }^{{}^{3}/{}_{2}}}(\dfrac{\pi }{2}-x)+{{\sin }^{{}^{3}/{}_{2}}}(\dfrac{\pi }{2}-x)}} \\
& \text{ }=\int\limits_{0}^{{}^{\pi }/{}_{2}}{\dfrac{{{\cos }^{{}^{3}/{}_{2}}}xdx}{{{\sin }^{{}^{3}/{}_{2}}}x+{{\cos }^{{}^{3}/{}_{2}}}x}} \\
\end{align}\]
So, we can write
\[\begin{align}
& I+I=\int\limits_{0}^{{}^{\pi }/{}_{2}}{\dfrac{{{\sin }^{{}^{3}/{}_{2}}}xdx}{{{\cos }^{{}^{3}/{}_{2}}}x+{{\sin }^{{}^{3}/{}_{2}}}x}}+\int\limits_{0}^{{}^{\pi }/{}_{2}}{\dfrac{{{\cos }^{{}^{3}/{}_{2}}}xdx}{{{\sin }^{{}^{3}/{}_{2}}}x+{{\cos }^{{}^{3}/{}_{2}}}x}} \\
& \Rightarrow 2I=\int\limits_{0}^{{}^{\pi }/{}_{2}}{\dfrac{{{\sin }^{{}^{3}/{}_{2}}}x+{{\cos }^{{}^{3}/{}_{2}}}x}{{{\cos }^{{}^{3}/{}_{2}}}x+{{\sin }^{{}^{3}/{}_{2}}}x}}dx \\
& \Rightarrow 2I=\int\limits_{0}^{{}^{\pi }/{}_{2}}{[1]dx} \\
& \therefore I=\dfrac{1}{2}\left[ x \right]_{0}^{{}^{\pi }/{}_{2}}=\dfrac{\pi }{4} \\
\end{align}\]
Option ‘D’ is correct
Note: In this question, the interval for the given integral is in the form of $[0,a]$. So, we can apply $x=a-x$ in the function of the given integral. By evaluating this, we get the required integral. Here we can also use a direct formula since the function in the given integral has trigonometric functions. The direct formula used for solving this type of integral is $\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{f(\sin x)}{f(\sin x)+f(\cos x)}dx=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{f(\cos x)}{f(\sin x)+f(\cos x)}dx=\dfrac{\pi }{4}}}$. Here we need to remember that $\sin \left( \dfrac{\pi }{2}-\theta \right)=\cos \theta $ and $\cos \left( \dfrac{\pi }{2}-\theta \right)=\sin \theta $. By using this, we can easily evaluate the given integral.
Recently Updated Pages
Geometry of Complex Numbers – Topics, Reception, Audience and Related Readings

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

Atomic Structure - Electrons, Protons, Neutrons and Atomic Models

Displacement-Time Graph and Velocity-Time Graph for JEE

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Degree of Dissociation and Its Formula With Solved Example for JEE

Instantaneous Velocity - Formula based Examples for JEE

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics

JEE Advanced 2025 Notes
