Question

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 1

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.